{"id":638,"date":"2021-07-23T09:20:18","date_gmt":"2021-07-23T13:20:18","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/phase-transitions\/"},"modified":"2022-06-23T09:10:24","modified_gmt":"2022-06-23T13:10:24","slug":"phase-transitions","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/phase-transitions\/","title":{"raw":"10.3 Phase Transitions","rendered":"10.3 Phase Transitions"},"content":{"raw":"<div class=\"textbox textbox--learning-objectives\">\r\n<h3><strong>Learning Objectives<\/strong><\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Define phase transitions and phase transition temperatures<\/li>\r\n \t<li>Explain the relation between phase transition temperatures and intermolecular attractive forces<\/li>\r\n \t<li>Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp10874160\">We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth\u2019s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored.<\/p>\r\n\r\n<div id=\"fs-idm35636464\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Vaporization and Condensation<\/strong><\/h3>\r\n<p id=\"fs-idm100542016\">When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called <strong>condensation<\/strong>. When the rate of condensation becomes equal to the rate of <strong>vaporization<\/strong>, neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be <em data-effect=\"italics\">in equilibrium<\/em> with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a <span data-type=\"term\">dynamic equilibrium<\/span>, the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid\u2019s <strong>vapor pressure<\/strong> (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress1\">(Figure)<\/a>, and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_10_03_VapPress1\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from the liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue.<\/div>\r\n<span id=\"fs-idm103231072\" data-type=\"media\" data-alt=\"Three images are shown and labeled \u201ca,\u201d \u201cb,\u201d and \u201cc.\u201d Each image shows a round bulb connected on the right to a tube that is horizontal, then is bent vertically, curves, and then is vertical again to make a u-shape. A valve is located in the horizontal portion of the tube. Image a depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid. The phrase, \u201cMolecules escape surface and form vapor\u201d is written below the bulb, and a gray liquid in the u-shaped portion of the tube is shown at equal heights on the right and left sides. Image b depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. A gray liquid in the u-shaped portion of the tube is shown slightly higher on the right side than on the left side. Image c depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. There are more molecules present in c than in b. The phrase \u201cEquilibrium reached, vapor pressure determined,\u201d is written below the bulb and a gray liquid in the u-shaped portion of the tube is shown higher on the right side. A horizontal line is drawn level with each of these liquid levels and the distance between the lines is labeled with a double-headed arrow. This section is labeled with the phrase, \u201cVapor pressure.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_VapPress1-1.jpg\" alt=\"Three images are shown and labeled \u201ca,\u201d \u201cb,\u201d and \u201cc.\u201d Each image shows a round bulb connected on the right to a tube that is horizontal, then is bent vertically, curves, and then is vertical again to make a u-shape. A valve is located in the horizontal portion of the tube. Image a depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid. The phrase, \u201cMolecules escape surface and form vapor\u201d is written below the bulb, and a gray liquid in the u-shaped portion of the tube is shown at equal heights on the right and left sides. Image b depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. A gray liquid in the u-shaped portion of the tube is shown slightly higher on the right side than on the left side. Image c depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. There are more molecules present in c than in b. The phrase \u201cEquilibrium reached, vapor pressure determined,\u201d is written below the bulb and a gray liquid in the u-shaped portion of the tube is shown higher on the right side. A horizontal line is drawn level with each of these liquid levels and the distance between the lines is labeled with a double-headed arrow. This section is labeled with the phrase, \u201cVapor pressure.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idm63266864\">The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring \u201crecapture\u201d of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces.<\/p>\r\n\r\n<div id=\"fs-idm12389904\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm149114816\"><strong>Explaining Vapor Pressure in Terms of IMFs:<\/strong><\/p>\r\nGiven the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs:\r\n\r\n<span id=\"fs-idm178035424\" data-type=\"media\" data-alt=\"Four Lewis structures are shown. The first structure, labeled \u201cethanol,\u201d shows a carbon bonded to three hydrogen atoms that is single bonded to a second carbon that is bonded to two hydrogen atoms and a hydroxyl group. The second structure, labeled \u201cethylene glycol, shows two carbon atoms, single bonded to one another, single bonded each to two hydrogen atoms, and each single bonded to a hydroxyl group. The third image, labeled \u201cdiethyl ether,\u201d shows an oxygen atom single bonded on both sides to a carbon that is bonded to two hydrogens, and a second carbon, that is itself bonded to three hydrogen atoms. The fourth image, labeled \u201cwater,\u201d shows an oxygen atom that is single bonded on both sides to hydrogen atoms.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_Ethanol_img-1.jpg\" alt=\"Four Lewis structures are shown. The first structure, labeled \u201cethanol,\u201d shows a carbon bonded to three hydrogen atoms that is single bonded to a second carbon that is bonded to two hydrogen atoms and a hydroxyl group. The second structure, labeled \u201cethylene glycol, shows two carbon atoms, single bonded to one another, single bonded each to two hydrogen atoms, and each single bonded to a hydroxyl group. The third image, labeled \u201cdiethyl ether,\u201d shows an oxygen atom single bonded on both sides to a carbon that is bonded to two hydrogens, and a second carbon, that is itself bonded to three hydrogen atoms. The fourth image, labeled \u201cwater,\u201d shows an oxygen atom that is single bonded on both sides to hydrogen atoms.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n&nbsp;\r\n<p id=\"fs-idm12378064\"><strong>Solution:<\/strong><\/p>\r\nDiethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two \u2212OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest.\r\n\r\n&nbsp;\r\n<p id=\"fs-idm224164560\"><strong>Check Your Learning:<\/strong><\/p>\r\nAt 20 \u00b0C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:\r\n<table id=\"fs-idm60218336\" class=\"column-header medium unnumbered\" summary=\"This table has two rows and five columns. The first column is a header column, and it labels each row: \u201cCompound,\u201d and \u201cVapor Pressure at 25 degrees C.\u201d To the right of the \u201cCompound\u201d column are the following: methanol, C H subscript 3 O H; ethanol, C subscript 2 H subscript 5 O H; propanol C subscript 3 H subscript 7 O H; and butanol C subscript 4 H subscript 9 O H. To the right of the \u201cVapor Pressure at 25 degrees C\u201d column are the following: 11.9 k P a, 5.95 k P a, 2.67 k P a, and 0.56 k P a.\" data-label=\"\">\r\n<tbody valign=\"middle\">\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">Compound<\/td>\r\n<td data-align=\"left\">methanol CH<sub>3<\/sub>OH<\/td>\r\n<td data-align=\"left\">ethanol C<sub>2<\/sub>H<sub>5<\/sub>OH<\/td>\r\n<td data-align=\"left\">propanol C<sub>3<\/sub>H<sub>7<\/sub>OH<\/td>\r\n<td data-align=\"left\">butanol C<sub>4<\/sub>H<sub>9<\/sub>OH<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">Vapor Pressure at 20 \u00b0C<\/td>\r\n<td data-align=\"left\">11.9 kPa<\/td>\r\n<td data-align=\"left\">5.95 kPa<\/td>\r\n<td data-align=\"left\">2.67 kPa<\/td>\r\n<td data-align=\"left\">0.56 kPa<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-idm91362784\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm44101792\">All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:<span data-type=\"newline\">\r\n<\/span> P<sub>methanol<\/sub> &gt; P<sub>ethanol<\/sub> &gt; P<sub>propanol<\/sub> &gt; P<sub>butanol<\/sub>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm73489328\">As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress3\">(Figure)<\/a>. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_10_03_VapPress3\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, KE, to escape from the liquid into the gas phase.<\/div>\r\n<span id=\"fs-idm211797072\" data-type=\"media\" data-alt=\"A graph is shown where the y-axis is labeled \u201cNumber of molecules\u201d and the x-axis is labeled \u201cKinetic Energy.\u201d Two lines are graphed and a vertical dotted line, labeled \u201cMinimum K E needed to escape,\u201d is drawn halfway across the x-axis. The first line move sharply upward and has a high peak near the left side of the x-axis. It drops just as steeply and ends about 60 percent of the way across the x-axis. This line is labeled \u201cLow T.\u201d A second line, labeled \u201cHigh T,\u201d begins at the same point as the first, but does not go to such a high point, is wider, and ends slightly further to the right on the x-axis.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_VapPress3-1.jpg\" alt=\"A graph is shown where the y-axis is labeled \u201cNumber of molecules\u201d and the x-axis is labeled \u201cKinetic Energy.\u201d Two lines are graphed and a vertical dotted line, labeled \u201cMinimum K E needed to escape,\u201d is drawn halfway across the x-axis. The first line move sharply upward and has a high peak near the left side of the x-axis. It drops just as steeply and ends about 60 percent of the way across the x-axis. This line is labeled \u201cLow T.\u201d A second line, labeled \u201cHigh T,\u201d begins at the same point as the first, but does not go to such a high point, is wider, and ends slightly further to the right on the x-axis.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm153006576\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Boiling Points<\/strong><\/h3>\r\n<p id=\"fs-idm70709936\">When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The <strong>boiling point<\/strong> of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth\u2019s atmosphere. The<strong> normal boiling point<\/strong> of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress2\">(Figure)<\/a> shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid\u2019s boiling point on surrounding pressure.