{"id":718,"date":"2021-07-23T09:20:31","date_gmt":"2021-07-23T13:20:31","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/integrated-rate-laws\/"},"modified":"2023-01-20T13:57:54","modified_gmt":"2023-01-20T18:57:54","slug":"integrated-rate-laws","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/integrated-rate-laws\/","title":{"raw":"12.4 Integrated Rate Laws","rendered":"12.4 Integrated Rate Laws"},"content":{"raw":"<strong style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em;background-color: #cbd4b6;color: #000000\">Learning Objectives<\/strong>\r\n<div class=\"textbox textbox--learning-objectives\">\r\n\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Explain the form and function of an integrated rate law<\/li>\r\n \t<li>Perform integrated rate law calculations for zero-, first-, and second-order reactions<\/li>\r\n \t<li>Define half-life and carry out related calculations<\/li>\r\n \t<li>Identify the order of a reaction from concentration\/time data<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp89497264\">The rate laws discussed thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called <strong>integrated rate laws<\/strong>. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.<\/p>\r\n<p id=\"fs-idp93158656\">Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.<\/p>\r\n\r\n<div id=\"fs-idm140678368\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>First-Order Reactions<\/strong><\/h3>\r\nIntegration of the rate law for a simple first-order reaction (rate = <em data-effect=\"italics\">k<\/em>[<em>A<\/em>]) results in an equation describing how the reactant concentration varies with time:\r\n<div id=\"fs-idm380890320\" data-type=\"equation\"><img class=\"wp-image-1727 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a.png\" alt=\"\" width=\"133\" height=\"64\" \/><\/div>\r\n<p id=\"fs-idm63176528\">where [<em data-effect=\"italics\">A<\/em>]<sub><em data-effect=\"italics\">t<\/em><\/sub> is the concentration of <em data-effect=\"italics\">A<\/em> at any time <em data-effect=\"italics\">t<\/em>, [<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub> is the initial concentration of <em data-effect=\"italics\">A<\/em>, and <em data-effect=\"italics\">k<\/em> is the first-order rate constant.<\/p>\r\n<p id=\"fs-idm512172928\">For mathematical convenience, this equation may be rearranged to a format showing a linear dependence of concentration in time:<\/p>\r\n\r\n<div id=\"fs-idm358923408\" style=\"text-align: center\" data-type=\"equation\">ln[A]<sub>t<\/sub> = ln[A]<sub>0<\/sub>\u2212kt<\/div>\r\n<div id=\"fs-idm148528000\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm87528896\"><strong>The Integrated Rate Law for a First-Order Reaction:<\/strong><\/p>\r\nThe rate constant for the first-order decomposition of cyclobutane, C<sub>4<\/sub>H<sub>8<\/sub> at 500 \u00b0C is 9.2 \u00d7 10<sup>\u22123<\/sup> s<sup>\u22121<\/sup>:\r\n<div id=\"fs-idp60442464\" style=\"text-align: center\" data-type=\"equation\">C<sub>4<\/sub>H<sub>8<\/sub> \u27f6 2C<sub>2<\/sub>H<sub>4<\/sub><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm139703904\">How long will it take for 80.0% of a sample of C<sub>4<\/sub>H<sub>8<\/sub> to decompose?<\/p>\r\n<p id=\"fs-idm133511056\"><strong>Solution:<\/strong><\/p>\r\nSince the relative change in reactant concentration is provided, a convenient format for the integrated rate law is:\r\n<div id=\"fs-idp11818384\" data-type=\"equation\"><img class=\"wp-image-1727 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a.png\" alt=\"\" width=\"133\" height=\"64\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm149413632\">The initial concentration of C<sub>4<\/sub>H<sub>8<\/sub>, [<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub>, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let <em data-effect=\"italics\">x<\/em> be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of <em data-effect=\"italics\">x<\/em> or 0.200<em data-effect=\"italics\">x.<\/em> Rearranging the rate law to isolate <em data-effect=\"italics\">t<\/em> and substituting the provided quantities yields:<\/p>\r\n\r\n<div id=\"fs-idm92065248\" data-type=\"equation\"><img class=\"wp-image-1728 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4b-300x185.png\" alt=\"\" width=\"238\" height=\"147\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp38768624\"><strong>Check Your Learning:<\/strong><\/p>\r\nIodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:\r\n<div id=\"fs-idm115576112\" style=\"text-align: center\" data-type=\"equation\">I-131 \u27f6 Xe-131 + electron<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm29159424\">The decay is first-order with a rate constant of 0.138 d<sup>\u22121<\/sup>. How many days will it take for 90% of the iodine\u2212131 in a 0.500 <em data-effect=\"italics\">M<\/em> solution of this substance to decay to Xe-131?<\/p>\r\n\r\n<div id=\"fs-idm89005952\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm22449888\">16.7 days<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm132947408\">In the next example exercise, a linear format for the integrated rate law will be convenient:<\/p>\r\n\r\n<div id=\"fs-idm88512368\" style=\"text-align: center\" data-type=\"equation\">ln[<em>A<\/em>]<sub>t<\/sub> = -<em>kt<\/em> + ln[<em>A<\/em>]<sub>0<\/sub><\/div>\r\n<div style=\"text-align: center\" data-type=\"equation\">\u00a0<em>y<\/em> = <em>mx<\/em> + <em>b<\/em><\/div>\r\n<p id=\"fs-idm134557872\">A plot of ln[<em data-effect=\"italics\">A<\/em>]<sub><em data-effect=\"italics\">t<\/em><\/sub> versus <em data-effect=\"italics\">t<\/em> for a first-order reaction is a straight line with a slope of \u2212<em data-effect=\"italics\">k<\/em> and a <em data-effect=\"italics\">y<\/em>-intercept of ln[<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub>. If a set of rate data are plotted in this fashion but do <em data-effect=\"italics\">not<\/em> result in a straight line, the reaction is not first order in <em data-effect=\"italics\">A<\/em>.<\/p>\r\n\r\n<div id=\"fs-idp35909968\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm76041456\"><strong>Graphical Determination of Reaction Order and Rate Constant:<\/strong><\/p>\r\nThe following data is for the decomposition of H<sub>2<\/sub>O<sub>2<\/sub> at 40<sup>o<\/sup>C:\r\n\r\n<\/div>\r\n<p style=\"text-align: center\">2H<sub>2<\/sub>O<sub>2<\/sub> \u27f6 2H<sub>2<\/sub>O + O<sub>2<\/sub><\/p>\r\n\r\n<div id=\"fs-idp35909968\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n\r\nShow that the data can be represented by a first-order rate law by graphing ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time. Determine the rate constant for the decomposition of H<sub>2<\/sub>O<sub>2<\/sub> from these data.\r\n<table id=\"fs-idp114347648\" class=\"medium unnumbered\" summary=\"This table contains four columns and six rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( h ),\u201d \u201c[ H subscript 2 O subscript 2 ] ( M ),\u201d and \u201cl n [ H subscript 2 O subscript 2 ].\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the column, \u201cTime ( h )\u201d are the numbers 0, 6.00, 12.00, 18.00, and 24.00. Under the column \u201c[ H subscript 2 O subscript 2 ] ( M ),\u201d are the numbers 1.000, 0.500, 0.250, 0.125, and 0.0625. Under the column, \u201cl n [ H subscript 2 O subscript 2 ],\u201d are the numbers: 0.0, negative 0.693, negative 1.386, negative 2.079, and negative 2.772.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th data-align=\"left\">Trial<\/th>\r\n<th data-align=\"left\">Time (h)<\/th>\r\n<th data-align=\"left\">[H<sub>2<\/sub>O<sub>2<\/sub>] (<em data-effect=\"italics\">M<\/em>)<\/th>\r\n<th data-align=\"left\">ln[H<sub>2<\/sub>O<sub>2<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">1<\/td>\r\n<td data-align=\"left\">0.00<\/td>\r\n<td data-align=\"left\">1.000<\/td>\r\n<td data-align=\"left\">0.000<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">2<\/td>\r\n<td data-align=\"left\">6.00<\/td>\r\n<td data-align=\"left\">0.500<\/td>\r\n<td data-align=\"left\">\u22120.693<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">3<\/td>\r\n<td data-align=\"left\">12.00<\/td>\r\n<td data-align=\"left\">0.250<\/td>\r\n<td data-align=\"left\">\u22121.386<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">4<\/td>\r\n<td data-align=\"left\">18.00<\/td>\r\n<td data-align=\"left\">0.125<\/td>\r\n<td data-align=\"left\">\u22122.079<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">5<\/td>\r\n<td data-align=\"left\">24.00<\/td>\r\n<td data-align=\"left\">0.0625<\/td>\r\n<td data-align=\"left\">\u22122.772<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"CNX_Chem_12_04_FrstOKin\" class=\"scaled-down\">\r\n<div>\r\n<p id=\"fs-idm136711920\"><strong>Solution:<\/strong><\/p>\r\nThe plot of ln[H<sub>2<\/sub>O<sub>2<\/sub>] is shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_FrstOKin\">(Figure)<\/a>.\r\n\r\n<\/div>\r\n<div class=\"bc-figcaption figcaption\">A linear relationship between ln[H<sub>2<\/sub>O<sub>2<\/sub>] and time suggests the decomposition of hydrogen peroxide is a first-order reaction.<\/div>\r\n<span id=\"fs-idm90898176\" data-type=\"media\" data-alt=\"A graph is shown with the label \u201cTime ( h )\u201d on the x-axis and \u201cl n [ H subscript 2 O subscript 2 ]\u201d on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_FrstOKin-2.jpg\" alt=\"A graph is shown with the label \u201cTime ( h )\u201d on the x-axis and \u201cl n [ H subscript 2 O subscript 2 ]\u201d on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idm88773168\">The plot of ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time is linear, indicating that the reaction may be described by a first-order rate law.<\/p>\r\n<p id=\"fs-idm97546480\">According to the linear format of the first-order integrated rate law, the rate constant is given by the negative of this plot\u2019s slope.<\/p>\r\n\r\n<div id=\"fs-idp49567824\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1729 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4c-300x48.png\" alt=\"\" width=\"300\" height=\"48\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm145444128\">The slope of this line may be derived from two values of ln[H<sub>2<\/sub>O<sub>2<\/sub>] at different values of <em data-effect=\"italics\">t<\/em> (one near each end of the line is preferable). For example, the value of ln[H<sub>2<\/sub>O<sub>2<\/sub>] when <em data-effect=\"italics\">t<\/em> is 0.00 h is 0.000; the value when <em data-effect=\"italics\">t<\/em> = 24.00 h is \u22122.772.<\/p>\r\n\r\n<div id=\"fs-idm14805520\" data-type=\"equation\"><img class=\"size-medium wp-image-1730 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4d-300x100.png\" alt=\"\" width=\"300\" height=\"100\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp92343136\"><strong>Check Your Learning:<\/strong><\/p>\r\nGraph the following data to determine whether the reaction <em>A<\/em> \u27f6 <em>B<\/em> + <em>C<\/em> is first order.\r\n<table id=\"fs-idm149704608\" class=\"medium unnumbered\" summary=\"This table has three columns and six rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and, \u201c[ A ].\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201cTime ( s )\u201d column are the numbers: 4.0, 8.0, 12.0, 16.0, and 20.0. Under the \u201c [ A ]\u201d column are the numbers: 0.220, 0.144, 0.110, 0.088, and 0.074.