{"id":707,"date":"2021-07-23T09:20:30","date_gmt":"2021-07-23T13:20:30","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/rate-laws\/"},"modified":"2022-06-23T09:16:39","modified_gmt":"2022-06-23T13:16:39","slug":"rate-laws","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/aperrott\/chapter\/rate-laws\/","title":{"raw":"12.3 Rate Laws","rendered":"12.3 Rate Laws"},"content":{"raw":"<div class=\"textbox textbox--learning-objectives\">\r\n<h3><strong>Learning Objectives<\/strong><\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Explain the form and function of a rate law<\/li>\r\n \t<li>Use rate laws to calculate reaction rates<\/li>\r\n \t<li>Use rate and concentration data to identify reaction orders and derive rate laws<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm238441168\">As described in the previous module, the rate of a reaction is often affected by the concentrations of reactants. <strong>Rate laws <\/strong>(sometimes called <em data-effect=\"italics\">differential rate laws<\/em>) are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation<\/p>\r\n\r\n<div id=\"fs-idm248316432\" style=\"text-align: center\" data-type=\"equation\"><em>aA<\/em> + <em>bB<\/em> \u27f6 <em>products<\/em><\/div>\r\n<p id=\"fs-idm656673904\">where <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">b<\/em> are stoichiometric coefficients. The rate law for this reaction is written as:<\/p>\r\n\r\n<div id=\"fs-idm276589968\" style=\"text-align: center\" data-type=\"equation\">rate =<em>k<\/em>[A]<em><sup>m<\/sup><\/em>[B]<em><sup>n<\/sup><\/em><\/div>\r\n<p id=\"fs-idp18257184\">in which [<em data-effect=\"italics\">A<\/em>] and [<em data-effect=\"italics\">B<\/em>] represent the molar concentrations of reactants, and <em data-effect=\"italics\">k<\/em> is the <strong>rate constant<\/strong>, which is specific for a particular reaction at a particular temperature. The exponents <em data-effect=\"italics\">m<\/em> and <em data-effect=\"italics\">n<\/em> are the <strong>reaction orders<\/strong> and are typically positive integers, though they can be fractions, negative, or zero. The rate constant <em data-effect=\"italics\">k<\/em> and the reaction orders <em data-effect=\"italics\">m<\/em> and <em data-effect=\"italics\">n<\/em> must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant <em data-effect=\"italics\">k<\/em> is independent of the reactant concentrations, but it does vary with temperature.<\/p>\r\n<p id=\"fs-idm115903376\">The reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is <em data-effect=\"italics\">m<\/em> order with respect to <em data-effect=\"italics\">A<\/em> and <em data-effect=\"italics\">n<\/em> order with respect to <em data-effect=\"italics\">B<\/em>. For example, if <em data-effect=\"italics\">m<\/em> = 1 and <em data-effect=\"italics\">n<\/em> = 2, the reaction is first order in <em data-effect=\"italics\">A<\/em> and second order in <em data-effect=\"italics\">B<\/em>. The <strong>overall reaction order<\/strong> is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall (1 + 2 = 3). A few specific examples are shown below to further illustrate this concept.<\/p>\r\n<p id=\"fs-idm217470464\">The rate law:<\/p>\r\n\r\n<div id=\"fs-idm204886480\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[H<sub>2<\/sub>O<sub>2<\/sub>]<\/div>\r\n<p id=\"fs-idm388320\">describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:<\/p>\r\n\r\n<div id=\"fs-idm185553664\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[C<sub>4<\/sub>H<sub>6<\/sub>]<sup>2<\/sup><\/div>\r\n<p id=\"fs-idm105991808\">describes a reaction that is second order in C<sub>4<\/sub>H<sub>6<\/sub> and second order overall. The rate law:<\/p>\r\n\r\n<div id=\"fs-idm41933536\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[H<sup>+<\/sup>][OH<sup>-<\/sup>]<\/div>\r\n<p id=\"fs-idm221631328\">describes a reaction that is first order in H<sup>+<\/sup>, first order in OH<sup>\u2212<\/sup>, and second order overall.<\/p>\r\n\r\n<div id=\"fs-idm235878704\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm114205984\"><strong>Writing Rate Laws from Reaction Orders:<\/strong><\/p>\r\nAn experiment shows that the reaction of nitrogen dioxide with carbon monoxide:\r\n<div id=\"fs-idm189808080\" style=\"text-align: center\" data-type=\"equation\">NO<sub>2<\/sub>(<em>g<\/em>) + CO(<em>g<\/em>) \u27f6 NO(<em>g<\/em>) + CO<sub>2<\/sub>(<em>g<\/em>)<\/div>\r\n<p id=\"fs-idm213018368\">is second order in NO<sub>2<\/sub> and zero order in CO at 100 \u00b0C. What is the rate law for the reaction?