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_10_03_VapPress2\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.)<\/div>\r\n<span id=\"fs-idm188874672\" data-type=\"media\" data-alt=\"A graph is shown where the x-axis is labeled \u201cTemperature ( degree sign, C )\u201d and has values of 200 to 1000 in increments of 200 and the y-axis is labeled \u201cPressure ( k P a )\u201d and has values of 20 to 120 in increments of 20. A horizontal dotted line extends across the graph at point 780 on the y-axis while three vertical dotted lines extend from points 35, 78, and 100 to meet the horizontal dotted line. Four lines are graphed. The first line, labeled \u201cethyl ether,\u201d begins at the point \u201c0 , 200\u201d and extends in a slight curve to point \u201c45, 1000\u201d while the second line, labeled \u201cethanol\u201d, extends from point \u201c0, 20\u201d to point \u201c88, 1000\u201d in a more extreme curve. The third line, labeled \u201cwater,\u201d begins at the point \u201c0, 0\u201d and extends in a curve to point \u201c108, 1000\u201d while the fourth line, labeled \u201cethylene glycol,\u201d extends from point \u201c80, 0\u201d to point \u201c140, 100\u201d in a very shallow curve.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_VapPress2-1.jpg\" alt=\"A graph is shown where the x-axis is labeled \u201cTemperature ( degree sign, C )\u201d and has values of 200 to 1000 in increments of 200 and the y-axis is labeled \u201cPressure ( k P a )\u201d and has values of 20 to 120 in increments of 20. A horizontal dotted line extends across the graph at point 780 on the y-axis while three vertical dotted lines extend from points 35, 78, and 100 to meet the horizontal dotted line. Four lines are graphed. The first line, labeled \u201cethyl ether,\u201d begins at the point \u201c0 , 200\u201d and extends in a slight curve to point \u201c45, 1000\u201d while the second line, labeled \u201cethanol\u201d, extends from point \u201c0, 20\u201d to point \u201c88, 1000\u201d in a more extreme curve. The third line, labeled \u201cwater,\u201d begins at the point \u201c0, 0\u201d and extends in a curve to point \u201c108, 1000\u201d while the fourth line, labeled \u201cethylene glycol,\u201d extends from point \u201c80, 0\u201d to point \u201c140, 100\u201d in a very shallow curve.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-idm272706256\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm208677136\"><strong>A Boiling Point at Reduced Pressure:<\/strong><\/p>\r\nA typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress2\">(Figure)<\/a> to determine the boiling point of water at this elevation.\r\n\r\n&nbsp;\r\n<p id=\"fs-idm10799984\"><strong>Solution:<\/strong><\/p>\r\nThe graph of the vapor pressure of water versus temperature in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress2\">(Figure)<\/a> indicates that the vapor pressure of water is 68 kPa at about 90 \u00b0C. Thus, at about 90 \u00b0C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.\r\n\r\n&nbsp;\r\n<p id=\"fs-idm153672912\"><strong>Check Your Learning:<\/strong><\/p>\r\nThe boiling point of ethyl ether was measured to be 10 \u00b0C at a base camp on the slopes of Mount Everest. Use <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress2\">(Figure)<\/a> to determine the approximate atmospheric pressure at the camp.\r\n\r\n&nbsp;\r\n<div id=\"fs-idm212817424\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm95497088\">Approximately 40 kPa (0.4 atm)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm287754592\">The quantitative relation between a substance\u2019s vapor pressure and its temperature is described by the <span data-type=\"term\">Clausius-Clapeyron equation<\/span>:<\/p>\r\n\r\n<div id=\"fs-idm100818032\" data-type=\"equation\"><img class=\"wp-image-1612 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3a.png\" alt=\"\" width=\"171\" height=\"37\" \/><\/div>\r\n<p id=\"fs-idm95462576\">where \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub> is the enthalpy of vaporization for the liquid, <em data-effect=\"italics\">R<\/em> is the gas constant, and <em data-effect=\"italics\">A<\/em> is a constant whose value depends on the chemical identity of the substance. Temperature T must be in Kelvin in this equation. This equation is often rearranged into logarithmic form to yield the linear equation:<\/p>\r\n\r\n<div id=\"fs-idm13139376\" data-type=\"equation\"><img class=\" wp-image-1614 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3b.png\" alt=\"\" width=\"206\" height=\"47\" \/><\/div>\r\n<p id=\"fs-idm13032112\">This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature T<sub>1<\/sub>, the vapor pressure is P<sub>1<\/sub>, and at temperature T<sub>2<\/sub>, the vapor pressure is P<sub>2<\/sub>, the corresponding linear equations are:<\/p>\r\n\r\n<div id=\"fs-idm220001056\" data-type=\"equation\"><img class=\" wp-image-1615 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3c-300x27.png\" alt=\"\" width=\"623\" height=\"56\" \/><\/div>\r\n<p id=\"fs-idm51875808\">Since the constant, <em data-effect=\"italics\">A<\/em>, is the same, these two equations may be rearranged to isolate ln <em data-effect=\"italics\">A<\/em> and then set them equal to one another:<\/p>\r\n\r\n<div id=\"fs-idm142013200\" data-type=\"equation\"><img class=\"alignnone wp-image-1616 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3d-300x59.png\" alt=\"\" width=\"259\" height=\"51\" \/><\/div>\r\n<p id=\"fs-idm72918608\">which can be combined into:<\/p>\r\n\r\n<div id=\"fs-idm218464864\" data-type=\"equation\"><img class=\"alignnone wp-image-1617 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png\" alt=\"\" width=\"257\" height=\"54\" \/><\/div>\r\n<div id=\"fs-idm96861632\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm305163312\"><strong>Estimating Enthalpy of Vaporization:<\/strong><\/p>\r\nIsooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 \u00b0C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 \u00b0C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane.\r\n\r\n&nbsp;\r\n<p id=\"fs-idm195543920\"><strong>Solution:<\/strong><\/p>\r\nThe enthalpy of vaporization, \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub>, can be determined by using the Clausius-Clapeyron equation:\r\n<div id=\"fs-idp17288896\" data-type=\"equation\"><strong><img class=\"alignnone wp-image-1617 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png\" alt=\"\" width=\"257\" height=\"54\" \/><\/strong><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm95291536\">Since we have two vapor pressure-temperature values (<em data-effect=\"italics\">T<\/em><sub>1<\/sub> = 34.0 \u00b0C = 307.2 K, <em data-effect=\"italics\">P<\/em><sub>1<\/sub> = 10.0 kPa and <em data-effect=\"italics\">T<\/em><sub>2<\/sub> = 98.8 \u00b0C = 372.0 K, <em data-effect=\"italics\">P<\/em><sub>2<\/sub> = 100 kPa), we can substitute them into this equation and solve for \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub>. Rearranging the Clausius-Clapeyron equation and solving for \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub> yields:<\/p>\r\n\r\n<div id=\"fs-idm44175456\" data-type=\"equation\"><img class=\"wp-image-1618 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3f-300x34.png\" alt=\"\" width=\"538\" height=\"61\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm201782400\">Note that the pressure can be in any units, so long as they agree for both <em data-effect=\"italics\">P<\/em> values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.<\/p>\r\n&nbsp;\r\n<p id=\"fs-idm101600256\"><strong>Check Your Learning:<\/strong><\/p>\r\nAt 20.0 \u00b0C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 \u00b0C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.\r\n<div id=\"fs-idp12869536\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm146844672\">41,360 J\/mol or 41.4 kJ\/mol<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm143693744\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm467728\"><strong>Estimating Temperature (or Vapor Pressure):<\/strong><\/p>\r\nFor benzene (C<sub>6<\/sub>H<sub>6<\/sub>), the normal boiling point is 80.1 \u00b0C and the enthalpy of vaporization is 30.8 kJ\/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?\r\n\r\n&nbsp;\r\n<p id=\"fs-idm220027984\"><strong>Solution:<\/strong><\/p>\r\nIf the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, \u0394<em data-effect=\"italics\">H<\/em><sub>vap,<\/sub> then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation:\r\n<div id=\"fs-idm92115136\" data-type=\"equation\"><strong><img class=\"alignnone wp-image-1617 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png\" alt=\"\" width=\"257\" height=\"54\" \/><\/strong><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm198425552\">Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (<em data-effect=\"italics\">T<\/em><sub>1<\/sub> = 80.1 \u00b0C = 353.3 K, <em data-effect=\"italics\">P<\/em><sub>1<\/sub> = 101.3 kPa, \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub> = 30.8 kJ\/mol) and want to find the temperature (<em data-effect=\"italics\">T<\/em><sub>2<\/sub>) that corresponds to vapor pressure <em data-effect=\"italics\">P<\/em><sub>2<\/sub> = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for <em data-effect=\"italics\">T<\/em><sub>2<\/sub>. Rearranging the Clausius-Clapeyron equation and solving for <em data-effect=\"italics\">T<\/em><sub>2<\/sub> yields:<\/p>\r\n\r\n<div id=\"fs-idm103279184\" data-type=\"equation\"><img class=\"wp-image-1619 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3g-300x33.png\" alt=\"\" width=\"509\" height=\"56\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm164810384\"><strong>Check Your Learning:<\/strong><\/p>\r\nFor acetone (CH<sub>3<\/sub>)<sub>2<\/sub>CO, the normal boiling point is 56.5 \u00b0C and the enthalpy of vaporization is 31.3 kJ\/mol. What is the vapor pressure of acetone at 25.0 \u00b0C?