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th data-align=\"left\">Trial<\/th>\r\n<th data-align=\"left\">Time (s)<\/th>\r\n<th data-align=\"left\">[<em data-effect=\"italics\">A<\/em>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">1<\/td>\r\n<td data-align=\"left\">4.0<\/td>\r\n<td data-align=\"left\">0.220<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">2<\/td>\r\n<td data-align=\"left\">8.0<\/td>\r\n<td data-align=\"left\">0.144<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">3<\/td>\r\n<td data-align=\"left\">12.0<\/td>\r\n<td data-align=\"left\">0.110<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">4<\/td>\r\n<td data-align=\"left\">16.0<\/td>\r\n<td data-align=\"left\">0.088<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">5<\/td>\r\n<td data-align=\"left\">20.0<\/td>\r\n<td data-align=\"left\">0.074<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-idm206842256\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp73329760\">The plot of ln[<em data-effect=\"italics\">A<\/em>]<sub><em data-effect=\"italics\">t<\/em><\/sub> vs. <em data-effect=\"italics\">t<\/em> is not linear, indicating the reaction is not first order:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n<span id=\"fs-idm135178736\" class=\"scaled-down\" data-type=\"media\" data-alt=\"A graph, labeled above as \u201cl n [ A ] vs. Time\u201d is shown. The x-axis is labeled, \u201cTime ( s )\u201d and the y-axis is labeled, \u201cl n [ A ].\u201d The x-axis shows markings at 5, 10, 15, 20, and 25 hours. The y-axis shows markings at negative 3, negative 2, negative 1, and 0. A slight curve is drawn connecting five points at coordinates of approximately (4, negative 1.5), (8, negative 2), (12, negative 2.2), (16, negative 2.4), and (20, negative 2.6).\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_CYL1_img-2.jpg\" alt=\"A graph, labeled above as \u201cl n [ A ] vs. Time\u201d is shown. The x-axis is labeled, \u201cTime ( s )\u201d and the y-axis is labeled, \u201cl n [ A ].\u201d The x-axis shows markings at 5, 10, 15, 20, and 25 hours. The y-axis shows markings at negative 3, negative 2, negative 1, and 0. A slight curve is drawn connecting five points at coordinates of approximately (4, negative 1.5), (8, negative 2), (12, negative 2.2), (16, negative 2.4), and (20, negative 2.6).\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm162181696\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Second-Order Reactions<\/strong><\/h3>\r\n<p id=\"fs-idm75401888\">The equations that relate the concentrations of reactants and the rate constant of second-order reactions can be fairly complicated. To illustrate the point with minimal complexity, only the simplest second-order reactions will be described here, namely, those whose rates depend on the concentration of just one reactant. For these types of reactions, the differential rate law is written as:<\/p>\r\n\r\n<div id=\"fs-idm149309552\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[A]<sup>2<\/sup><\/div>\r\n<p id=\"fs-idm55886848\">For these second-order reactions, the integrated rate law is:<\/p>\r\n\r\n<div id=\"fs-idm72168352\" data-type=\"equation\"><img class=\"wp-image-1731 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png\" alt=\"\" width=\"137\" height=\"56\" \/><\/div>\r\n<p id=\"fs-idm19194496\">where the terms in the equation have their usual meanings as defined earlier.<\/p>\r\n\r\n<div id=\"fs-idm85213520\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp6881136\"><strong>The Integrated Rate Law for a Second-Order Reaction:<\/strong><\/p>\r\nThe reaction of butadiene gas (C<sub>4<\/sub>H<sub>6<\/sub>) to yield C<sub>8<\/sub>H<sub>12<\/sub> gas is described by the equation:\r\n<div id=\"fs-idm46947984\" style=\"text-align: center\" data-type=\"equation\">2C<sub>4<\/sub>H<sub>6<\/sub>(<em>g<\/em>) \u27f6 C<sub>8<\/sub>H<sub>12<\/sub>(<em>g<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idp68803280\">This \u201cdimerization\u201d reaction is second order with a rate constant equal to 5.76 \u00d7 10<sup>\u22122<\/sup> L mol<sup>\u22121<\/sup> min<sup>\u22121<\/sup> under certain conditions. If the initial concentration of butadiene is 0.200 <em data-effect=\"italics\">M<\/em>, what is the concentration after 10.0 min?<\/p>\r\n<p id=\"fs-idm47025104\"><strong>Solution:<\/strong><\/p>\r\nFor a second-order reaction, the integrated rate law is written\r\n<div id=\"fs-idm191622048\" data-type=\"equation\"><img class=\"aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png\" width=\"128\" height=\"52\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm25993952\">We know three variables in this equation: [<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub> = 0.200 mol\/L, <em data-effect=\"italics\">k<\/em> = 5.76 \u00d7 10<sup>\u22122<\/sup> L<sup>.<\/sup>mol<sup>-1.<\/sup>min, and <em data-effect=\"italics\">t<\/em> = 10.0 min. Therefore, we can solve for [<em data-effect=\"italics\">A<\/em>], the fourth variable:<\/p>\r\n\r\n<div id=\"fs-idp58774528\" data-type=\"equation\"><img class=\"alignnone wp-image-1733 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4f-300x87.png\" alt=\"\" width=\"314\" height=\"91\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm85057584\">Therefore 0.179 mol\/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol\/L that was originally present.<\/p>\r\n<p id=\"fs-idp90256\"><strong>Check Your Learning:<\/strong><\/p>\r\nIf the initial concentration of butadiene is 0.0200 <em data-effect=\"italics\">M<\/em>, what is the concentration remaining after 20.0 min?\r\n<div id=\"fs-idp27082688\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm23200256\">0.0195 mol\/L<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm88643152\">The integrated rate law for second-order reactions has the form of the equation of a straight line:<\/p>\r\n\r\n<div id=\"fs-idm140436832\" data-type=\"equation\"><img class=\"aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png\" width=\"128\" height=\"52\" \/><\/div>\r\n<div data-type=\"equation\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<em>y<\/em> = <em>mx<\/em> + <em>b<\/em><\/div>\r\n<p id=\"fs-idm67290896\">A plot of 1\/[<em>A<\/em>]<sub>t<\/sub> versus <em data-effect=\"italics\">t<\/em> for a second-order reaction is a straight line with a slope of <em data-effect=\"italics\">k<\/em> and a <em data-effect=\"italics\">y<\/em>-intercept of 1\/[<em>A<\/em>]<sub>0<\/sub>. If the plot is not a straight line, then the reaction is not second order.<\/p>\r\n\r\n<div id=\"fs-idp69093472\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm27496000\"><strong>Graphical Determination of Reaction Order and Rate Constant:<\/strong><\/p>\r\n\u00a0The data below are for the same reaction described in <a class=\"autogenerated-content\" href=\"#fs-idm85213520\">(Figure)<\/a>. Prepare and compare two appropriate data plots to identify the reaction as being either first or second order. After identifying the reaction order, estimate a value for the rate constant.\r\n<p id=\"fs-idp4978176\"><strong>Solution:<\/strong><\/p>\r\n\r\n<table id=\"fs-idm140502592\" class=\"medium unnumbered\" summary=\"This table contains three columns and six rows. The first row is a header row and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and \u201c[ C subscript 4 H subscript 6 ] ( M ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201cTime ( s )\u201d column are the numbers: 0, 1600, 3200, 4800, and 6200. Under the column \u201c[ C subscript 4 H subscript 6 ] ( M )\u201d are the numbers: 1.00 times ten to the negative 2; 5.04 times ten to the negative 3; 3.37 times ten to the negative 3; 2.53 times ten to the negative 3; and 2.08 times ten to the negative 3.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th data-align=\"left\">Trial<\/th>\r\n<th data-align=\"left\">Time (s)<\/th>\r\n<th data-align=\"left\">[C<sub>4<\/sub>H<sub>6<\/sub>] (<em data-effect=\"italics\">M<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">1<\/td>\r\n<td data-align=\"left\">0<\/td>\r\n<td data-align=\"left\">1.00 \u00d7 10<sup>\u22122<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">2<\/td>\r\n<td data-align=\"left\">1600<\/td>\r\n<td data-align=\"left\">5.04 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">3<\/td>\r\n<td data-align=\"left\">3200<\/td>\r\n<td data-align=\"left\">3.37 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">4<\/td>\r\n<td data-align=\"left\">4800<\/td>\r\n<td data-align=\"left\">2.53 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">5<\/td>\r\n<td data-align=\"left\">6200<\/td>\r\n<td data-align=\"left\">2.08 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idm190212352\">In order to distinguish a first-order reaction from a second-order reaction, prepare a plot of ln[C<sub>4<\/sub>H<sub>6<\/sub>]<sub><em data-effect=\"italics\">t<\/em><\/sub> versus <em data-effect=\"italics\">t<\/em> and compare it to a plot of 1\/[C<sub>4<\/sub>H<sub>6<\/sub>]<sub>t<\/sub> versus <em data-effect=\"italics\">t<\/em>. The values needed for these plots follow.<\/p>\r\n\r\n<table id=\"fs-idp62232800\" class=\"medium unnumbered\" summary=\"This table contains three columns and six rows. The first row is a header row and it labels each column, \u201cTime ( s ),\u201d \u201c1 over [ C subscript 4 H subscript 6 ] ( M superscript negative 1 ),\u201d and \u201cl n [ C subscript 4 H subscript 6 ].\u201d Under the column \u201cTime ( s )\u201d are the numbers: 0, 1600, 3200, 4800, and 6200. Under the \u201c1 over [ C subscript 4 H subscript 6 ] ( M superscript negative 1 )\u201d column are the numbers: 100, 198, 296, 395, and 481. Under the \u201cl n [ C subscript 4 H subscript 6 ]\u201d column are the numbers: negative 4.605, negative 5.289, negative 5.692, negative 5.978, and negative 6.175.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th data-align=\"left\">Time (s)<\/th>\r\n<th data-align=\"left\">1\/[C<sub>4<\/sub>H<sub>6<\/sub>] (M<sup>-1<\/sup>)<\/th>\r\n<th data-align=\"left\">ln[C<sub>4<\/sub>H<sub>6<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">0<\/td>\r\n<td data-align=\"left\">100<\/td>\r\n<td data-align=\"left\">\u22124.605<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">1600<\/td>\r\n<td data-align=\"left\">198<\/td>\r\n<td data-align=\"left\">\u22125.289<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">3200<\/td>\r\n<td data-align=\"left\">296<\/td>\r\n<td data-align=\"left\">\u22125.692<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">4800<\/td>\r\n<td data-align=\"left\">395<\/td>\r\n<td data-align=\"left\">\u22125.978<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">6200<\/td>\r\n<td data-align=\"left\">481<\/td>\r\n<td data-align=\"left\">\u22126.175<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp69115360\">The plots are shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_2OrdKin\">(Figure)<\/a>, which clearly shows the plot of ln[C<sub>4<\/sub>H<sub>6<\/sub>]<sub><em data-effect=\"italics\">t<\/em><\/sub> versus <em data-effect=\"italics\">t<\/em> is not linear, therefore the reaction is not first order. The plot of 1\/[C<sub>4<\/sub>H<sub>6<\/sub>]<em><sub>t<\/sub><\/em> versus <em data-effect=\"italics\">t<\/em> is linear, indicating that the reaction is second order.<\/p>\r\n&nbsp;\r\n<div id=\"CNX_Chem_12_04_2OrdKin\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">These two graphs show first- and second-order plots for the dimerization of C<sub>4<\/sub>H<sub>6<\/sub>. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.