<\/p>\r\n<p id=\"fs-idm222185424\"><strong>Solution:<\/strong><\/p>\r\nThe reaction will have the form:\r\n<div id=\"fs-idm212887136\" style=\"text-align: center\" data-type=\"equation\">rate =<em>k<\/em>[NO<sub>2<\/sub>]<em><sup>m<\/sup><\/em>[CO]<em><sup>n<\/sup><\/em><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm221521296\">The reaction is second order in NO<sub>2<\/sub>; thus <em data-effect=\"italics\">m<\/em> = 2. The reaction is zero order in CO; thus <em data-effect=\"italics\">n<\/em> = 0. The rate law is:<\/p>\r\n\r\n<div id=\"fs-idm49710800\" style=\"text-align: center\" data-type=\"equation\">rate =<em>k<\/em>[NO<sub>2<\/sub>]<sup>2<\/sup>[CO]<sup>0<\/sup> = <em>k<\/em>[NO<sub>2<\/sub>]<sup>2<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm250199824\">Remember that a number raised to the zero power is equal to 1, thus [CO]<sup>0<\/sup> = 1, which is why the CO concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of NO<sub>2<\/sub>. A later chapter section on reaction mechanisms will explain how a reactant\u2019s concentration can have no effect on a reaction rate despite being involved in the reaction.<\/p>\r\n<p id=\"fs-idm273206544\"><strong>Check Your Learning:<\/strong><\/p>\r\nThe rate law for the reaction:\r\n<div id=\"fs-idm122197664\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>(<em>g<\/em>) + 2NO(<em>g<\/em>) \u27f6N<sub>2<\/sub>O(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm122425904\">has been determined to be rate = <em data-effect=\"italics\">k<\/em>[NO]<sup>2<\/sup>[H<sub>2<\/sub>]. What are the orders with respect to each reactant, and what is the overall order of the reaction?<\/p>\r\n\r\n<div id=\"fs-idm168583904\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm141295376\">order in NO = 2; order in H<sub>2<\/sub> = 1; overall order = 3<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-idm117052240\"><strong>Check Your Learning:<\/strong><\/p>\r\nIn a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH<sub>3<\/sub>OH) and ethyl acetate (CH<sub>3<\/sub>CH<sub>2<\/sub>OCOCH<sub>3<\/sub>) as a sample reaction before studying the chemical reactions that produce biodiesel:\r\n<div id=\"fs-idm155639824\" style=\"text-align: center\" data-type=\"equation\">CH<sub>3<\/sub>OH + CH<sub>3<\/sub>CH<sub>2<\/sub>OCOCH<sub>3<\/sub> \u27f6 CH<sub>3<\/sub>OCOCH<sub>3<\/sub> + CH<sub>3<\/sub>CH<sub>2<\/sub>OH<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm213233440\">The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be:<\/p>\r\n\r\n<div id=\"fs-idm236204576\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[CH<sub>3<\/sub>OH]<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm219218784\">What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?<\/p>\r\n\r\n<div id=\"fs-idm253536096\" data-type=\"note\">\r\n<div data-type=\"title\">Answer:<\/div>\r\n<p id=\"fs-idm215326320\">order in CH<sub>3<\/sub>OH = 1; order in CH<sub>3<\/sub>CH<sub>2<\/sub>OCOCH<sub>3<\/sub> = 0; overall order = 1<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm147402384\">A common experimental approach to the determination of rate laws is the <strong>method of initial rates<\/strong>. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.<\/p>\r\n\r\n<div id=\"fs-idm234815200\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm162746544\"><strong>Determining a Rate Law from Initial Rates:<\/strong><\/p>\r\nOzone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (<a class=\"autogenerated-content\" href=\"#CNX_Chem_12_03_OzoneHole\">(Figure)<\/a>). One such reaction is the combination of nitric oxide, NO, with ozone, O<sub>3<\/sub>:\r\n\r\n&nbsp;\r\n<div id=\"CNX_Chem_12_03_OzoneHole\" class=\"scaled-down\">\r\n<div class=\"bc-figcaption figcaption\">A contour map showing stratospheric ozone concentration and the \u201cozone hole\u201d that occurs over Antarctica during its spring months. (credit: modification of work by NASA)<\/div>\r\n<span id=\"fs-idm204213280\" data-type=\"media\" data-alt=\"A view of Earth\u2019s southern hemisphere is shown. A nearly circular region of approximately half the diameter of the image is shown in shades of purple, with Antarctica appearing in a slightly lighter color than the surrounding ocean areas. Immediately outside this region is a narrow bright blue zone followed by a bright green zone. In the top half of the figure, the purple region extends slightly outward from the circle and the blue zone extends more outward to the right of the center as compared to the lower half of the image. In the upper half of the image, the majority of the space outside the purple region is shaded green, with a few small strips of interspersed blue regions. The lower half however shows the majority of the space outside the central purple zone in yellow, orange, and red. The red zones appear in the lower central and left regions outside the purple zone. To the lower right of this image is a color scale that is labeled \u201cTotal Ozone (Dobsone units).