\r\n<div id=\"fs-idm209654080\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm215055184\">30.1 kPa<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm196961088\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Enthalpy of Vaporization<\/strong><\/h3>\r\n<p id=\"fs-idm238047040\">Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub>. For example, the vaporization of water at standard temperature is represented by:<\/p>\r\n\r\n<div id=\"fs-idp7928688\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6H<sub>2<\/sub>O(<em>g<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>vap<\/sub>= 44.01 kJ<\/div>\r\n<p id=\"fs-idm149636464\">As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:<\/p>\r\n\r\n<div id=\"fs-idm190668720\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>O(<em>g<\/em>) \u27f6H<sub>2<\/sub>O(<em>l<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>con<\/sub> = -\u0394<em>H<\/em><sub>vap<\/sub>= -44.01 kJ<\/div>\r\n<div id=\"fs-idm57843872\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm70086528\"><strong>Using Enthalpy of Vaporization:<\/strong><\/p>\r\nOne way our body is cooled is by evaporation of the water in sweat (<a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_Evap\">(Figure)<\/a>). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at <em data-effect=\"italics\">T<\/em> = 37 \u00b0C (normal body temperature); \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub> = 43.46 kJ\/mol at 37 \u00b0C?\r\n\r\n&nbsp;\r\n<div id=\"CNX_Chem_10_03_Evap\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">Evaporation of sweat helps cool the body. (credit: \u201cKullez\u201d\/Flickr)<\/div>\r\n<span id=\"fs-idm183436304\" data-type=\"media\" data-alt=\"A person\u2019s shoulder and neck are shown and their skin is covered in beads of liquid.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_Evap-1.jpg\" alt=\"A person\u2019s shoulder and neck are shown and their skin is covered in beads of liquid.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-idm146648608\"><strong>Solution:<\/strong><\/p>\r\nWe start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:\r\n<div id=\"fs-idm38095424\" data-type=\"equation\"><img class=\"wp-image-1621 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3i-300x39.png\" alt=\"\" width=\"400\" height=\"52\" \/><\/div>\r\n<p id=\"fs-idm104953776\">Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.<\/p>\r\n&nbsp;\r\n<p id=\"fs-idm119901584\"><strong>Check Your Learning:<\/strong><\/p>\r\nHow much heat is required to evaporate 100.0 g of liquid ammonia, NH<sub>3<\/sub>, at its boiling point if its enthalpy of vaporization is 4.8 kJ\/mol?\r\n<div id=\"fs-idm64022976\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm13083920\">28 kJ<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm167698656\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Melting and Freezing<\/strong><\/h3>\r\n<p id=\"fs-idp14081632\">When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or <span data-type=\"term\">melting<\/span>. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (<a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_MeltingIce\">(Figure)<\/a>).<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_10_03_MeltingIce\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">(a) This beaker of ice has a temperature of \u221212.0 \u00b0C. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to 0 \u00b0C. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still 0 \u00b0C. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to 22.2 \u00b0C. (credit: modification of work by Mark Ott)<\/div>\r\n<span id=\"fs-idm137978064\" data-type=\"media\" data-alt=\"This figure shows four photos each labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and, \u201cd.\u201d Each photo shows a beaker with ice and a digital thermometer. The first photo shows ice cubes in the beaker, and the thermometer reads negative 12.0 degrees C. The second photo shows slightly melted ice, and the thermometer reads 0.0 degrees C. The third photo shows more water than ice in the beaker. The thermometer reads 0.0 degrees C. The fourth photo shows the ice completely melted, and the thermometer reads 22.2 degrees C.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_MeltingIce-1.jpg\" alt=\"This figure shows four photos each labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and, \u201cd.\u201d Each photo shows a beaker with ice and a digital thermometer. The first photo shows ice cubes in the beaker, and the thermometer reads negative 12.0 degrees C. The second photo shows slightly melted ice, and the thermometer reads 0.0 degrees C. The third photo shows more water than ice in the beaker. The thermometer reads 0.0 degrees C. The fourth photo shows the ice completely melted, and the thermometer reads 22.2 degrees C.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp12376064\">If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and <span data-type=\"term\">freezing<\/span> occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the <strong>melting point <\/strong>of the solid or the <strong>freezing point<\/strong> of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing).<\/p>\r\n<p id=\"fs-idm90105504\">The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.<\/p>\r\n<p id=\"fs-idm133861568\">The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, \u0394H<sub>fus<\/sub> of the substance. The enthalpy of fusion of ice is 6.0 kJ\/mol at 0 \u00b0C. Fusion (melting) is an endothermic process:<\/p>\r\n\r\n<div id=\"fs-idm185471776\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>O(<em>s<\/em>) \u27f6H<sub>2<\/sub>O(<em>l<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>fus<\/sub>= 6.01 kJ<\/div>\r\n<p id=\"fs-idm119984512\">The reciprocal process, freezing, is an exothermic process whose enthalpy change is \u22126.0 kJ\/mol at 0 \u00b0C:<\/p>\r\n\r\n<div id=\"fs-idm145529840\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6H<sub>2<\/sub>O(<em>s<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>frz<\/sub> = -\u0394<em>H<\/em><sub>fus<\/sub>= -6.01 kJ<\/div>\r\n<\/div>\r\n<div id=\"fs-idm97104816\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Sublimation and Deposition<\/strong><\/h3>\r\n<p id=\"fs-idm191153504\">Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as <strong>sublimation<\/strong>. At room temperature and standard pressure, a piece of dry ice (solid CO<sub>2<\/sub>) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (<a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_Sublimtn\">(Figure)<\/a>). The reverse of sublimation is called <strong>deposition<\/strong>, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_10_03_Sublimtn\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott)<\/div>\r\n<span id=\"fs-idm195457584\" data-type=\"media\" data-alt=\"This figure shows a test tube. In the bottom is a dark substance which breaks up into a purple gas at the top.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_Sublimtn-1.jpg\" alt=\"This figure shows a test tube. In the bottom is a dark substance which breaks up into a purple gas at the top.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idm208982384\">Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, \u0394H<sub>sub<\/sub>, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:<\/p>\r\n\r\n<div id=\"fs-idm92174656\" style=\"text-align: center\" data-type=\"equation\">CO<sub>2<\/sub>(<em>s<\/em>) \u27f6 CO<sub>2<\/sub>(<em>g<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>sub<\/sub>= 26.1 kJ<\/div>\r\n<p id=\"fs-idm110822080\">Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:<\/p>\r\n\r\n<div id=\"fs-idp16237312\" style=\"text-align: center\" data-type=\"equation\">CO<sub>2<\/sub>(<em>g<\/em>) \u27f6CO<sub>2<\/sub>(<em>s<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>dep<\/sub> = -\u0394<em>H<\/em><sub>sub<\/sub>= -26.1 kJ<\/div>\r\n<p id=\"fs-idm69412784\">Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess\u2019s Law. Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_PhaseChng\">(Figure)<\/a>. For example:<\/p>\r\n\r\n<div id=\"fs-idm81561872\" data-type=\"equation\">solid \u27f6 liquid\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394<em>H<\/em><sub>fus<\/sub><\/div>\r\n<div data-type=\"equation\">liquid \u27f6 gas\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 \u0394<em>H<\/em><sub>vap<\/sub><\/div>\r\n<div data-type=\"equation\">-------------------------<\/div>\r\n<div data-type=\"equation\">solid \u27f6 gas\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>sub<\/sub> = \u0394<em>H<\/em><sub>fus<\/sub> + \u0394<em>H<\/em><sub>vap<\/sub><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"CNX_Chem_10_03_PhaseChng\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation.<\/div>\r\n<span id=\"fs-idm63916320\" data-type=\"media\" data-alt=\"A diagram is shown with a vertical line drawn on the left side and labeled \u201cEnergy\u201d and three horizontal lines drawn near the bottom, lower third and top of the diagram. These three lines are labeled, from bottom to top, \u201cSolid,\u201d \u201cLiquid\u201d and \u201cGas.\u201d Near the middle of the diagram, a vertical, upward-facing arrow is drawn from the solid line to the gas line and labeled \u201cSublimation, delta sign, H, subscript sub.\u201d To the right of this arrow is a second vertical, upward-facing arrow that is drawn from the solid line to the liquid line and labeled \u201cFusion, delta sign, H, subscript fus.\u201d Above the second arrow is a third arrow drawn from the liquid line to the gas line and labeled, \u201cVaporization, delta sign, H, subscript vap.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_PhaseChng-1.jpg\" alt=\"A diagram is shown with a vertical line drawn on the left side and labeled \u201cEnergy\u201d and three horizontal lines drawn near the bottom, lower third and top of the diagram. These three lines are labeled, from bottom to top, \u201cSolid,\u201d \u201cLiquid\u201d and \u201cGas.\u201d Near the middle of the diagram, a vertical, upward-facing arrow is drawn from the solid line to the gas line and labeled \u201cSublimation, delta sign, H, subscript sub.\u201d To the right of this arrow is a second vertical, upward-facing arrow that is drawn from the solid line to the liquid line and labeled \u201cFusion, delta sign, H, subscript fus.\u201d Above the second arrow is a third arrow drawn from the liquid line to the gas line and labeled, \u201cVaporization, delta sign, H, subscript vap.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm93930720\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Heating and Cooling Curves<\/strong><\/h3>\r\n<p id=\"fs-idp17603264\">In the chapter on thermochemistry, the relation between the amount of heat absorbed or released by a substance, <em data-effect=\"italics\">q<\/em>, and its accompanying temperature change, \u0394<em data-effect=\"italics\">T<\/em>, was introduced:<\/p>\r\n\r\n<div id=\"fs-idm100720480\" style=\"text-align: center\" data-type=\"equation\"><em>q<\/em> = <em>mc<\/em>\u0394<em>T<\/em><\/div>\r\n<p id=\"fs-idm73087600\">where <em data-effect=\"italics\">m<\/em> is the mass of the substance and <em data-effect=\"italics\">c<\/em> is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_HeatCurve\">(Figure)<\/a> shows a typical heating curve.