<\/div>\r\n<span id=\"fs-idp16234496\" data-type=\"media\" data-alt=\"Two graphs are shown, each with the label \u201cTime ( s )\u201d on the x-axis. The graph on the left is labeled, \u201cl n [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The graph on the right is labeled \u201c1 divided by [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The x-axes for both graphs show markings at 3000 and 6000. The y-axis for the graph on the left shows markings at negative 6, negative 5, and negative 4. A decreasing slightly concave up curve is drawn through five points at coordinates that are (0, negative 4.605), (1600, negative 5.289), (3200, negative 5.692), (4800, negative 5.978), and (6200, negative 6.175). The y-axis for the graph on the right shows markings at 100, 300, and 500. An approximately linear increasing curve is drawn through five points at coordinates that are (0, 100), (1600, 198), (3200, 296), and (4800, 395), and (6200, 481).\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_2OrdKin-2.jpg\" alt=\"Two graphs are shown, each with the label \u201cTime ( s )\u201d on the x-axis. The graph on the left is labeled, \u201cl n [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The graph on the right is labeled \u201c1 divided by [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The x-axes for both graphs show markings at 3000 and 6000. The y-axis for the graph on the left shows markings at negative 6, negative 5, and negative 4. A decreasing slightly concave up curve is drawn through five points at coordinates that are (0, negative 4.605), (1600, negative 5.289), (3200, negative 5.692), (4800, negative 5.978), and (6200, negative 6.175). The y-axis for the graph on the right shows markings at 100, 300, and 500. An approximately linear increasing curve is drawn through five points at coordinates that are (0, 100), (1600, 198), (3200, 296), and (4800, 395), and (6200, 481).\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idm339836784\">According to the second-order integrated rate law, the rate constant is equal to the slope of the 1\/[<em>A<\/em>]<em><sub>t<\/sub><\/em> versus <em data-effect=\"italics\">t<\/em> plot. Using the data for <em data-effect=\"italics\">t<\/em> = 0 <em data-effect=\"italics\">s<\/em> and <em data-effect=\"italics\">t<\/em> = 6200 <em data-effect=\"italics\">s<\/em>, the rate constant is estimated as follows:<\/p>\r\n\r\n<div id=\"fs-idm363866800\" data-type=\"equation\"><img class=\"size-medium wp-image-1735 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4g-300x43.png\" alt=\"\" width=\"300\" height=\"43\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm56284464\"><strong>Check Your Learning:<\/strong><\/p>\r\nDo the following data fit a second-order rate law?\r\n<table id=\"fs-idm88760288\" class=\"medium unnumbered\" summary=\"This table contains three columns and seven rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and, \u201c[ A ] ( M ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, 5, and 6. Under the \u201cTime ( s )\u201d column are the numbers: 5, 10, 15, 20, 25, and 35. Under the \u201c[ A ] ( M )\u201d column are the numbers 0.952, 0.625, 0.465, 0.370, 0.308, and 0.230.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th data-align=\"left\">Trial<\/th>\r\n<th data-align=\"left\">Time (s)<\/th>\r\n<th data-align=\"left\">[<em data-effect=\"italics\">A<\/em>] (<em data-effect=\"italics\">M<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">1<\/td>\r\n<td data-align=\"left\">5<\/td>\r\n<td data-align=\"left\">0.952<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">2<\/td>\r\n<td data-align=\"left\">10<\/td>\r\n<td data-align=\"left\">0.625<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">3<\/td>\r\n<td data-align=\"left\">15<\/td>\r\n<td data-align=\"left\">0.465<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">4<\/td>\r\n<td data-align=\"left\">20<\/td>\r\n<td data-align=\"left\">0.370<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">5<\/td>\r\n<td data-align=\"left\">25<\/td>\r\n<td data-align=\"left\">0.308<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">6<\/td>\r\n<td data-align=\"left\">35<\/td>\r\n<td data-align=\"left\">0.230<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-idm72354384\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idp122966448\">Yes. The plot of 1\/[<em>A<\/em>]<em><sub>t <\/sub><\/em>vs. <em data-effect=\"italics\">t<\/em> is linear:<\/p>\r\n<span id=\"fs-idm144508704\" class=\"scaled-down\" data-type=\"media\" data-alt=\"A graph, with the title \u201c1 divided by [ A ] vs. Time\u201d is shown, with the label, \u201cTime ( s ),\u201d on the x-axis. The label \u201c1 divided by [ A ]\u201d appears left of the y-axis. The x-axis shows markings beginning at zero and continuing at intervals of 10 up to and including 40. The y-axis on the left shows markings beginning at 0 and increasing by intervals of 1 up to and including 5. A line with an increasing trend is drawn through six points at approximately (4, 1), (10, 1.5), (15, 2.2), (20, 2.8), (26, 3.4), and (36, 4.4).\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_CYL2_img-2.jpg\" alt=\"A graph, with the title \u201c1 divided by [ A ] vs. Time\u201d is shown, with the label, \u201cTime ( s ),\u201d on the x-axis. The label \u201c1 divided by [ A ]\u201d appears left of the y-axis. The x-axis shows markings beginning at zero and continuing at intervals of 10 up to and including 40. The y-axis on the left shows markings beginning at 0 and increasing by intervals of 1 up to and including 5. A line with an increasing trend is drawn through six points at approximately (4, 1), (10, 1.5), (15, 2.2), (20, 2.8), (26, 3.4), and (36, 4.4).\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm152558640\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Zero-Order Reactions<\/strong><\/h3>\r\n<p id=\"fs-idp474768\">For zero-order reactions, the differential rate law is:<\/p>\r\n\r\n<div id=\"fs-idp24382368\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em><\/div>\r\n<p id=\"fs-idm91448320\">A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactant(s). This may seem counterintuitive, since the reaction rate certainly can\u2019t be finite when the reactant concentration is zero. For purposes of this introductory text, it will suffice to note that zero-order kinetics are observed for some reactions only under certain specific conditions. These same reactions exhibit different kinetic behaviors when the specific conditions aren\u2019t met, and for this reason the more prudent term <em data-effect=\"italics\">pseudo-zero-order<\/em> is sometimes used.<\/p>\r\n<p id=\"fs-idp73236848\">The integrated rate law for a zero-order reaction is a linear function:<\/p>\r\n\r\n<div id=\"fs-idm130548304\" style=\"text-align: center\" data-type=\"equation\">[<em>A<\/em>]<sub><em>t<\/em><\/sub> = \u2212<em>kt<\/em> + [<em>A<\/em>]<sub>0<\/sub><\/div>\r\n<div style=\"text-align: center\" data-type=\"equation\"><em>y<\/em> = <em>mx<\/em> + <em>b<\/em><\/div>\r\n<p id=\"fs-idm99514160\">A plot of [<em data-effect=\"italics\">A<\/em>] versus <em data-effect=\"italics\">t<\/em> for a zero-order reaction is a straight line with a slope of <em data-effect=\"italics\">\u2212k<\/em> and a <em data-effect=\"italics\">y<\/em>-intercept of [<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub>. <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_AmDecomK\">(Figure)<\/a> shows a plot of [NH<sub>3<\/sub>] versus <em data-effect=\"italics\">t<\/em> for the thermal decomposition of ammonia at the surface of two different heated solids. The decomposition reaction exhibits first-order behavior at a quartz (SiO<sub>2<\/sub>) surface, as suggested by the exponentially decaying plot of concentration versus time. On a tungsten surface, however, the plot is linear, indicating zero-order kinetics.<\/p>\r\n\r\n<div id=\"fs-idm339943456\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm358086992\"><strong>Graphical Determination of Zero-Order Rate Constant:<\/strong><\/p>\r\nUse the data plot in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_AmDecomK\">(Figure)<\/a> to graphically estimate the zero-order rate constant for ammonia decomposition at a tungsten surface.\r\n<p id=\"fs-idm382849456\"><strong>Solution:<\/strong><\/p>\r\nThe integrated rate law for zero-order kinetics describes a linear plot of reactant concentration, [<em data-effect=\"italics\">A<\/em>]<sub><em data-effect=\"italics\">t<\/em><\/sub>, versus time, <em data-effect=\"italics\">t<\/em>, with a slope equal to the negative of the rate constant, \u2212<em data-effect=\"italics\">k<\/em>. Following the mathematical approach of previous examples, the slope of the linear data plot (for decomposition on W) is estimated from the graph. Using the ammonia concentrations at <em data-effect=\"italics\">t<\/em> = 0 and <em data-effect=\"italics\">t<\/em> = 1000 s:\r\n<div id=\"fs-idm375658448\" data-type=\"equation\"><img class=\" wp-image-1736 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4h-300x25.png\" alt=\"\" width=\"528\" height=\"44\" \/><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm376604432\"><strong>Check Your Learning:<\/strong><\/p>\r\nThe zero-order plot in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_AmDecomK\">(Figure)<\/a> shows an initial ammonia concentration of 0.0028 mol L<sup>\u22121<\/sup> decreasing linearly with time for 1000 s. Assuming no change in this zero-order behavior, at what time (min) will the concentration reach 0.0001 mol L<sup>\u22121<\/sup>?\r\n\r\n&nbsp;\r\n<div id=\"fs-idm379795136\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm508285088\">35 min<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"CNX_Chem_12_04_AmDecomK\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">The decomposition of NH<sub>3<\/sub> on a tungsten (W) surface is a zero-order reaction, whereas on a quartz (SiO<sub>2<\/sub>) surface, the reaction is first order.<\/div>\r\n<span id=\"fs-idp157593168\" data-type=\"media\" data-alt=\"A graph is shown with the label, \u201cTime ( s ),\u201d on the x-axis and, \u201c[ N H subscript 3 ] M,\u201d on the y-axis. The x-axis shows a single value of 1000 marked near the right end of the axis. The vertical axis shows markings at 1.0 times 10 superscript negative 3, 2.0 times 10 superscript negative 3, and 3.0 times 10 superscript negative 3. A decreasing linear trend line is drawn through six points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (200, 2.6 times 10 superscript negative 3), (400, 2.3 times 10 superscript negative 3), (600, 2.0 times 10 superscript negative 3), (800, 1.8 times 10 superscript negative 3), and (1000, 1.6 times 10 superscript negative 3). This line is labeled \u201cDecomposition on W.\u201d A decreasing slightly concave up curve is similarly drawn through eight points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (100, 2.5 times 10 superscript negative 3), (200, 2.1 times 10 superscript negative 3), (300, 1.9 times 10 superscript negative 3), (400, 1.6 times 10 superscript negative 3), (500, 1.4 times 10 superscript negative 3), and (750, 1.1 times 10 superscript negative 3), ending at about (1000, 0.7 times 10 superscript negative 3). This curve is labeled \u201cDecomposition on S i O subscript 2.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_AmDecomK-2.jpg\" alt=\"A graph is shown with the label, \u201cTime ( s ),\u201d on the x-axis and, \u201c[ N H subscript 3 ] M,\u201d on the y-axis. The x-axis shows a single value of 1000 marked near the right end of the axis. The vertical axis shows markings at 1.0 times 10 superscript negative 3, 2.0 times 10 superscript negative 3, and 3.0 times 10 superscript negative 3. A decreasing linear trend line is drawn through six points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (200, 2.6 times 10 superscript negative 3), (400, 2.3 times 10 superscript negative 3), (600, 2.0 times 10 superscript negative 3), (800, 1.8 times 10 superscript negative 3), and (1000, 1.6 times 10 superscript negative 3). This line is labeled \u201cDecomposition on W.