\u201d This scale begins at 0 and increases by 100\u2019s up to 700. At the left end of the scale, the value 0 shows a very deep purple color, 100 is indigo, 200 is blue, 300 is green, 400 is a yellow-orange, 500 is red, 600 is pink, and 700 is white.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_03_OzoneHole-2.jpg\" alt=\"A view of Earth\u2019s southern hemisphere is shown. A nearly circular region of approximately half the diameter of the image is shown in shades of purple, with Antarctica appearing in a slightly lighter color than the surrounding ocean areas. Immediately outside this region is a narrow bright blue zone followed by a bright green zone. In the top half of the figure, the purple region extends slightly outward from the circle and the blue zone extends more outward to the right of the center as compared to the lower half of the image. In the upper half of the image, the majority of the space outside the purple region is shaded green, with a few small strips of interspersed blue regions. The lower half however shows the majority of the space outside the central purple zone in yellow, orange, and red. The red zones appear in the lower central and left regions outside the purple zone. To the lower right of this image is a color scale that is labeled \u201cTotal Ozone (Dobsone units).\u201d This scale begins at 0 and increases by 100\u2019s up to 700. At the left end of the scale, the value 0 shows a very deep purple color, 100 is indigo, 200 is blue, 300 is green, 400 is a yellow-orange, 500 is red, 600 is pink, and 700 is white.\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\n<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div id=\"fs-idm215850976\" style=\"text-align: center\" data-type=\"equation\">NO(<em>g<\/em>) + O<sub>3<\/sub>(<em>g<\/em>) \u27f6 NO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm116899456\">This reaction has been studied in the laboratory, and the following rate data were determined at 25 \u00b0C.<\/p>\r\n\r\n<table id=\"fs-idm205685856\" class=\"medium unnumbered\" summary=\"This table has four columns and six rows. The first row is a header row, and it labels each column: \u201cTrial,\u201d \u201c[ N O ] ( mol \/ L),\u201d \u201c[ O subscript 3 ] ( mol \/ L ),\u201d and \u201ccapital delta [ N O subscript 2 ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201c[ N O ] ( mol \/ L)\u201d column are the numbers: 1.00 times ten to the negative six power; 1.00 times ten to the negative six power; 1.00 times ten to the negative six power; 2.00 times ten to the negative six power; and 3.00 times ten to the negative six power. Under the \u201c[ O subscript 3 ] ( mol \/ L )\u201d column are the numbers: 3.00 times ten to the negative six; 6.00 times ten to the negative six; 9.00 times ten to the negative six; 9.00 times ten to the negative six; and 9.00 times ten to the negative six. Under the column \u201ccapital delta [ N O subscript 2 ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 )\u201d are the numbers: 6.60 times ten to the negative five; 1.32 times ten to the negative four; 1.98 times ten to the negative four; 3.96 times ten to the negative four; and 5.94 times ten to the negative four.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th data-align=\"left\">Trial<\/th>\r\n<th data-align=\"left\">[NO] (mol\/L)<\/th>\r\n<th data-align=\"left\">[O<sub>3<\/sub>] (mol\/L)<\/th>\r\n<th data-align=\"left\"><img class=\"alignnone wp-image-1711\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3a.png\" alt=\"\" width=\"154\" height=\"34\" \/><\/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\">1.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">3.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">6.60 \u00d7 10<sup>\u22125<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">2<\/td>\r\n<td data-align=\"left\">1.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">6.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">1.32 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">3<\/td>\r\n<td data-align=\"left\">1.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">9.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">1.98 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">4<\/td>\r\n<td data-align=\"left\">2.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">9.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">3.96 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">5<\/td>\r\n<td data-align=\"left\">3.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">9.00 \u00d7 10<sup>\u22126<\/sup><\/td>\r\n<td data-align=\"left\">5.94 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idm234408416\">Determine the rate law and the rate constant for the reaction at 25 \u00b0C.