<\/p>\r\n<p id=\"fs-idm178482560\">Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water\u2019s temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_10_03_HeatCurve\" class=\"bc-figure figure\">\r\n<div class=\"bc-figcaption figcaption\">A typical heating curve for a substance depicts changes in temperature that result as the substance absorbs increasing amounts of heat. Plateaus in the curve (regions of constant temperature) are exhibited when the substance undergoes phase transitions.<\/div>\r\n<span id=\"fs-idm120231984\" data-type=\"media\" data-alt=\"A graph is shown where the x-axis is labeled \u201cAmount of heat added\u201d and the y-axis is labeled \u201cTemperature ( degree sign C )\u201d and has values of negative 10 to 100 in increments of 20. A right-facing horizontal arrow extends from point \u201c0, 0\u201d to the right side of the graph. A line graph begins at the lower left of the graph and moves to point \u201c0\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O ( s ).\u201d The line then flattens and travels horizontally for a small distance. This segment is labeled \u201cSolid begins to melt\u201d on its left side and \u201cAll solid melted\u201d on its right side. The line then goes steeply upward in a linear fashion until it hits point \u201c100\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O,( l ).\u201d The line then flattens and travels horizontally for a moderate distance. This segment is labeled \u201cLiquid begins to boil\u201d on its left side and \u201cAll liquid evaporated\u201d on its right side. The line then rises to a point above \u201c100\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O ( g ).\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_HeatCurve-1.jpg\" alt=\"A graph is shown where the x-axis is labeled \u201cAmount of heat added\u201d and the y-axis is labeled \u201cTemperature ( degree sign C )\u201d and has values of negative 10 to 100 in increments of 20. A right-facing horizontal arrow extends from point \u201c0, 0\u201d to the right side of the graph. A line graph begins at the lower left of the graph and moves to point \u201c0\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O ( s ).\u201d The line then flattens and travels horizontally for a small distance. This segment is labeled \u201cSolid begins to melt\u201d on its left side and \u201cAll solid melted\u201d on its right side. The line then goes steeply upward in a linear fashion until it hits point \u201c100\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O,( l ).\u201d The line then flattens and travels horizontally for a moderate distance. This segment is labeled \u201cLiquid begins to boil\u201d on its left side and \u201cAll liquid evaporated\u201d on its right side. The line then rises to a point above \u201c100\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O ( g ).\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-idm143749376\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm171202608\"><strong>Total Heat Needed to Change Temperature and Phase for a Substance:<\/strong><\/p>\r\nHow much heat is required to convert 135 g of ice at \u221215 \u00b0C into water vapor at 120 \u00b0C?\r\n\r\n&nbsp;\r\n<p id=\"fs-idm136425392\"><strong>Solution:<\/strong><\/p>\r\nThe transition described involves the following steps:\r\n<ol id=\"fs-idm226216704\" type=\"1\">\r\n \t<li>Heat ice from \u221215 \u00b0C to 0 \u00b0C<\/li>\r\n \t<li>Melt ice<\/li>\r\n \t<li>Heat water from 0 \u00b0C to 100 \u00b0C<\/li>\r\n \t<li>Boil water<\/li>\r\n \t<li>Heat steam from 100 \u00b0C to 120 \u00b0C<\/li>\r\n<\/ol>\r\n<p id=\"fs-idp5662336\">The heat needed to change the temperature of a given substance (with no change in phase) is: <em data-effect=\"italics\">q<\/em> = <em data-effect=\"italics\">m<\/em> \u00d7 <em data-effect=\"italics\">c<\/em> \u00d7 \u0394<em data-effect=\"italics\">T<\/em> (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by <em data-effect=\"italics\">q<\/em> = <em data-effect=\"italics\">n<\/em> \u00d7 \u0394<em data-effect=\"italics\">H<\/em>.<\/p>\r\n<p id=\"fs-idm193222960\">Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have:<\/p>\r\n<img class=\"wp-image-1622 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3j-300x119.png\" alt=\"\" width=\"494\" height=\"196\" \/>\r\n<p id=\"fs-idm91013680\">Converting the quantities in J to kJ permits them to be summed, yielding the total heat required:<\/p>\r\n\r\n<div id=\"fs-idm153490352\" data-type=\"equation\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 4.23 kJ + 45.0 kJ + 56.5 kJ + 305 kJ = 4.97 kJ = 416 kJ<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm144427968\"><strong>Check Your Learning:<\/strong><\/p>\r\nWhat is the total amount of heat released when 94.0 g water at 80.0 \u00b0C cools to form ice at \u221230.0 \u00b0C?\r\n<div id=\"fs-idm100519184\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm101204704\">68.7 kJ<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm147145776\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idm196349008\">Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp14058304\" class=\"key-equations\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Equations<\/strong><\/h3>\r\n<\/div>\r\n<div id=\"fs-idm73098240\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idm136735584\" data-type=\"exercise\">\r\n<div id=\"fs-idm76672896\" data-type=\"solution\">\r\n<h3 data-type=\"title\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <img class=\"alignnone size-medium wp-image-1617\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png\" alt=\"\" width=\"300\" height=\"63\" \/><\/h3>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"glossary\">\r\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\r\n<dl id=\"fs-idm193814624\">\r\n \t<dt>boiling point<\/dt>\r\n \t<dd id=\"fs-idm174536272\">temperature at which the vapor pressure of a liquid equals the pressure of the gas above it<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp13200944\">\r\n \t<dt>Clausius-Clapeyron equation<\/dt>\r\n \t<dd id=\"fs-idm153169776\">mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm193371200\">\r\n \t<dt>condensation<\/dt>\r\n \t<dd id=\"fs-idm201416208\">change from a gaseous to a liquid state<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm7778192\">\r\n \t<dt>deposition<\/dt>\r\n \t<dd id=\"fs-idm73497952\">change from a gaseous state directly to a solid state<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm101715056\">\r\n \t<dt>dynamic equilibrium<\/dt>\r\n \t<dd id=\"fs-idm64775616\">state of a system in which reciprocal processes are occurring at equal rates<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm144682752\">\r\n \t<dt>freezing<\/dt>\r\n \t<dd id=\"fs-idm50094928\">change from a liquid state to a solid state<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idp5444368\">\r\n \t<dt>freezing point<\/dt>\r\n \t<dd id=\"fs-idp203536\">temperature at which the solid and liquid phases of a substance are in equilibrium; see also <em data-effect=\"italics\">melting point<\/em><\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm69563216\">\r\n \t<dt>melting<\/dt>\r\n \t<dd id=\"fs-idm164437712\">change from a solid state to a liquid state<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm143011312\">\r\n \t<dt>melting point<\/dt>\r\n \t<dd id=\"fs-idm128669248\">temperature at which the solid and liquid phases of a substance are in equilibrium; see also <em data-effect=\"italics\">freezing point<\/em><\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm205028976\">\r\n \t<dt>normal boiling point<\/dt>\r\n \t<dd id=\"fs-idm139655296\">temperature at which a liquid\u2019s vapor pressure equals 1 atm (760 torr)<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm146567360\">\r\n \t<dt>sublimation<\/dt>\r\n \t<dd id=\"fs-idp17385840\">change from solid state directly to gaseous state<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm195421616\">\r\n \t<dt>vapor pressure<\/dt>\r\n \t<dd id=\"fs-idm149104240\">(also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm93994416\">\r\n \t<dt>vaporization<\/dt>\r\n \t<dd id=\"fs-idm90249920\">change from liquid state to gaseous state<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox textbox--learning-objectives\">\n<h3><strong>Learning Objectives<\/strong><\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Define phase transitions and phase transition temperatures<\/li>\n<li>Explain the relation between phase transition temperatures and intermolecular attractive forces<\/li>\n<li>Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp10874160\">We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth\u2019s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored.<\/p>\n<div id=\"fs-idm35636464\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Vaporization and Condensation<\/strong><\/h3>\n<p id=\"fs-idm100542016\">When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called <strong>condensation<\/strong>. When the rate of condensation becomes equal to the rate of <strong>vaporization<\/strong>, neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be <em data-effect=\"italics\">in equilibrium<\/em> with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a <span data-type=\"term\">dynamic equilibrium<\/span>, the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid\u2019s <strong>vapor pressure<\/strong> (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress1\">(Figure)<\/a>, and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_10_03_VapPress1\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from the liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue.<\/div>\n<p><span id=\"fs-idm103231072\" data-type=\"media\" data-alt=\"Three images are shown and labeled \u201ca,\u201d \u201cb,\u201d and \u201cc.