\u201d A decreasing slightly concave up curve is similarly drawn through eight points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (100, 2.5 times 10 superscript negative 3), (200, 2.1 times 10 superscript negative 3), (300, 1.9 times 10 superscript negative 3), (400, 1.6 times 10 superscript negative 3), (500, 1.4 times 10 superscript negative 3), and (750, 1.1 times 10 superscript negative 3), ending at about (1000, 0.7 times 10 superscript negative 3). This curve is labeled \u201cDecomposition on S i O subscript 2.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp6500464\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>The Half-Life of a Reaction<\/strong><\/h3>\r\n<p id=\"fs-idm99562880\">The <span data-type=\"term\">half-life of a reaction (<em data-effect=\"italics\">t<\/em><sub>1\/2<\/sub>)<\/span> is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (<a class=\"autogenerated-content\" href=\"\/contents\/19ee5333-b3d6-42c0-9ee6-ca6fcf3b1c62#CNX_Chem_12_01_KDataH2O2\">(Figure)<\/a>) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases from 1.000 <em data-effect=\"italics\">M<\/em> to 0.500 <em data-effect=\"italics\">M<\/em>. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 <em data-effect=\"italics\">M<\/em> to 0.250 <em data-effect=\"italics\">M<\/em>; during the third half-life, it decreases from 0.250 <em data-effect=\"italics\">M<\/em> to 0.125 <em data-effect=\"italics\">M<\/em>. The concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.<\/p>\r\n\r\n<div id=\"fs-idm97307168\" class=\"bc-section section\" data-depth=\"2\">\r\n<h4 data-type=\"title\"><strong>First-Order Reactions<\/strong><\/h4>\r\n<p id=\"fs-idp162624\">An equation relating the half-life of a first-order reaction to its rate constant may be derived from the integrated rate law as follows:<\/p>\r\n\r\n<div id=\"fs-idm58156032\" data-type=\"equation\"><img class=\" wp-image-1737 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4i-300x143.png\" alt=\"\" width=\"206\" height=\"98\" \/><\/div>\r\n<p id=\"fs-idp33747664\">Invoking the definition of half-life, symbolized <em>t<\/em><sub>1\/2<\/sub>, requires that the concentration of <em data-effect=\"italics\">A<\/em> at this point is one-half its initial concentration: <em>t<\/em>= <em>t<\/em><sub>1\/2, <\/sub>[<em>A<\/em>]<sub>t<\/sub> = (1\/2)[A]<sub>0<\/sub>.<\/p>\r\n<p id=\"fs-idp123084256\">Substituting these terms into the rearranged integrated rate law and simplifying yields the equation for half-life:<\/p>\r\n\r\n<div id=\"fs-idp22354416\" data-type=\"equation\"><img class=\" wp-image-1738 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4j-300x128.png\" alt=\"\" width=\"279\" height=\"119\" \/><\/div>\r\n<p id=\"fs-idm124479504\">This equation describes an expected inverse relation between the half-life of the reaction and its rate constant, <em data-effect=\"italics\">k<\/em>. Faster reactions exhibit larger rate constants and correspondingly shorter half-lives. Slower reactions exhibit smaller rate constants and longer half-lives.<\/p>\r\n\r\n<div id=\"fs-idm35443680\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm33355984\"><strong>Calculation of a First-order Rate Constant using Half-Life:<\/strong><\/p>\r\nCalculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 \u00b0C, using the data given in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_HPerDcmp\">(Figure)<\/a>.\r\n\r\n&nbsp;\r\n<div id=\"CNX_Chem_12_04_HPerDcmp\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">The decomposition of H<sub>2<\/sub>O<sub>2<\/sub> (2H<sub>2<\/sub>O<sub>2<\/sub> \u27f6 2H<sub>2<\/sub>O + O<sub>2<\/sub>) at 40 \u00b0C is illustrated. The intensity of the color symbolizes the concentration of H<sub>2<\/sub>O<sub>2<\/sub> at the indicated times; H<sub>2<\/sub>O<sub>2<\/sub> is actually colorless.<\/div>\r\n<span id=\"fs-idm83598000\" data-type=\"media\" data-alt=\"A diagram of 5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, \u201c1.000 M, 0 s, and ( 0 h ).\u201d The second beaker contains a slightly lighter green substance and is labeled below as, \u201c0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).\u201d The third beaker contains an even lighter green substance and is labeled below as, \u201c0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).\u201d The fourth beaker contains a green tinted substance and is labeled below as, \u201c0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).\u201d The fifth beaker contains a colorless substance and is labeled below as, \u201c0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_HPerDcmp-2.jpg\" alt=\"A diagram of 5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, \u201c1.000 M, 0 s, and ( 0 h ).\u201d The second beaker contains a slightly lighter green substance and is labeled below as, \u201c0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).\u201d The third beaker contains an even lighter green substance and is labeled below as, \u201c0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).\u201d The fourth beaker contains a green tinted substance and is labeled below as, \u201c0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).\u201d The fifth beaker contains a colorless substance and is labeled below as, \u201c0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).\u201d\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-idp122074992\"><strong>Solution:<\/strong><\/p>\r\nInspecting the concentration\/time data in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_HPerDcmp\">(Figure)<\/a> shows the half-life for the decomposition of H<sub>2<\/sub>O<sub>2<\/sub> is 2.16 \u00d7 10<sup>4<\/sup> s:\r\n<div id=\"fs-idm2063184\" data-type=\"equation\"><img class=\"alignnone wp-image-1740 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4k-300x59.png\" alt=\"\" width=\"310\" height=\"61\" \/><\/div>\r\n&nbsp;\r\n<p id=\"fs-idp33836864\"><strong>Check Your Learning:<\/strong><\/p>\r\nThe first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d<sup>\u22121<\/sup>. What is the half-life for this decay?\r\n<div id=\"fs-idp5361280\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm101854176\">5.02 d.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm130842880\" class=\"bc-section section\" data-depth=\"2\">\r\n<p id=\"fs-idm22269120\">Equations for the half-lives of zero- and second-order reactions can be derived, but will not be covered in this course.\u00a0 Equations for both differential and integrated rate laws and the corresponding half-life for first-order reactions are summarized in <a class=\"autogenerated-content\" href=\"#fs-idm117482272\">(Figure)<\/a>.<\/p>\r\n<img class=\"wp-image-1741 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4l-300x152.png\" alt=\"\" width=\"588\" height=\"298\" \/><strong style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Key Concepts and Summary<\/strong>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm63099680\" class=\"summary\" data-depth=\"1\">\r\n<p id=\"fs-idm85329552\">Integrated rate laws are mathematically derived from differential rate laws, and they describe the time dependence of reactant and product concentrations.<\/p>\r\n<p id=\"fs-idp258559744\">The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. A reaction\u2019s half-life varies with rate constant and, for some reaction orders, reactant concentration. The half-life of a first-order reaction is independent of concentration.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm150818384\" class=\"key-equations\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Equations<\/strong><\/h3>\r\n<ul id=\"fs-idm45705808\" data-bullet-style=\"bullet\">\r\n \t<li>integrated rate law for zero-order reactions: [<em>A<\/em>]<em><sub>t<\/sub><\/em> = \u2212<em>kt<\/em> + [<em>A<\/em>]<sub>0<\/sub>,<\/li>\r\n \t<li>integrated rate law for first-order reactions: ln[<em>A<\/em>]<em><sub>t<\/sub><\/em> = \u2212<em>kt<\/em> + ln[<em>A<\/em>]<sub>0<\/sub>,<\/li>\r\n \t<li>half-life for a first-order reaction: <em>t<\/em><sub>1\/2<\/sub> = 0.693\/<em>k<\/em>,<\/li>\r\n \t<li>integrated rate law for second-order reactions: <img class=\"alignnone wp-image-1731\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png\" alt=\"\" width=\"85\" height=\"35\" \/>,<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-idm92366464\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idm43018880\" data-type=\"exercise\">\r\n<div id=\"fs-idm43018624\" data-type=\"problem\">\r\n<p id=\"fs-idp18442736\"><\/p>\r\n\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-idm46849712\">\r\n \t<dt>half-life of a reaction (<em data-effect=\"italics\">t<\/em><sub>l\/2<\/sub>)<\/dt>\r\n \t<dd id=\"fs-idm46848192\">time required for half of a given amount of reactant to be consumed<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm46847680\">\r\n \t<dt>integrated rate law<\/dt>\r\n \t<dd id=\"fs-idm46847040\">equation that relates the concentration of a reactant to elapsed time of reaction<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<p><strong style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em;background-color: #cbd4b6;color: #000000\">Learning Objectives<\/strong><\/p>\n<div class=\"textbox textbox--learning-objectives\">\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Explain the form and function of an integrated rate law<\/li>\n<li>Perform integrated rate law calculations for zero-, first-, and second-order reactions<\/li>\n<li>Define half-life and carry out related calculations<\/li>\n<li>Identify the order of a reaction from concentration\/time data<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp89497264\">The rate laws discussed thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called <strong>integrated rate laws<\/strong>. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.<\/p>\n<p id=\"fs-idp93158656\">Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-, second-, and zero-order reactions.<\/p>\n<div id=\"fs-idm140678368\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>First-Order Reactions<\/strong><\/h3>\n<p>Integration of the rate law for a simple first-order reaction (rate = <em data-effect=\"italics\">k<\/em>[<em>A<\/em>]) results in an equation describing how the reactant concentration varies with time:<\/p>\n<div id=\"fs-idm380890320\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1727 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a.png\" alt=\"\" width=\"133\" height=\"64\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a.png 281w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a-65x31.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a-225x108.png 225w\" sizes=\"auto, (max-width: 133px) 100vw, 133px\" \/><\/div>\n<p id=\"fs-idm63176528\">where [<em data-effect=\"italics\">A<\/em>]<sub><em data-effect=\"italics\">t<\/em><\/sub> is the concentration of <em data-effect=\"italics\">A<\/em> at any time <em data-effect=\"italics\">t<\/em>, [<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub> is the initial concentration of <em data-effect=\"italics\">A<\/em>, and <em data-effect=\"italics\">k<\/em> is the first-order rate constant.<\/p>\n<p id=\"fs-idm512172928\">For mathematical convenience, this equation may be rearranged to a format showing a linear dependence of concentration in time:<\/p>\n<div id=\"fs-idm358923408\" style=\"text-align: center\" data-type=\"equation\">ln[A]<sub>t<\/sub> = ln[A]<sub>0<\/sub>\u2212kt<\/div>\n<div id=\"fs-idm148528000\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm87528896\"><strong>The Integrated Rate Law for a First-Order Reaction:<\/strong><\/p>\n<p>The rate constant for the first-order decomposition of cyclobutane, C<sub>4<\/sub>H<sub>8<\/sub> at 500 \u00b0C is 9.2 \u00d7 10<sup>\u22123<\/sup> s<sup>\u22121<\/sup>:<\/p>\n<div id=\"fs-idp60442464\" style=\"text-align: center\" data-type=\"equation\">C<sub>4<\/sub>H<sub>8<\/sub> \u27f6 2C<sub>2<\/sub>H<sub>4<\/sub><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm139703904\">How long will it take for 80.0% of a sample of C<sub>4<\/sub>H<sub>8<\/sub> to decompose?