\u201d Each image shows a round bulb connected on the right to a tube that is horizontal, then is bent vertically, curves, and then is vertical again to make a u-shape. A valve is located in the horizontal portion of the tube. Image a depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid. The phrase, \u201cMolecules escape surface and form vapor\u201d is written below the bulb, and a gray liquid in the u-shaped portion of the tube is shown at equal heights on the right and left sides. Image b depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. A gray liquid in the u-shaped portion of the tube is shown slightly higher on the right side than on the left side. Image c depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. There are more molecules present in c than in b. The phrase \u201cEquilibrium reached, vapor pressure determined,\u201d is written below the bulb and a gray liquid in the u-shaped portion of the tube is shown higher on the right side. A horizontal line is drawn level with each of these liquid levels and the distance between the lines is labeled with a double-headed arrow. This section is labeled with the phrase, \u201cVapor pressure.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_VapPress1-1.jpg\" alt=\"Three images are shown and labeled \u201ca,\u201d \u201cb,\u201d and \u201cc.\u201d Each image shows a round bulb connected on the right to a tube that is horizontal, then is bent vertically, curves, and then is vertical again to make a u-shape. A valve is located in the horizontal portion of the tube. Image a depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid. The phrase, \u201cMolecules escape surface and form vapor\u201d is written below the bulb, and a gray liquid in the u-shaped portion of the tube is shown at equal heights on the right and left sides. Image b depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. A gray liquid in the u-shaped portion of the tube is shown slightly higher on the right side than on the left side. Image c depicts a liquid in the bulb, labeled, \u201cLiquid,\u201d and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. There are more molecules present in c than in b. The phrase \u201cEquilibrium reached, vapor pressure determined,\u201d is written below the bulb and a gray liquid in the u-shaped portion of the tube is shown higher on the right side. A horizontal line is drawn level with each of these liquid levels and the distance between the lines is labeled with a double-headed arrow. This section is labeled with the phrase, \u201cVapor pressure.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idm63266864\">The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring \u201crecapture\u201d of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces.<\/p>\n<div id=\"fs-idm12389904\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm149114816\"><strong>Explaining Vapor Pressure in Terms of IMFs:<\/strong><\/p>\n<p>Given the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs:<\/p>\n<p><span id=\"fs-idm178035424\" data-type=\"media\" data-alt=\"Four Lewis structures are shown. The first structure, labeled \u201cethanol,\u201d shows a carbon bonded to three hydrogen atoms that is single bonded to a second carbon that is bonded to two hydrogen atoms and a hydroxyl group. The second structure, labeled \u201cethylene glycol, shows two carbon atoms, single bonded to one another, single bonded each to two hydrogen atoms, and each single bonded to a hydroxyl group. The third image, labeled \u201cdiethyl ether,\u201d shows an oxygen atom single bonded on both sides to a carbon that is bonded to two hydrogens, and a second carbon, that is itself bonded to three hydrogen atoms. The fourth image, labeled \u201cwater,\u201d shows an oxygen atom that is single bonded on both sides to hydrogen atoms.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_Ethanol_img-1.jpg\" alt=\"Four Lewis structures are shown. The first structure, labeled \u201cethanol,\u201d shows a carbon bonded to three hydrogen atoms that is single bonded to a second carbon that is bonded to two hydrogen atoms and a hydroxyl group. The second structure, labeled \u201cethylene glycol, shows two carbon atoms, single bonded to one another, single bonded each to two hydrogen atoms, and each single bonded to a hydroxyl group. The third image, labeled \u201cdiethyl ether,\u201d shows an oxygen atom single bonded on both sides to a carbon that is bonded to two hydrogens, and a second carbon, that is itself bonded to three hydrogen atoms. The fourth image, labeled \u201cwater,\u201d shows an oxygen atom that is single bonded on both sides to hydrogen atoms.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm12378064\"><strong>Solution:<\/strong><\/p>\n<p>Diethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two \u2212OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm224164560\"><strong>Check Your Learning:<\/strong><\/p>\n<p>At 20 \u00b0C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:<\/p>\n<table id=\"fs-idm60218336\" class=\"column-header medium unnumbered\" summary=\"This table has two rows and five columns. The first column is a header column, and it labels each row: \u201cCompound,\u201d and \u201cVapor Pressure at 25 degrees C.\u201d To the right of the \u201cCompound\u201d column are the following: methanol, C H subscript 3 O H; ethanol, C subscript 2 H subscript 5 O H; propanol C subscript 3 H subscript 7 O H; and butanol C subscript 4 H subscript 9 O H. To the right of the \u201cVapor Pressure at 25 degrees C\u201d column are the following: 11.9 k P a, 5.95 k P a, 2.67 k P a, and 0.56 k P a.\" data-label=\"\">\n<tbody valign=\"middle\">\n<tr valign=\"middle\">\n<td data-align=\"left\">Compound<\/td>\n<td data-align=\"left\">methanol CH<sub>3<\/sub>OH<\/td>\n<td data-align=\"left\">ethanol C<sub>2<\/sub>H<sub>5<\/sub>OH<\/td>\n<td data-align=\"left\">propanol C<sub>3<\/sub>H<sub>7<\/sub>OH<\/td>\n<td data-align=\"left\">butanol C<sub>4<\/sub>H<sub>9<\/sub>OH<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">Vapor Pressure at 20 \u00b0C<\/td>\n<td data-align=\"left\">11.9 kPa<\/td>\n<td data-align=\"left\">5.95 kPa<\/td>\n<td data-align=\"left\">2.67 kPa<\/td>\n<td data-align=\"left\">0.56 kPa<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-idm91362784\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm44101792\">All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:<span data-type=\"newline\"><br \/>\n<\/span> P<sub>methanol<\/sub> &gt; P<sub>ethanol<\/sub> &gt; P<sub>propanol<\/sub> &gt; P<sub>butanol<\/sub>.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm73489328\">As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress3\">(Figure)<\/a>. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_10_03_VapPress3\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, KE, to escape from the liquid into the gas phase.<\/div>\n<p><span id=\"fs-idm211797072\" data-type=\"media\" data-alt=\"A graph is shown where the y-axis is labeled \u201cNumber of molecules\u201d and the x-axis is labeled \u201cKinetic Energy.\u201d Two lines are graphed and a vertical dotted line, labeled \u201cMinimum K E needed to escape,\u201d is drawn halfway across the x-axis. The first line move sharply upward and has a high peak near the left side of the x-axis. It drops just as steeply and ends about 60 percent of the way across the x-axis. This line is labeled \u201cLow T.\u201d A second line, labeled \u201cHigh T,\u201d begins at the same point as the first, but does not go to such a high point, is wider, and ends slightly further to the right on the x-axis.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_VapPress3-1.jpg\" alt=\"A graph is shown where the y-axis is labeled \u201cNumber of molecules\u201d and the x-axis is labeled \u201cKinetic Energy.\u201d Two lines are graphed and a vertical dotted line, labeled \u201cMinimum K E needed to escape,\u201d is drawn halfway across the x-axis. The first line move sharply upward and has a high peak near the left side of the x-axis. It drops just as steeply and ends about 60 percent of the way across the x-axis. This line is labeled \u201cLow T.\u201d A second line, labeled \u201cHigh T,\u201d begins at the same point as the first, but does not go to such a high point, is wider, and ends slightly further to the right on the x-axis.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm153006576\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Boiling Points<\/strong><\/h3>\n<p id=\"fs-idm70709936\">When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The <strong>boiling point<\/strong> of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth\u2019s atmosphere. The<strong> normal boiling point<\/strong> of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress2\">(Figure)<\/a> shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid\u2019s boiling point on surrounding pressure.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_10_03_VapPress2\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.)<\/div>\n<p><span id=\"fs-idm188874672\" data-type=\"media\" data-alt=\"A graph is shown where the x-axis is labeled \u201cTemperature ( degree sign, C )\u201d and has values of 200 to 1000 in increments of 200 and the y-axis is labeled \u201cPressure ( k P a )\u201d and has values of 20 to 120 in increments of 20. A horizontal dotted line extends across the graph at point 780 on the y-axis while three vertical dotted lines extend from points 35, 78, and 100 to meet the horizontal dotted line. Four lines are graphed. The first line, labeled \u201cethyl ether,\u201d begins at the point \u201c0 , 200\u201d and extends in a slight curve to point \u201c45, 1000\u201d while the second line, labeled \u201cethanol\u201d, extends from point \u201c0, 20\u201d to point \u201c88, 1000\u201d in a more extreme curve. The third line, labeled \u201cwater,\u201d begins at the point \u201c0, 0\u201d and extends in a curve to point \u201c108, 1000\u201d while the fourth line, labeled \u201cethylene glycol,\u201d extends from point \u201c80, 0\u201d to point \u201c140, 100\u201d in a very shallow curve.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_VapPress2-1.jpg\" alt=\"A graph is shown where the x-axis is labeled \u201cTemperature ( degree sign, C )\u201d and has values of 200 to 1000 in increments of 200 and the y-axis is labeled \u201cPressure ( k P a )\u201d and has values of 20 to 120 in increments of 20. A horizontal dotted line extends across the graph at point 780 on the y-axis while three vertical dotted lines extend from points 35, 78, and 100 to meet the horizontal dotted line. Four lines are graphed. The first line, labeled \u201cethyl ether,\u201d begins at the point \u201c0 , 200\u201d and extends in a slight curve to point \u201c45, 1000\u201d while the second line, labeled \u201cethanol\u201d, extends from point \u201c0, 20\u201d to point \u201c88, 1000\u201d in a more extreme curve. The third line, labeled \u201cwater,\u201d begins at the point \u201c0, 0\u201d and extends in a curve to point \u201c108, 1000\u201d while the fourth line, labeled \u201cethylene glycol,\u201d extends from point \u201c80, 0\u201d to point \u201c140, 100\u201d in a very shallow curve.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<div id=\"fs-idm272706256\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm208677136\"><strong>A Boiling Point at Reduced Pressure:<\/strong><\/p>\n<p>A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress2\">(Figure)<\/a> to determine the boiling point of water at this elevation.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm10799984\"><strong>Solution:<\/strong><\/p>\n<p>The graph of the vapor pressure of water versus temperature in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress2\">(Figure)<\/a> indicates that the vapor pressure of water is 68 kPa at about 90 \u00b0C. Thus, at about 90 \u00b0C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm153672912\"><strong>Check Your Learning:<\/strong><\/p>\n<p>The boiling point of ethyl ether was measured to be 10 \u00b0C at a base camp on the slopes of Mount Everest. Use <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_VapPress2\">(Figure)<\/a> to determine the approximate atmospheric pressure at the camp.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm212817424\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm95497088\">Approximately 40 kPa (0.4 atm)<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm287754592\">The quantitative relation between a substance\u2019s vapor pressure and its temperature is described by the <span data-type=\"term\">Clausius-Clapeyron equation<\/span>:<\/p>\n<div id=\"fs-idm100818032\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1612 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3a.png\" alt=\"\" width=\"171\" height=\"37\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3a.png 194w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3a-65x14.png 65w\" sizes=\"auto, (max-width: 171px) 100vw, 171px\" \/><\/div>\n<p id=\"fs-idm95462576\">where \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub> is the enthalpy of vaporization for the liquid, <em data-effect=\"italics\">R<\/em> is the gas constant, and <em data-effect=\"italics\">A<\/em> is a constant whose value depends on the chemical identity of the substance. Temperature T must be in Kelvin in this equation. This equation is often rearranged into logarithmic form to yield the linear equation:<\/p>\n<div id=\"fs-idm13139376\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1614 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3b.png\" alt=\"\" width=\"206\" height=\"47\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3b.png 277w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3b-65x15.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3b-225x51.png 225w\" sizes=\"auto, (max-width: 206px) 100vw, 206px\" \/><\/div>\n<p id=\"fs-idm13032112\">This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature T<sub>1<\/sub>, the vapor pressure is P<sub>1<\/sub>, and at temperature T<sub>2<\/sub>, the vapor pressure is P<sub>2<\/sub>, the corresponding linear equations are:<\/p>\n<div id=\"fs-idm220001056\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1615 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3c-300x27.png\" alt=\"\" width=\"623\" height=\"56\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3c-300x27.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3c-768x68.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3c-65x6.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3c-225x20.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3c-350x31.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3c.png 844w\" sizes=\"auto, (max-width: 623px) 100vw, 623px\" \/><\/div>\n<p id=\"fs-idm51875808\">Since the constant, <em data-effect=\"italics\">A<\/em>, is the same, these two equations may be rearranged to isolate ln <em data-effect=\"italics\">A<\/em> and then set them equal to one another:<\/p>\n<div id=\"fs-idm142013200\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1616 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3d-300x59.png\" alt=\"\" width=\"259\" height=\"51\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3d-300x59.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3d-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3d-225x44.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3d-350x69.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3d.png 388w\" sizes=\"auto, (max-width: 259px) 100vw, 259px\" \/><\/div>\n<p id=\"fs-idm72918608\">which can be combined into:<\/p>\n<div id=\"fs-idm218464864\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1617 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png\" alt=\"\" width=\"257\" height=\"54\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-65x14.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-225x47.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-350x73.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e.png 386w\" sizes=\"auto, (max-width: 257px) 100vw, 257px\" \/><\/div>\n<div id=\"fs-idm96861632\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm305163312\"><strong>Estimating Enthalpy of Vaporization:<\/strong><\/p>\n<p>Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 \u00b0C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 \u00b0C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm195543920\"><strong>Solution:<\/strong><\/p>\n<p>The enthalpy of vaporization, \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub>, can be determined by using the Clausius-Clapeyron equation:<\/p>\n<div id=\"fs-idp17288896\" data-type=\"equation\"><strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1617 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png\" alt=\"\" width=\"257\" height=\"54\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-65x14.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-225x47.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-350x73.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e.png 386w\" sizes=\"auto, (max-width: 257px) 100vw, 257px\" \/><\/strong><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm95291536\">Since we have two vapor pressure-temperature values (<em data-effect=\"italics\">T<\/em><sub>1<\/sub> = 34.0 \u00b0C = 307.2 K, <em data-effect=\"italics\">P<\/em><sub>1<\/sub> = 10.0 kPa and <em data-effect=\"italics\">T<\/em><sub>2<\/sub> = 98.8 \u00b0C = 372.0 K, <em data-effect=\"italics\">P<\/em><sub>2<\/sub> = 100 kPa), we can substitute them into this equation and solve for \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub>. Rearranging the Clausius-Clapeyron equation and solving for \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub> yields:<\/p>\n<div id=\"fs-idm44175456\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1618 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3f-300x34.png\" alt=\"\" width=\"538\" height=\"61\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3f-300x34.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3f-768x86.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3f-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3f-225x25.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3f-350x39.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3f.png 988w\" sizes=\"auto, (max-width: 538px) 100vw, 538px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm201782400\">Note that the pressure can be in any units, so long as they agree for both <em data-effect=\"italics\">P<\/em> values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm101600256\"><strong>Check Your Learning:<\/strong><\/p>\n<p>At 20.0 \u00b0C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 \u00b0C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.<\/p>\n<div id=\"fs-idp12869536\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm146844672\">41,360 J\/mol or 41.4 kJ\/mol<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm143693744\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm467728\"><strong>Estimating Temperature (or Vapor Pressure):<\/strong><\/p>\n<p>For benzene (C<sub>6<\/sub>H<sub>6<\/sub>), the normal boiling point is 80.1 \u00b0C and the enthalpy of vaporization is 30.8 kJ\/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm220027984\"><strong>Solution:<\/strong><\/p>\n<p>If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, \u0394<em data-effect=\"italics\">H<\/em><sub>vap,<\/sub> then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation:<\/p>\n<div id=\"fs-idm92115136\" data-type=\"equation\"><strong><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1617 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png\" alt=\"\" width=\"257\" height=\"54\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-65x14.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-225x47.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-350x73.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e.png 386w\" sizes=\"auto, (max-width: 257px) 100vw, 257px\" \/><\/strong><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm198425552\">Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (<em data-effect=\"italics\">T<\/em><sub>1<\/sub> = 80.1 \u00b0C = 353.3 K, <em data-effect=\"italics\">P<\/em><sub>1<\/sub> = 101.3 kPa, \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub> = 30.8 kJ\/mol) and want to find the temperature (<em data-effect=\"italics\">T<\/em><sub>2<\/sub>) that corresponds to vapor pressure <em data-effect=\"italics\">P<\/em><sub>2<\/sub> = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for <em data-effect=\"italics\">T<\/em><sub>2<\/sub>. Rearranging the Clausius-Clapeyron equation and solving for <em data-effect=\"italics\">T<\/em><sub>2<\/sub> yields:<\/p>\n<div id=\"fs-idm103279184\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1619 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3g-300x33.png\" alt=\"\" width=\"509\" height=\"56\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3g-300x33.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3g-1024x112.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3g-768x84.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3g-65x7.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3g-225x25.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3g-350x38.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3g.png 1070w\" sizes=\"auto, (max-width: 509px) 100vw, 509px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm164810384\"><strong>Check Your Learning:<\/strong><\/p>\n<p>For acetone (CH<sub>3<\/sub>)<sub>2<\/sub>CO, the normal boiling point is 56.5 \u00b0C and the enthalpy of vaporization is 31.3 kJ\/mol. What is the vapor pressure of acetone at 25.0 \u00b0C?