<\/p>\n<p id=\"fs-idm133511056\"><strong>Solution:<\/strong><\/p>\n<p>Since the relative change in reactant concentration is provided, a convenient format for the integrated rate law is:<\/p>\n<div id=\"fs-idp11818384\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1727 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a.png\" alt=\"\" width=\"133\" height=\"64\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a.png 281w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a-65x31.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4a-225x108.png 225w\" sizes=\"auto, (max-width: 133px) 100vw, 133px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm149413632\">The initial concentration of C<sub>4<\/sub>H<sub>8<\/sub>, [<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub>, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let <em data-effect=\"italics\">x<\/em> be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of <em data-effect=\"italics\">x<\/em> or 0.200<em data-effect=\"italics\">x.<\/em> Rearranging the rate law to isolate <em data-effect=\"italics\">t<\/em> and substituting the provided quantities yields:<\/p>\n<div id=\"fs-idm92065248\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1728 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4b-300x185.png\" alt=\"\" width=\"238\" height=\"147\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4b-300x185.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4b-65x40.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4b-225x139.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4b-350x216.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4b.png 411w\" sizes=\"auto, (max-width: 238px) 100vw, 238px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp38768624\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Iodine-131 is a radioactive isotope that is used to diagnose and treat some forms of thyroid cancer. Iodine-131 decays to xenon-131 according to the equation:<\/p>\n<div id=\"fs-idm115576112\" style=\"text-align: center\" data-type=\"equation\">I-131 \u27f6 Xe-131 + electron<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm29159424\">The decay is first-order with a rate constant of 0.138 d<sup>\u22121<\/sup>. How many days will it take for 90% of the iodine\u2212131 in a 0.500 <em data-effect=\"italics\">M<\/em> solution of this substance to decay to Xe-131?<\/p>\n<div id=\"fs-idm89005952\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm22449888\">16.7 days<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm132947408\">In the next example exercise, a linear format for the integrated rate law will be convenient:<\/p>\n<div id=\"fs-idm88512368\" style=\"text-align: center\" data-type=\"equation\">ln[<em>A<\/em>]<sub>t<\/sub> = &#8211;<em>kt<\/em> + ln[<em>A<\/em>]<sub>0<\/sub><\/div>\n<div style=\"text-align: center\" data-type=\"equation\">\u00a0<em>y<\/em> = <em>mx<\/em> + <em>b<\/em><\/div>\n<p id=\"fs-idm134557872\">A plot of ln[<em data-effect=\"italics\">A<\/em>]<sub><em data-effect=\"italics\">t<\/em><\/sub> versus <em data-effect=\"italics\">t<\/em> for a first-order reaction is a straight line with a slope of \u2212<em data-effect=\"italics\">k<\/em> and a <em data-effect=\"italics\">y<\/em>-intercept of ln[<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub>. If a set of rate data are plotted in this fashion but do <em data-effect=\"italics\">not<\/em> result in a straight line, the reaction is not first order in <em data-effect=\"italics\">A<\/em>.<\/p>\n<div id=\"fs-idp35909968\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm76041456\"><strong>Graphical Determination of Reaction Order and Rate Constant:<\/strong><\/p>\n<p>The following data is for the decomposition of H<sub>2<\/sub>O<sub>2<\/sub> at 40<sup>o<\/sup>C:<\/p>\n<\/div>\n<p style=\"text-align: center\">2H<sub>2<\/sub>O<sub>2<\/sub> \u27f6 2H<sub>2<\/sub>O + O<sub>2<\/sub><\/p>\n<div id=\"fs-idp35909968\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p>Show that the data can be represented by a first-order rate law by graphing ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time. Determine the rate constant for the decomposition of H<sub>2<\/sub>O<sub>2<\/sub> from these data.<\/p>\n<table id=\"fs-idp114347648\" class=\"medium unnumbered\" summary=\"This table contains four columns and six rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( h ),\u201d \u201c[ H subscript 2 O subscript 2 ] ( M ),\u201d and \u201cl n [ H subscript 2 O subscript 2 ].\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the column, \u201cTime ( h )\u201d are the numbers 0, 6.00, 12.00, 18.00, and 24.00. Under the column \u201c[ H subscript 2 O subscript 2 ] ( M ),\u201d are the numbers 1.000, 0.500, 0.250, 0.125, and 0.0625. Under the column, \u201cl n [ H subscript 2 O subscript 2 ],\u201d are the numbers: 0.0, negative 0.693, negative 1.386, negative 2.079, and negative 2.772.\" data-label=\"\">\n<thead>\n<tr valign=\"middle\">\n<th data-align=\"left\">Trial<\/th>\n<th data-align=\"left\">Time (h)<\/th>\n<th data-align=\"left\">[H<sub>2<\/sub>O<sub>2<\/sub>] (<em data-effect=\"italics\">M<\/em>)<\/th>\n<th data-align=\"left\">ln[H<sub>2<\/sub>O<sub>2<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">1<\/td>\n<td data-align=\"left\">0.00<\/td>\n<td data-align=\"left\">1.000<\/td>\n<td data-align=\"left\">0.000<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">2<\/td>\n<td data-align=\"left\">6.00<\/td>\n<td data-align=\"left\">0.500<\/td>\n<td data-align=\"left\">\u22120.693<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">3<\/td>\n<td data-align=\"left\">12.00<\/td>\n<td data-align=\"left\">0.250<\/td>\n<td data-align=\"left\">\u22121.386<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">4<\/td>\n<td data-align=\"left\">18.00<\/td>\n<td data-align=\"left\">0.125<\/td>\n<td data-align=\"left\">\u22122.079<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">5<\/td>\n<td data-align=\"left\">24.00<\/td>\n<td data-align=\"left\">0.0625<\/td>\n<td data-align=\"left\">\u22122.772<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"CNX_Chem_12_04_FrstOKin\" class=\"scaled-down\">\n<div>\n<p id=\"fs-idm136711920\"><strong>Solution:<\/strong><\/p>\n<p>The plot of ln[H<sub>2<\/sub>O<sub>2<\/sub>] is shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_FrstOKin\">(Figure)<\/a>.<\/p>\n<\/div>\n<div class=\"bc-figcaption figcaption\">A linear relationship between ln[H<sub>2<\/sub>O<sub>2<\/sub>] and time suggests the decomposition of hydrogen peroxide is a first-order reaction.<\/div>\n<p><span id=\"fs-idm90898176\" data-type=\"media\" data-alt=\"A graph is shown with the label \u201cTime ( h )\u201d on the x-axis and \u201cl n [ H subscript 2 O subscript 2 ]\u201d on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_FrstOKin-2.jpg\" alt=\"A graph is shown with the label \u201cTime ( h )\u201d on the x-axis and \u201cl n [ H subscript 2 O subscript 2 ]\u201d on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idm88773168\">The plot of ln[H<sub>2<\/sub>O<sub>2<\/sub>] versus time is linear, indicating that the reaction may be described by a first-order rate law.<\/p>\n<p id=\"fs-idm97546480\">According to the linear format of the first-order integrated rate law, the rate constant is given by the negative of this plot\u2019s slope.<\/p>\n<div id=\"fs-idp49567824\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1729 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4c-300x48.png\" alt=\"\" width=\"300\" height=\"48\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4c-300x48.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4c-65x10.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4c-225x36.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4c-350x56.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4c.png 693w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm145444128\">The slope of this line may be derived from two values of ln[H<sub>2<\/sub>O<sub>2<\/sub>] at different values of <em data-effect=\"italics\">t<\/em> (one near each end of the line is preferable). For example, the value of ln[H<sub>2<\/sub>O<sub>2<\/sub>] when <em data-effect=\"italics\">t<\/em> is 0.00 h is 0.000; the value when <em data-effect=\"italics\">t<\/em> = 24.00 h is \u22122.772.<\/p>\n<div id=\"fs-idm14805520\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1730 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4d-300x100.png\" alt=\"\" width=\"300\" height=\"100\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4d-300x100.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4d-65x22.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4d-225x75.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4d-350x116.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4d.png 749w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp92343136\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Graph the following data to determine whether the reaction <em>A<\/em> \u27f6 <em>B<\/em> + <em>C<\/em> is first order.<\/p>\n<table id=\"fs-idm149704608\" class=\"medium unnumbered\" summary=\"This table has three columns and six rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and, \u201c[ A ].\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201cTime ( s )\u201d column are the numbers: 4.0, 8.0, 12.0, 16.0, and 20.0. Under the \u201c [ A ]\u201d column are the numbers: 0.220, 0.144, 0.110, 0.088, and 0.074.\" data-label=\"\">\n<thead>\n<tr valign=\"middle\">\n<th data-align=\"left\">Trial<\/th>\n<th data-align=\"left\">Time (s)<\/th>\n<th data-align=\"left\">[<em data-effect=\"italics\">A<\/em>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">1<\/td>\n<td data-align=\"left\">4.0<\/td>\n<td data-align=\"left\">0.220<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">2<\/td>\n<td data-align=\"left\">8.0<\/td>\n<td data-align=\"left\">0.144<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">3<\/td>\n<td data-align=\"left\">12.0<\/td>\n<td data-align=\"left\">0.110<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">4<\/td>\n<td data-align=\"left\">16.0<\/td>\n<td data-align=\"left\">0.088<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">5<\/td>\n<td data-align=\"left\">20.0<\/td>\n<td data-align=\"left\">0.074<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-idm206842256\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp73329760\">The plot of ln[<em data-effect=\"italics\">A<\/em>]<sub><em data-effect=\"italics\">t<\/em><\/sub> vs. <em data-effect=\"italics\">t<\/em> is not linear, indicating the reaction is not first order:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p><span id=\"fs-idm135178736\" class=\"scaled-down\" data-type=\"media\" data-alt=\"A graph, labeled above as \u201cl n [ A ] vs. Time\u201d is shown. The x-axis is labeled, \u201cTime ( s )\u201d and the y-axis is labeled, \u201cl n [ A ].\u201d The x-axis shows markings at 5, 10, 15, 20, and 25 hours. The y-axis shows markings at negative 3, negative 2, negative 1, and 0. A slight curve is drawn connecting five points at coordinates of approximately (4, negative 1.5), (8, negative 2), (12, negative 2.2), (16, negative 2.4), and (20, negative 2.6).