<\/p>\n<div id=\"fs-idm209654080\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm215055184\">30.1 kPa<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm196961088\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Enthalpy of Vaporization<\/strong><\/h3>\n<p id=\"fs-idm238047040\">Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub>. For example, the vaporization of water at standard temperature is represented by:<\/p>\n<div id=\"fs-idp7928688\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6H<sub>2<\/sub>O(<em>g<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>vap<\/sub>= 44.01 kJ<\/div>\n<p id=\"fs-idm149636464\">As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:<\/p>\n<div id=\"fs-idm190668720\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>O(<em>g<\/em>) \u27f6H<sub>2<\/sub>O(<em>l<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>con<\/sub> = -\u0394<em>H<\/em><sub>vap<\/sub>= -44.01 kJ<\/div>\n<div id=\"fs-idm57843872\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm70086528\"><strong>Using Enthalpy of Vaporization:<\/strong><\/p>\n<p>One way our body is cooled is by evaporation of the water in sweat (<a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_Evap\">(Figure)<\/a>). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at <em data-effect=\"italics\">T<\/em> = 37 \u00b0C (normal body temperature); \u0394<em data-effect=\"italics\">H<\/em><sub>vap<\/sub> = 43.46 kJ\/mol at 37 \u00b0C?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_10_03_Evap\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">Evaporation of sweat helps cool the body. (credit: \u201cKullez\u201d\/Flickr)<\/div>\n<p><span id=\"fs-idm183436304\" data-type=\"media\" data-alt=\"A person\u2019s shoulder and neck are shown and their skin is covered in beads of liquid.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_Evap-1.jpg\" alt=\"A person\u2019s shoulder and neck are shown and their skin is covered in beads of liquid.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-idm146648608\"><strong>Solution:<\/strong><\/p>\n<p>We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:<\/p>\n<div id=\"fs-idm38095424\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1621 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3i-300x39.png\" alt=\"\" width=\"400\" height=\"52\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3i-300x39.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3i-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3i-225x30.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3i-350x46.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3i.png 647w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/div>\n<p id=\"fs-idm104953776\">Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm119901584\"><strong>Check Your Learning:<\/strong><\/p>\n<p>How much heat is required to evaporate 100.0 g of liquid ammonia, NH<sub>3<\/sub>, at its boiling point if its enthalpy of vaporization is 4.8 kJ\/mol?<\/p>\n<div id=\"fs-idm64022976\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm13083920\">28 kJ<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm167698656\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Melting and Freezing<\/strong><\/h3>\n<p id=\"fs-idp14081632\">When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or <span data-type=\"term\">melting<\/span>. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (<a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_MeltingIce\">(Figure)<\/a>).<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_10_03_MeltingIce\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">(a) This beaker of ice has a temperature of \u221212.0 \u00b0C. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to 0 \u00b0C. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still 0 \u00b0C. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to 22.2 \u00b0C. (credit: modification of work by Mark Ott)<\/div>\n<p><span id=\"fs-idm137978064\" data-type=\"media\" data-alt=\"This figure shows four photos each labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and, \u201cd.\u201d Each photo shows a beaker with ice and a digital thermometer. The first photo shows ice cubes in the beaker, and the thermometer reads negative 12.0 degrees C. The second photo shows slightly melted ice, and the thermometer reads 0.0 degrees C. The third photo shows more water than ice in the beaker. The thermometer reads 0.0 degrees C. The fourth photo shows the ice completely melted, and the thermometer reads 22.2 degrees C.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_MeltingIce-1.jpg\" alt=\"This figure shows four photos each labeled, \u201ca,\u201d \u201cb,\u201d \u201cc,\u201d and, \u201cd.\u201d Each photo shows a beaker with ice and a digital thermometer. The first photo shows ice cubes in the beaker, and the thermometer reads negative 12.0 degrees C. The second photo shows slightly melted ice, and the thermometer reads 0.0 degrees C. The third photo shows more water than ice in the beaker. The thermometer reads 0.0 degrees C. The fourth photo shows the ice completely melted, and the thermometer reads 22.2 degrees C.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp12376064\">If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal processes of melting and <span data-type=\"term\">freezing<\/span> occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the <strong>melting point <\/strong>of the solid or the <strong>freezing point<\/strong> of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing).<\/p>\n<p id=\"fs-idm90105504\">The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.<\/p>\n<p id=\"fs-idm133861568\">The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, \u0394H<sub>fus<\/sub> of the substance. The enthalpy of fusion of ice is 6.0 kJ\/mol at 0 \u00b0C. Fusion (melting) is an endothermic process:<\/p>\n<div id=\"fs-idm185471776\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>O(<em>s<\/em>) \u27f6H<sub>2<\/sub>O(<em>l<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>fus<\/sub>= 6.01 kJ<\/div>\n<p id=\"fs-idm119984512\">The reciprocal process, freezing, is an exothermic process whose enthalpy change is \u22126.0 kJ\/mol at 0 \u00b0C:<\/p>\n<div id=\"fs-idm145529840\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>O(<em>l<\/em>) \u27f6H<sub>2<\/sub>O(<em>s<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>frz<\/sub> = -\u0394<em>H<\/em><sub>fus<\/sub>= -6.01 kJ<\/div>\n<\/div>\n<div id=\"fs-idm97104816\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Sublimation and Deposition<\/strong><\/h3>\n<p id=\"fs-idm191153504\">Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as <strong>sublimation<\/strong>. At room temperature and standard pressure, a piece of dry ice (solid CO<sub>2<\/sub>) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (<a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_Sublimtn\">(Figure)<\/a>). The reverse of sublimation is called <strong>deposition<\/strong>, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_10_03_Sublimtn\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott)<\/div>\n<p><span id=\"fs-idm195457584\" data-type=\"media\" data-alt=\"This figure shows a test tube. In the bottom is a dark substance which breaks up into a purple gas at the top.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_Sublimtn-1.jpg\" alt=\"This figure shows a test tube. In the bottom is a dark substance which breaks up into a purple gas at the top.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idm208982384\">Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, \u0394H<sub>sub<\/sub>, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:<\/p>\n<div id=\"fs-idm92174656\" style=\"text-align: center\" data-type=\"equation\">CO<sub>2<\/sub>(<em>s<\/em>) \u27f6 CO<sub>2<\/sub>(<em>g<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>sub<\/sub>= 26.1 kJ<\/div>\n<p id=\"fs-idm110822080\">Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:<\/p>\n<div id=\"fs-idp16237312\" style=\"text-align: center\" data-type=\"equation\">CO<sub>2<\/sub>(<em>g<\/em>) \u27f6CO<sub>2<\/sub>(<em>s<\/em>)\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>dep<\/sub> = -\u0394<em>H<\/em><sub>sub<\/sub>= -26.1 kJ<\/div>\n<p id=\"fs-idm69412784\">Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess\u2019s Law. Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_PhaseChng\">(Figure)<\/a>. For example:<\/p>\n<div id=\"fs-idm81561872\" data-type=\"equation\">solid \u27f6 liquid\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u0394<em>H<\/em><sub>fus<\/sub><\/div>\n<div data-type=\"equation\">liquid \u27f6 gas\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 \u0394<em>H<\/em><sub>vap<\/sub><\/div>\n<div data-type=\"equation\">&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-<\/div>\n<div data-type=\"equation\">solid \u27f6 gas\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u0394<em>H<\/em><sub>sub<\/sub> = \u0394<em>H<\/em><sub>fus<\/sub> + \u0394<em>H<\/em><sub>vap<\/sub><\/div>\n<div data-type=\"equation\"><\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"CNX_Chem_10_03_PhaseChng\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation.<\/div>\n<p><span id=\"fs-idm63916320\" data-type=\"media\" data-alt=\"A diagram is shown with a vertical line drawn on the left side and labeled \u201cEnergy\u201d and three horizontal lines drawn near the bottom, lower third and top of the diagram. These three lines are labeled, from bottom to top, \u201cSolid,\u201d \u201cLiquid\u201d and \u201cGas.\u201d Near the middle of the diagram, a vertical, upward-facing arrow is drawn from the solid line to the gas line and labeled \u201cSublimation, delta sign, H, subscript sub.\u201d To the right of this arrow is a second vertical, upward-facing arrow that is drawn from the solid line to the liquid line and labeled \u201cFusion, delta sign, H, subscript fus.\u201d Above the second arrow is a third arrow drawn from the liquid line to the gas line and labeled, \u201cVaporization, delta sign, H, subscript vap.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_PhaseChng-1.jpg\" alt=\"A diagram is shown with a vertical line drawn on the left side and labeled \u201cEnergy\u201d and three horizontal lines drawn near the bottom, lower third and top of the diagram. These three lines are labeled, from bottom to top, \u201cSolid,\u201d \u201cLiquid\u201d and \u201cGas.\u201d Near the middle of the diagram, a vertical, upward-facing arrow is drawn from the solid line to the gas line and labeled \u201cSublimation, delta sign, H, subscript sub.\u201d To the right of this arrow is a second vertical, upward-facing arrow that is drawn from the solid line to the liquid line and labeled \u201cFusion, delta sign, H, subscript fus.\u201d Above the second arrow is a third arrow drawn from the liquid line to the gas line and labeled, \u201cVaporization, delta sign, H, subscript vap.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm93930720\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Heating and Cooling Curves<\/strong><\/h3>\n<p id=\"fs-idp17603264\">In the chapter on thermochemistry, the relation between the amount of heat absorbed or released by a substance, <em data-effect=\"italics\">q<\/em>, and its accompanying temperature change, \u0394<em data-effect=\"italics\">T<\/em>, was introduced:<\/p>\n<div id=\"fs-idm100720480\" style=\"text-align: center\" data-type=\"equation\"><em>q<\/em> = <em>mc<\/em>\u0394<em>T<\/em><\/div>\n<p id=\"fs-idm73087600\">where <em data-effect=\"italics\">m<\/em> is the mass of the substance and <em data-effect=\"italics\">c<\/em> is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. <a class=\"autogenerated-content\" href=\"#CNX_Chem_10_03_HeatCurve\">(Figure)<\/a> shows a typical heating curve.<\/p>\n<p id=\"fs-idm178482560\">Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water\u2019s temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid). This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_10_03_HeatCurve\" class=\"bc-figure figure\">\n<div class=\"bc-figcaption figcaption\">A typical heating curve for a substance depicts changes in temperature that result as the substance absorbs increasing amounts of heat. Plateaus in the curve (regions of constant temperature) are exhibited when the substance undergoes phase transitions.<\/div>\n<p><span id=\"fs-idm120231984\" data-type=\"media\" data-alt=\"A graph is shown where the x-axis is labeled \u201cAmount of heat added\u201d and the y-axis is labeled \u201cTemperature ( degree sign C )\u201d and has values of negative 10 to 100 in increments of 20. A right-facing horizontal arrow extends from point \u201c0, 0\u201d to the right side of the graph. A line graph begins at the lower left of the graph and moves to point \u201c0\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O ( s ).\u201d The line then flattens and travels horizontally for a small distance. This segment is labeled \u201cSolid begins to melt\u201d on its left side and \u201cAll solid melted\u201d on its right side. The line then goes steeply upward in a linear fashion until it hits point \u201c100\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O,( l ).\u201d The line then flattens and travels horizontally for a moderate distance. This segment is labeled \u201cLiquid begins to boil\u201d on its left side and \u201cAll liquid evaporated\u201d on its right side. The line then rises to a point above \u201c100\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O ( g ).\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_10_03_HeatCurve-1.jpg\" alt=\"A graph is shown where the x-axis is labeled \u201cAmount of heat added\u201d and the y-axis is labeled \u201cTemperature ( degree sign C )\u201d and has values of negative 10 to 100 in increments of 20. A right-facing horizontal arrow extends from point \u201c0, 0\u201d to the right side of the graph. A line graph begins at the lower left of the graph and moves to point \u201c0\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O ( s ).\u201d The line then flattens and travels horizontally for a small distance. This segment is labeled \u201cSolid begins to melt\u201d on its left side and \u201cAll solid melted\u201d on its right side. The line then goes steeply upward in a linear fashion until it hits point \u201c100\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O,( l ).\u201d The line then flattens and travels horizontally for a moderate distance. This segment is labeled \u201cLiquid begins to boil\u201d on its left side and \u201cAll liquid evaporated\u201d on its right side. The line then rises to a point above \u201c100\u201d on the y-axis. This segment of the line is labeled \u201cH, subscript 2, O ( g ).\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-idm143749376\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm171202608\"><strong>Total Heat Needed to Change Temperature and Phase for a Substance:<\/strong><\/p>\n<p>How much heat is required to convert 135 g of ice at \u221215 \u00b0C into water vapor at 120 \u00b0C?<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm136425392\"><strong>Solution:<\/strong><\/p>\n<p>The transition described involves the following steps:<\/p>\n<ol id=\"fs-idm226216704\" type=\"1\">\n<li>Heat ice from \u221215 \u00b0C to 0 \u00b0C<\/li>\n<li>Melt ice<\/li>\n<li>Heat water from 0 \u00b0C to 100 \u00b0C<\/li>\n<li>Boil water<\/li>\n<li>Heat steam from 100 \u00b0C to 120 \u00b0C<\/li>\n<\/ol>\n<p id=\"fs-idp5662336\">The heat needed to change the temperature of a given substance (with no change in phase) is: <em data-effect=\"italics\">q<\/em> = <em data-effect=\"italics\">m<\/em> \u00d7 <em data-effect=\"italics\">c<\/em> \u00d7 \u0394<em data-effect=\"italics\">T<\/em> (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by <em data-effect=\"italics\">q<\/em> = <em data-effect=\"italics\">n<\/em> \u00d7 \u0394<em data-effect=\"italics\">H<\/em>.<\/p>\n<p id=\"fs-idm193222960\">Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1622 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3j-300x119.png\" alt=\"\" width=\"494\" height=\"196\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3j-300x119.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3j-768x304.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3j-65x26.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3j-225x89.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3j-350x138.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3j.png 829w\" sizes=\"auto, (max-width: 494px) 100vw, 494px\" \/><\/p>\n<p id=\"fs-idm91013680\">Converting the quantities in J to kJ permits them to be summed, yielding the total heat required:<\/p>\n<div id=\"fs-idm153490352\" data-type=\"equation\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 4.23 kJ + 45.0 kJ + 56.5 kJ + 305 kJ = 4.97 kJ = 416 kJ<\/div>\n<div data-type=\"equation\"><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm144427968\"><strong>Check Your Learning:<\/strong><\/p>\n<p>What is the total amount of heat released when 94.0 g water at 80.0 \u00b0C cools to form ice at \u221230.0 \u00b0C?<\/p>\n<div id=\"fs-idm100519184\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm101204704\">68.7 kJ<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm147145776\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idm196349008\">Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.<\/p>\n<\/div>\n<div id=\"fs-idp14058304\" class=\"key-equations\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Equations<\/strong><\/h3>\n<\/div>\n<div id=\"fs-idm73098240\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idm136735584\" data-type=\"exercise\">\n<div id=\"fs-idm76672896\" data-type=\"solution\">\n<h3 data-type=\"title\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1617\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png\" alt=\"\" width=\"300\" height=\"63\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-300x63.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-65x14.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-225x47.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e-350x73.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/10.3e.png 386w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/h3>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\"><strong>Glossary<\/strong><\/h3>\n<dl id=\"fs-idm193814624\">\n<dt>boiling point<\/dt>\n<dd id=\"fs-idm174536272\">temperature at which the vapor pressure of a liquid equals the pressure of the gas above it<\/dd>\n<\/dl>\n<dl id=\"fs-idp13200944\">\n<dt>Clausius-Clapeyron equation<\/dt>\n<dd id=\"fs-idm153169776\">mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance<\/dd>\n<\/dl>\n<dl id=\"fs-idm193371200\">\n<dt>condensation<\/dt>\n<dd id=\"fs-idm201416208\">change from a gaseous to a liquid state<\/dd>\n<\/dl>\n<dl id=\"fs-idm7778192\">\n<dt>deposition<\/dt>\n<dd id=\"fs-idm73497952\">change from a gaseous state directly to a solid state<\/dd>\n<\/dl>\n<dl id=\"fs-idm101715056\">\n<dt>dynamic equilibrium<\/dt>\n<dd id=\"fs-idm64775616\">state of a system in which reciprocal processes are occurring at equal rates<\/dd>\n<\/dl>\n<dl id=\"fs-idm144682752\">\n<dt>freezing<\/dt>\n<dd id=\"fs-idm50094928\">change from a liquid state to a solid state<\/dd>\n<\/dl>\n<dl id=\"fs-idp5444368\">\n<dt>freezing point<\/dt>\n<dd id=\"fs-idp203536\">temperature at which the solid and liquid phases of a substance are in equilibrium; see also <em data-effect=\"italics\">melting point<\/em><\/dd>\n<\/dl>\n<dl id=\"fs-idm69563216\">\n<dt>melting<\/dt>\n<dd id=\"fs-idm164437712\">change from a solid state to a liquid state<\/dd>\n<\/dl>\n<dl id=\"fs-idm143011312\">\n<dt>melting point<\/dt>\n<dd id=\"fs-idm128669248\">temperature at which the solid and liquid phases of a substance are in equilibrium; see also <em data-effect=\"italics\">freezing point<\/em><\/dd>\n<\/dl>\n<dl id=\"fs-idm205028976\">\n<dt>normal boiling point<\/dt>\n<dd id=\"fs-idm139655296\">temperature at which a liquid\u2019s vapor pressure equals 1 atm (760 torr)<\/dd>\n<\/dl>\n<dl id=\"fs-idm146567360\">\n<dt>sublimation<\/dt>\n<dd id=\"fs-idp17385840\">change from solid state directly to gaseous state<\/dd>\n<\/dl>\n<dl id=\"fs-idm195421616\">\n<dt>vapor pressure<\/dt>\n<dd id=\"fs-idm149104240\">(also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature<\/dd>\n<\/dl>\n<dl id=\"fs-idm93994416\">\n<dt>vaporization<\/dt>\n<dd id=\"fs-idm90249920\">change from liquid state to gaseous state<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-638","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":598,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/638","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/users\/1392"}],"version-history":[{"count":7,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/638\/revisions"}],"predecessor-version":[{"id":2151,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/638\/revisions\/2151"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/parts\/598"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/638\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=638"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=638"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=638"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=638"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}