\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_CYL1_img-2.jpg\" alt=\"A graph, labeled above as \u201cl n [ A ] vs. Time\u201d is shown. The x-axis is labeled, \u201cTime ( s )\u201d and the y-axis is labeled, \u201cl n [ A ].\u201d The x-axis shows markings at 5, 10, 15, 20, and 25 hours. The y-axis shows markings at negative 3, negative 2, negative 1, and 0. A slight curve is drawn connecting five points at coordinates of approximately (4, negative 1.5), (8, negative 2), (12, negative 2.2), (16, negative 2.4), and (20, negative 2.6).\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm162181696\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Second-Order Reactions<\/strong><\/h3>\n<p id=\"fs-idm75401888\">The equations that relate the concentrations of reactants and the rate constant of second-order reactions can be fairly complicated. To illustrate the point with minimal complexity, only the simplest second-order reactions will be described here, namely, those whose rates depend on the concentration of just one reactant. For these types of reactions, the differential rate law is written as:<\/p>\n<div id=\"fs-idm149309552\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[A]<sup>2<\/sup><\/div>\n<p id=\"fs-idm55886848\">For these second-order reactions, the integrated rate law is:<\/p>\n<div id=\"fs-idm72168352\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1731 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png\" alt=\"\" width=\"137\" height=\"56\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png 286w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e-65x26.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e-225x91.png 225w\" sizes=\"auto, (max-width: 137px) 100vw, 137px\" \/><\/div>\n<p id=\"fs-idm19194496\">where the terms in the equation have their usual meanings as defined earlier.<\/p>\n<div id=\"fs-idm85213520\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp6881136\"><strong>The Integrated Rate Law for a Second-Order Reaction:<\/strong><\/p>\n<p>The reaction of butadiene gas (C<sub>4<\/sub>H<sub>6<\/sub>) to yield C<sub>8<\/sub>H<sub>12<\/sub> gas is described by the equation:<\/p>\n<div id=\"fs-idm46947984\" style=\"text-align: center\" data-type=\"equation\">2C<sub>4<\/sub>H<sub>6<\/sub>(<em>g<\/em>) \u27f6 C<sub>8<\/sub>H<sub>12<\/sub>(<em>g<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idp68803280\">This \u201cdimerization\u201d reaction is second order with a rate constant equal to 5.76 \u00d7 10<sup>\u22122<\/sup> L mol<sup>\u22121<\/sup> min<sup>\u22121<\/sup> under certain conditions. If the initial concentration of butadiene is 0.200 <em data-effect=\"italics\">M<\/em>, what is the concentration after 10.0 min?<\/p>\n<p id=\"fs-idm47025104\"><strong>Solution:<\/strong><\/p>\n<p>For a second-order reaction, the integrated rate law is written<\/p>\n<div id=\"fs-idm191622048\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png\" width=\"128\" height=\"52\" alt=\"image\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm25993952\">We know three variables in this equation: [<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub> = 0.200 mol\/L, <em data-effect=\"italics\">k<\/em> = 5.76 \u00d7 10<sup>\u22122<\/sup> L<sup>.<\/sup>mol<sup>-1.<\/sup>min, and <em data-effect=\"italics\">t<\/em> = 10.0 min. Therefore, we can solve for [<em data-effect=\"italics\">A<\/em>], the fourth variable:<\/p>\n<div id=\"fs-idp58774528\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1733 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4f-300x87.png\" alt=\"\" width=\"314\" height=\"91\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4f-300x87.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4f-768x222.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4f-65x19.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4f-225x65.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4f-350x101.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4f.png 915w\" sizes=\"auto, (max-width: 314px) 100vw, 314px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm85057584\">Therefore 0.179 mol\/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol\/L that was originally present.<\/p>\n<p id=\"fs-idp90256\"><strong>Check Your Learning:<\/strong><\/p>\n<p>If the initial concentration of butadiene is 0.0200 <em data-effect=\"italics\">M<\/em>, what is the concentration remaining after 20.0 min?<\/p>\n<div id=\"fs-idp27082688\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm23200256\">0.0195 mol\/L<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm88643152\">The integrated rate law for second-order reactions has the form of the equation of a straight line:<\/p>\n<div id=\"fs-idm140436832\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png\" width=\"128\" height=\"52\" alt=\"image\" \/><\/div>\n<div data-type=\"equation\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<em>y<\/em> = <em>mx<\/em> + <em>b<\/em><\/div>\n<p id=\"fs-idm67290896\">A plot of 1\/[<em>A<\/em>]<sub>t<\/sub> versus <em data-effect=\"italics\">t<\/em> for a second-order reaction is a straight line with a slope of <em data-effect=\"italics\">k<\/em> and a <em data-effect=\"italics\">y<\/em>-intercept of 1\/[<em>A<\/em>]<sub>0<\/sub>. If the plot is not a straight line, then the reaction is not second order.<\/p>\n<div id=\"fs-idp69093472\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm27496000\"><strong>Graphical Determination of Reaction Order and Rate Constant:<\/strong><\/p>\n<p>\u00a0The data below are for the same reaction described in <a class=\"autogenerated-content\" href=\"#fs-idm85213520\">(Figure)<\/a>. Prepare and compare two appropriate data plots to identify the reaction as being either first or second order. After identifying the reaction order, estimate a value for the rate constant.<\/p>\n<p id=\"fs-idp4978176\"><strong>Solution:<\/strong><\/p>\n<table id=\"fs-idm140502592\" class=\"medium unnumbered\" summary=\"This table contains three columns and six rows. The first row is a header row and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and \u201c[ C subscript 4 H subscript 6 ] ( M ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201cTime ( s )\u201d column are the numbers: 0, 1600, 3200, 4800, and 6200. Under the column \u201c[ C subscript 4 H subscript 6 ] ( M )\u201d are the numbers: 1.00 times ten to the negative 2; 5.04 times ten to the negative 3; 3.37 times ten to the negative 3; 2.53 times ten to the negative 3; and 2.08 times ten to the negative 3.\" data-label=\"\">\n<thead>\n<tr valign=\"middle\">\n<th data-align=\"left\">Trial<\/th>\n<th data-align=\"left\">Time (s)<\/th>\n<th data-align=\"left\">[C<sub>4<\/sub>H<sub>6<\/sub>] (<em data-effect=\"italics\">M<\/em>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">1<\/td>\n<td data-align=\"left\">0<\/td>\n<td data-align=\"left\">1.00 \u00d7 10<sup>\u22122<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">2<\/td>\n<td data-align=\"left\">1600<\/td>\n<td data-align=\"left\">5.04 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">3<\/td>\n<td data-align=\"left\">3200<\/td>\n<td data-align=\"left\">3.37 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">4<\/td>\n<td data-align=\"left\">4800<\/td>\n<td data-align=\"left\">2.53 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">5<\/td>\n<td data-align=\"left\">6200<\/td>\n<td data-align=\"left\">2.08 \u00d7 10<sup>\u22123<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idm190212352\">In order to distinguish a first-order reaction from a second-order reaction, prepare a plot of ln[C<sub>4<\/sub>H<sub>6<\/sub>]<sub><em data-effect=\"italics\">t<\/em><\/sub> versus <em data-effect=\"italics\">t<\/em> and compare it to a plot of 1\/[C<sub>4<\/sub>H<sub>6<\/sub>]<sub>t<\/sub> versus <em data-effect=\"italics\">t<\/em>. The values needed for these plots follow.<\/p>\n<table id=\"fs-idp62232800\" class=\"medium unnumbered\" summary=\"This table contains three columns and six rows. The first row is a header row and it labels each column, \u201cTime ( s ),\u201d \u201c1 over [ C subscript 4 H subscript 6 ] ( M superscript negative 1 ),\u201d and \u201cl n [ C subscript 4 H subscript 6 ].\u201d Under the column \u201cTime ( s )\u201d are the numbers: 0, 1600, 3200, 4800, and 6200. Under the \u201c1 over [ C subscript 4 H subscript 6 ] ( M superscript negative 1 )\u201d column are the numbers: 100, 198, 296, 395, and 481. Under the \u201cl n [ C subscript 4 H subscript 6 ]\u201d column are the numbers: negative 4.605, negative 5.289, negative 5.692, negative 5.978, and negative 6.175.\" data-label=\"\">\n<thead>\n<tr valign=\"middle\">\n<th data-align=\"left\">Time (s)<\/th>\n<th data-align=\"left\">1\/[C<sub>4<\/sub>H<sub>6<\/sub>] (M<sup>-1<\/sup>)<\/th>\n<th data-align=\"left\">ln[C<sub>4<\/sub>H<sub>6<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">0<\/td>\n<td data-align=\"left\">100<\/td>\n<td data-align=\"left\">\u22124.605<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">1600<\/td>\n<td data-align=\"left\">198<\/td>\n<td data-align=\"left\">\u22125.289<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">3200<\/td>\n<td data-align=\"left\">296<\/td>\n<td data-align=\"left\">\u22125.692<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">4800<\/td>\n<td data-align=\"left\">395<\/td>\n<td data-align=\"left\">\u22125.978<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">6200<\/td>\n<td data-align=\"left\">481<\/td>\n<td data-align=\"left\">\u22126.175<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp69115360\">The plots are shown in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_2OrdKin\">(Figure)<\/a>, which clearly shows the plot of ln[C<sub>4<\/sub>H<sub>6<\/sub>]<sub><em data-effect=\"italics\">t<\/em><\/sub> versus <em data-effect=\"italics\">t<\/em> is not linear, therefore the reaction is not first order. The plot of 1\/[C<sub>4<\/sub>H<sub>6<\/sub>]<em><sub>t<\/sub><\/em> versus <em data-effect=\"italics\">t<\/em> is linear, indicating that the reaction is second order.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_12_04_2OrdKin\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">These two graphs show first- and second-order plots for the dimerization of C<sub>4<\/sub>H<sub>6<\/sub>. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.<\/div>\n<p><span id=\"fs-idp16234496\" data-type=\"media\" data-alt=\"Two graphs are shown, each with the label \u201cTime ( s )\u201d on the x-axis. The graph on the left is labeled, \u201cl n [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The graph on the right is labeled \u201c1 divided by [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The x-axes for both graphs show markings at 3000 and 6000. The y-axis for the graph on the left shows markings at negative 6, negative 5, and negative 4. A decreasing slightly concave up curve is drawn through five points at coordinates that are (0, negative 4.605), (1600, negative 5.289), (3200, negative 5.692), (4800, negative 5.978), and (6200, negative 6.175). The y-axis for the graph on the right shows markings at 100, 300, and 500. An approximately linear increasing curve is drawn through five points at coordinates that are (0, 100), (1600, 198), (3200, 296), and (4800, 395), and (6200, 481).\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_2OrdKin-2.jpg\" alt=\"Two graphs are shown, each with the label \u201cTime ( s )\u201d on the x-axis. The graph on the left is labeled, \u201cl n [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The graph on the right is labeled \u201c1 divided by [ C subscript 4 H subscript 6 ],\u201d on the y-axis. The x-axes for both graphs show markings at 3000 and 6000. The y-axis for the graph on the left shows markings at negative 6, negative 5, and negative 4. A decreasing slightly concave up curve is drawn through five points at coordinates that are (0, negative 4.605), (1600, negative 5.289), (3200, negative 5.692), (4800, negative 5.978), and (6200, negative 6.175). The y-axis for the graph on the right shows markings at 100, 300, and 500. An approximately linear increasing curve is drawn through five points at coordinates that are (0, 100), (1600, 198), (3200, 296), and (4800, 395), and (6200, 481).\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idm339836784\">According to the second-order integrated rate law, the rate constant is equal to the slope of the 1\/[<em>A<\/em>]<em><sub>t<\/sub><\/em> versus <em data-effect=\"italics\">t<\/em> plot. Using the data for <em data-effect=\"italics\">t<\/em> = 0 <em data-effect=\"italics\">s<\/em> and <em data-effect=\"italics\">t<\/em> = 6200 <em data-effect=\"italics\">s<\/em>, the rate constant is estimated as follows:<\/p>\n<div id=\"fs-idm363866800\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-1735 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4g-300x43.png\" alt=\"\" width=\"300\" height=\"43\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4g-300x43.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4g-768x109.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4g-65x9.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4g-225x32.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4g-350x50.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4g.png 809w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm56284464\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Do the following data fit a second-order rate law?<\/p>\n<table id=\"fs-idm88760288\" class=\"medium unnumbered\" summary=\"This table contains three columns and seven rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201cTime ( s ),\u201d and, \u201c[ A ] ( M ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, 5, and 6. Under the \u201cTime ( s )\u201d column are the numbers: 5, 10, 15, 20, 25, and 35. Under the \u201c[ A ] ( M )\u201d column are the numbers 0.952, 0.625, 0.465, 0.370, 0.308, and 0.230.\" data-label=\"\">\n<thead>\n<tr valign=\"middle\">\n<th data-align=\"left\">Trial<\/th>\n<th data-align=\"left\">Time (s)<\/th>\n<th data-align=\"left\">[<em data-effect=\"italics\">A<\/em>] (<em data-effect=\"italics\">M<\/em>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">1<\/td>\n<td data-align=\"left\">5<\/td>\n<td data-align=\"left\">0.952<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">2<\/td>\n<td data-align=\"left\">10<\/td>\n<td data-align=\"left\">0.625<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">3<\/td>\n<td data-align=\"left\">15<\/td>\n<td data-align=\"left\">0.465<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">4<\/td>\n<td data-align=\"left\">20<\/td>\n<td data-align=\"left\">0.370<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">5<\/td>\n<td data-align=\"left\">25<\/td>\n<td data-align=\"left\">0.308<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">6<\/td>\n<td data-align=\"left\">35<\/td>\n<td data-align=\"left\">0.230<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-idm72354384\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idp122966448\">Yes. The plot of 1\/[<em>A<\/em>]<em><sub>t <\/sub><\/em>vs. <em data-effect=\"italics\">t<\/em> is linear:<\/p>\n<p><span id=\"fs-idm144508704\" class=\"scaled-down\" data-type=\"media\" data-alt=\"A graph, with the title \u201c1 divided by [ A ] vs. Time\u201d is shown, with the label, \u201cTime ( s ),\u201d on the x-axis. The label \u201c1 divided by [ A ]\u201d appears left of the y-axis. The x-axis shows markings beginning at zero and continuing at intervals of 10 up to and including 40. The y-axis on the left shows markings beginning at 0 and increasing by intervals of 1 up to and including 5. A line with an increasing trend is drawn through six points at approximately (4, 1), (10, 1.5), (15, 2.2), (20, 2.8), (26, 3.4), and (36, 4.4).\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_CYL2_img-2.jpg\" alt=\"A graph, with the title \u201c1 divided by [ A ] vs. Time\u201d is shown, with the label, \u201cTime ( s ),\u201d on the x-axis. The label \u201c1 divided by [ A ]\u201d appears left of the y-axis. The x-axis shows markings beginning at zero and continuing at intervals of 10 up to and including 40. The y-axis on the left shows markings beginning at 0 and increasing by intervals of 1 up to and including 5. A line with an increasing trend is drawn through six points at approximately (4, 1), (10, 1.5), (15, 2.2), (20, 2.8), (26, 3.4), and (36, 4.4).\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm152558640\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Zero-Order Reactions<\/strong><\/h3>\n<p id=\"fs-idp474768\">For zero-order reactions, the differential rate law is:<\/p>\n<div id=\"fs-idp24382368\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em><\/div>\n<p id=\"fs-idm91448320\">A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactant(s). This may seem counterintuitive, since the reaction rate certainly can\u2019t be finite when the reactant concentration is zero. For purposes of this introductory text, it will suffice to note that zero-order kinetics are observed for some reactions only under certain specific conditions. These same reactions exhibit different kinetic behaviors when the specific conditions aren\u2019t met, and for this reason the more prudent term <em data-effect=\"italics\">pseudo-zero-order<\/em> is sometimes used.<\/p>\n<p id=\"fs-idp73236848\">The integrated rate law for a zero-order reaction is a linear function:<\/p>\n<div id=\"fs-idm130548304\" style=\"text-align: center\" data-type=\"equation\">[<em>A<\/em>]<sub><em>t<\/em><\/sub> = \u2212<em>kt<\/em> + [<em>A<\/em>]<sub>0<\/sub><\/div>\n<div style=\"text-align: center\" data-type=\"equation\"><em>y<\/em> = <em>mx<\/em> + <em>b<\/em><\/div>\n<p id=\"fs-idm99514160\">A plot of [<em data-effect=\"italics\">A<\/em>] versus <em data-effect=\"italics\">t<\/em> for a zero-order reaction is a straight line with a slope of <em data-effect=\"italics\">\u2212k<\/em> and a <em data-effect=\"italics\">y<\/em>-intercept of [<em data-effect=\"italics\">A<\/em>]<sub>0<\/sub>. <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_AmDecomK\">(Figure)<\/a> shows a plot of [NH<sub>3<\/sub>] versus <em data-effect=\"italics\">t<\/em> for the thermal decomposition of ammonia at the surface of two different heated solids. The decomposition reaction exhibits first-order behavior at a quartz (SiO<sub>2<\/sub>) surface, as suggested by the exponentially decaying plot of concentration versus time. On a tungsten surface, however, the plot is linear, indicating zero-order kinetics.<\/p>\n<div id=\"fs-idm339943456\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm358086992\"><strong>Graphical Determination of Zero-Order Rate Constant:<\/strong><\/p>\n<p>Use the data plot in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_AmDecomK\">(Figure)<\/a> to graphically estimate the zero-order rate constant for ammonia decomposition at a tungsten surface.<\/p>\n<p id=\"fs-idm382849456\"><strong>Solution:<\/strong><\/p>\n<p>The integrated rate law for zero-order kinetics describes a linear plot of reactant concentration, [<em data-effect=\"italics\">A<\/em>]<sub><em data-effect=\"italics\">t<\/em><\/sub>, versus time, <em data-effect=\"italics\">t<\/em>, with a slope equal to the negative of the rate constant, \u2212<em data-effect=\"italics\">k<\/em>. Following the mathematical approach of previous examples, the slope of the linear data plot (for decomposition on W) is estimated from the graph. Using the ammonia concentrations at <em data-effect=\"italics\">t<\/em> = 0 and <em data-effect=\"italics\">t<\/em> = 1000 s:<\/p>\n<div id=\"fs-idm375658448\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1736 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4h-300x25.png\" alt=\"\" width=\"528\" height=\"44\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4h-300x25.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4h-1024x84.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4h-768x63.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4h-65x5.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4h-225x18.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4h-350x29.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4h.png 1172w\" sizes=\"auto, (max-width: 528px) 100vw, 528px\" \/><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm376604432\"><strong>Check Your Learning:<\/strong><\/p>\n<p>The zero-order plot in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_AmDecomK\">(Figure)<\/a> shows an initial ammonia concentration of 0.0028 mol L<sup>\u22121<\/sup> decreasing linearly with time for 1000 s. Assuming no change in this zero-order behavior, at what time (min) will the concentration reach 0.0001 mol L<sup>\u22121<\/sup>?<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm379795136\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm508285088\">35 min<\/p>\n<\/div>\n<\/div>\n<div id=\"CNX_Chem_12_04_AmDecomK\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">The decomposition of NH<sub>3<\/sub> on a tungsten (W) surface is a zero-order reaction, whereas on a quartz (SiO<sub>2<\/sub>) surface, the reaction is first order.<\/div>\n<p><span id=\"fs-idp157593168\" data-type=\"media\" data-alt=\"A graph is shown with the label, \u201cTime ( s ),\u201d on the x-axis and, \u201c[ N H subscript 3 ] M,\u201d on the y-axis. The x-axis shows a single value of 1000 marked near the right end of the axis. The vertical axis shows markings at 1.0 times 10 superscript negative 3, 2.0 times 10 superscript negative 3, and 3.0 times 10 superscript negative 3. A decreasing linear trend line is drawn through six points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (200, 2.6 times 10 superscript negative 3), (400, 2.3 times 10 superscript negative 3), (600, 2.0 times 10 superscript negative 3), (800, 1.8 times 10 superscript negative 3), and (1000, 1.6 times 10 superscript negative 3). This line is labeled \u201cDecomposition on W.\u201d A decreasing slightly concave up curve is similarly drawn through eight points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (100, 2.5 times 10 superscript negative 3), (200, 2.1 times 10 superscript negative 3), (300, 1.9 times 10 superscript negative 3), (400, 1.6 times 10 superscript negative 3), (500, 1.4 times 10 superscript negative 3), and (750, 1.1 times 10 superscript negative 3), ending at about (1000, 0.7 times 10 superscript negative 3). This curve is labeled \u201cDecomposition on S i O subscript 2.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_AmDecomK-2.jpg\" alt=\"A graph is shown with the label, \u201cTime ( s ),\u201d on the x-axis and, \u201c[ N H subscript 3 ] M,\u201d on the y-axis. The x-axis shows a single value of 1000 marked near the right end of the axis. The vertical axis shows markings at 1.0 times 10 superscript negative 3, 2.0 times 10 superscript negative 3, and 3.0 times 10 superscript negative 3. A decreasing linear trend line is drawn through six points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (200, 2.6 times 10 superscript negative 3), (400, 2.3 times 10 superscript negative 3), (600, 2.0 times 10 superscript negative 3), (800, 1.8 times 10 superscript negative 3), and (1000, 1.6 times 10 superscript negative 3). This line is labeled \u201cDecomposition on W.\u201d A decreasing slightly concave up curve is similarly drawn through eight points at the approximate coordinates: (0, 2.8 times 10 superscript negative 3), (100, 2.5 times 10 superscript negative 3), (200, 2.1 times 10 superscript negative 3), (300, 1.9 times 10 superscript negative 3), (400, 1.6 times 10 superscript negative 3), (500, 1.4 times 10 superscript negative 3), and (750, 1.1 times 10 superscript negative 3), ending at about (1000, 0.7 times 10 superscript negative 3). This curve is labeled \u201cDecomposition on S i O subscript 2.\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp6500464\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>The Half-Life of a Reaction<\/strong><\/h3>\n<p id=\"fs-idm99562880\">The <span data-type=\"term\">half-life of a reaction (<em data-effect=\"italics\">t<\/em><sub>1\/2<\/sub>)<\/span> is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (<a class=\"autogenerated-content\" href=\"\/contents\/19ee5333-b3d6-42c0-9ee6-ca6fcf3b1c62#CNX_Chem_12_01_KDataH2O2\">(Figure)<\/a>) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases from 1.000 <em data-effect=\"italics\">M<\/em> to 0.500 <em data-effect=\"italics\">M<\/em>. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 <em data-effect=\"italics\">M<\/em> to 0.250 <em data-effect=\"italics\">M<\/em>; during the third half-life, it decreases from 0.250 <em data-effect=\"italics\">M<\/em> to 0.125 <em data-effect=\"italics\">M<\/em>. The concentration of H<sub>2<\/sub>O<sub>2<\/sub> decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.<\/p>\n<div id=\"fs-idm97307168\" class=\"bc-section section\" data-depth=\"2\">\n<h4 data-type=\"title\"><strong>First-Order Reactions<\/strong><\/h4>\n<p id=\"fs-idp162624\">An equation relating the half-life of a first-order reaction to its rate constant may be derived from the integrated rate law as follows:<\/p>\n<div id=\"fs-idm58156032\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1737 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4i-300x143.png\" alt=\"\" width=\"206\" height=\"98\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4i-300x143.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4i-65x31.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4i-225x107.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4i-350x166.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4i.png 385w\" sizes=\"auto, (max-width: 206px) 100vw, 206px\" \/><\/div>\n<p id=\"fs-idp33747664\">Invoking the definition of half-life, symbolized <em>t<\/em><sub>1\/2<\/sub>, requires that the concentration of <em data-effect=\"italics\">A<\/em> at this point is one-half its initial concentration: <em>t<\/em>= <em>t<\/em><sub>1\/2, <\/sub>[<em>A<\/em>]<sub>t<\/sub> = (1\/2)[A]<sub>0<\/sub>.<\/p>\n<p id=\"fs-idp123084256\">Substituting these terms into the rearranged integrated rate law and simplifying yields the equation for half-life:<\/p>\n<div id=\"fs-idp22354416\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1738 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4j-300x128.png\" alt=\"\" width=\"279\" height=\"119\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4j-300x128.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4j-65x28.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4j-225x96.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4j-350x149.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4j.png 529w\" sizes=\"auto, (max-width: 279px) 100vw, 279px\" \/><\/div>\n<p id=\"fs-idm124479504\">This equation describes an expected inverse relation between the half-life of the reaction and its rate constant, <em data-effect=\"italics\">k<\/em>. Faster reactions exhibit larger rate constants and correspondingly shorter half-lives. Slower reactions exhibit smaller rate constants and longer half-lives.<\/p>\n<div id=\"fs-idm35443680\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm33355984\"><strong>Calculation of a First-order Rate Constant using Half-Life:<\/strong><\/p>\n<p>Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 \u00b0C, using the data given in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_HPerDcmp\">(Figure)<\/a>.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_12_04_HPerDcmp\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">The decomposition of H<sub>2<\/sub>O<sub>2<\/sub> (2H<sub>2<\/sub>O<sub>2<\/sub> \u27f6 2H<sub>2<\/sub>O + O<sub>2<\/sub>) at 40 \u00b0C is illustrated. The intensity of the color symbolizes the concentration of H<sub>2<\/sub>O<sub>2<\/sub> at the indicated times; H<sub>2<\/sub>O<sub>2<\/sub> is actually colorless.<\/div>\n<p><span id=\"fs-idm83598000\" data-type=\"media\" data-alt=\"A diagram of 5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, \u201c1.000 M, 0 s, and ( 0 h ).\u201d The second beaker contains a slightly lighter green substance and is labeled below as, \u201c0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).\u201d The third beaker contains an even lighter green substance and is labeled below as, \u201c0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).\u201d The fourth beaker contains a green tinted substance and is labeled below as, \u201c0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).\u201d The fifth beaker contains a colorless substance and is labeled below as, \u201c0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_04_HPerDcmp-2.jpg\" alt=\"A diagram of 5 beakers is shown, each approximately half-filled with colored substances. Beneath each beaker are three rows of text. The first beaker contains a bright green substance and is labeled below as, \u201c1.000 M, 0 s, and ( 0 h ).\u201d The second beaker contains a slightly lighter green substance and is labeled below as, \u201c0.500 M, 2.16 times 10 superscript 4 s, and ( 6 h ).\u201d The third beaker contains an even lighter green substance and is labeled below as, \u201c0.250 M, 4.32 times 10 superscript 4 s, and ( 12 h ).\u201d The fourth beaker contains a green tinted substance and is labeled below as, \u201c0.125 M, 6.48 times 10 superscript 4 s, and ( 18 h ).\u201d The fifth beaker contains a colorless substance and is labeled below as, \u201c0.0625 M, 8.64 times 10 superscript 4 s, and ( 24 h ).\u201d\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<p id=\"fs-idp122074992\"><strong>Solution:<\/strong><\/p>\n<p>Inspecting the concentration\/time data in <a class=\"autogenerated-content\" href=\"#CNX_Chem_12_04_HPerDcmp\">(Figure)<\/a> shows the half-life for the decomposition of H<sub>2<\/sub>O<sub>2<\/sub> is 2.16 \u00d7 10<sup>4<\/sup> s:<\/p>\n<div id=\"fs-idm2063184\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1740 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4k-300x59.png\" alt=\"\" width=\"310\" height=\"61\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4k-300x59.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4k-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4k-225x45.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4k-350x69.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4k.png 707w\" sizes=\"auto, (max-width: 310px) 100vw, 310px\" \/><\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp33836864\"><strong>Check Your Learning:<\/strong><\/p>\n<p>The first-order radioactive decay of iodine-131 exhibits a rate constant of 0.138 d<sup>\u22121<\/sup>. What is the half-life for this decay?<\/p>\n<div id=\"fs-idp5361280\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm101854176\">5.02 d.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm130842880\" class=\"bc-section section\" data-depth=\"2\">\n<p id=\"fs-idm22269120\">Equations for the half-lives of zero- and second-order reactions can be derived, but will not be covered in this course.\u00a0 Equations for both differential and integrated rate laws and the corresponding half-life for first-order reactions are summarized in <a class=\"autogenerated-content\" href=\"#fs-idm117482272\">(Figure)<\/a>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1741 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4l-300x152.png\" alt=\"\" width=\"588\" height=\"298\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4l-300x152.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4l-1024x518.png 1024w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4l-768x389.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4l-65x33.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4l-225x114.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4l-350x177.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4l.png 1529w\" sizes=\"auto, (max-width: 588px) 100vw, 588px\" \/><strong style=\"font-family: 'Cormorant Garamond', serif;font-size: 1.602em\">Key Concepts and Summary<\/strong><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm63099680\" class=\"summary\" data-depth=\"1\">\n<p id=\"fs-idm85329552\">Integrated rate laws are mathematically derived from differential rate laws, and they describe the time dependence of reactant and product concentrations.<\/p>\n<p id=\"fs-idp258559744\">The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. A reaction\u2019s half-life varies with rate constant and, for some reaction orders, reactant concentration. The half-life of a first-order reaction is independent of concentration.<\/p>\n<\/div>\n<div id=\"fs-idm150818384\" class=\"key-equations\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Equations<\/strong><\/h3>\n<ul id=\"fs-idm45705808\" data-bullet-style=\"bullet\">\n<li>integrated rate law for zero-order reactions: [<em>A<\/em>]<em><sub>t<\/sub><\/em> = \u2212<em>kt<\/em> + [<em>A<\/em>]<sub>0<\/sub>,<\/li>\n<li>integrated rate law for first-order reactions: ln[<em>A<\/em>]<em><sub>t<\/sub><\/em> = \u2212<em>kt<\/em> + ln[<em>A<\/em>]<sub>0<\/sub>,<\/li>\n<li>half-life for a first-order reaction: <em>t<\/em><sub>1\/2<\/sub> = 0.693\/<em>k<\/em>,<\/li>\n<li>integrated rate law for second-order reactions: <img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1731\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png\" alt=\"\" width=\"85\" height=\"35\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e.png 286w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e-65x26.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.4e-225x91.png 225w\" sizes=\"auto, (max-width: 85px) 100vw, 85px\" \/>,<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-idm92366464\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idm43018880\" data-type=\"exercise\">\n<div id=\"fs-idm43018624\" data-type=\"problem\">\n<p id=\"fs-idp18442736\">\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-idm46849712\">\n<dt>half-life of a reaction (<em data-effect=\"italics\">t<\/em><sub>l\/2<\/sub>)<\/dt>\n<dd id=\"fs-idm46848192\">time required for half of a given amount of reactant to be consumed<\/dd>\n<\/dl>\n<dl id=\"fs-idm46847680\">\n<dt>integrated rate law<\/dt>\n<dd id=\"fs-idm46847040\">equation that relates the concentration of a reactant to elapsed time of reaction<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-718","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":695,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/718","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":11,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/718\/revisions"}],"predecessor-version":[{"id":2205,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/718\/revisions\/2205"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/parts\/695"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/718\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=718"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=718"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=718"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=718"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}