<\/p>\r\n<p id=\"fs-idm236104112\"><strong>Solution:<\/strong><\/p>\r\nThe rate law will have the form:\r\n<div id=\"fs-idm14888352\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[NO]<em><sup>m<\/sup><\/em>[O<sub>3<\/sub>]<sup><em>n<\/em><\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm187393984\">Determine the values of <em data-effect=\"italics\">m<\/em>, <em data-effect=\"italics\">n<\/em>, and <em data-effect=\"italics\">k<\/em> from the experimental data using the following three-part process:<\/p>\r\n\r\n<ol id=\"fs-idp46707232\" class=\"stepwise\" type=\"1\">\r\n \t<li>\r\n<p id=\"fs-idp221247360\"><em data-effect=\"italics\">Determine the value of<\/em> m <em data-effect=\"italics\">from the data in which [NO] varies and [O<sub>3<\/sub>] is constant.<\/em> In the last three experiments, [NO] varies while [O<sub>3<\/sub>] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and <em data-effect=\"italics\">m<\/em> in the rate law is equal to 1.<\/p>\r\n<\/li>\r\n \t<li>\r\n<p id=\"fs-idm59663552\"><em data-effect=\"italics\">Determine the value of<\/em> n <em data-effect=\"italics\">from data in which [O<sub>3<\/sub>] varies and [NO] is constant.<\/em> In the first three experiments, [NO] is constant and [O<sub>3<\/sub>] varies. The reaction rate changes in direct proportion to the change in [O<sub>3<\/sub>]. When [O<sub>3<\/sub>] doubles from trial 1 to 2, the rate doubles; when [O<sub>3<\/sub>] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O<sub>3<\/sub>], and <em data-effect=\"italics\">n<\/em> is equal to 1.The rate law is thus:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm236326672\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[NO]<em><sup>1<\/sup><\/em>[O<sub>3<\/sub>]<sup><em>1<\/em><\/sup><em> = k[NO][O<sub>3<\/sub>]\r\n<\/em><\/div><\/li>\r\n \t<li>\r\n<p id=\"fs-idm10220816\"><em data-effect=\"italics\">Determine the value of<\/em> k <em data-effect=\"italics\">from one set of concentrations and the corresponding rate<\/em>. The data from trial 1 are used below:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm208899344\" data-type=\"equation\"><img class=\"wp-image-1712 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3b-300x114.png\" alt=\"\" width=\"316\" height=\"120\" \/><\/div><\/li>\r\n<\/ol>\r\n<p id=\"fs-idm352713280\"><strong>Check Your Learning:<\/strong><\/p>\r\nAcetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:\r\n<div id=\"fs-idm196768496\" style=\"text-align: center\" data-type=\"equation\">CH<sub>3<\/sub>CHO(<em>g<\/em>) \u27f6 CH<sub>4<\/sub>(<em>g<\/em>) + CO(<em>g<\/em>)<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm167718688\">Determine the rate law and the rate constant for the reaction from the following experimental data:<\/p>\r\n\r\n<table id=\"fs-idm276791216\" class=\"medium unnumbered\" summary=\"This table has three columns and four rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201c[ C H subscript 3 C H O ] ( mol \/ L),\u201d and \u201cnegative capital delta [ C H subscript 3 C H O ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, and 3. Under the \u201c[ C H subscript 3 C H O ] ( mol \/ L)\u201d are the numbers: 1.75 times ten to the negative three; 3.50 times ten to the negative three; and 7.00 times ten to the negative three. Under the column \u201cnegative capital delta [ C H subscript 3 C H O ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 )\u201d are the numbers: 2.06 times ten to the negative 11; 8.24 times ten to the negative 11; and 3.30 times ten to the negative ten.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th data-align=\"left\">Trial<\/th>\r\n<th data-align=\"left\">[CH<sub>3<\/sub>CHO] (mol\/L)<\/th>\r\n<th data-align=\"left\"><img class=\"alignnone wp-image-1713\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3c.png\" alt=\"\" width=\"206\" height=\"40\" \/><\/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\">1.75 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<td data-align=\"left\">2.06 \u00d7 10<sup>\u221211<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">2<\/td>\r\n<td data-align=\"left\">3.50 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<td data-align=\"left\">8.24 \u00d7 10<sup>\u221211<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">3<\/td>\r\n<td data-align=\"left\">7.00 \u00d7 10<sup>\u22123<\/sup><\/td>\r\n<td data-align=\"left\">3.30 \u00d7 10<sup>\u221210<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-idm217041008\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm234756944\">rate = <em>k<\/em>[CH<sub>3<\/sub>CHO]<sup>2<\/sup> with <em>k<\/em> = 6.73 x 10<sup>-6<\/sup> L<sup>.<\/sup>mol<sup>-1.<\/sup>s<sup>-1<\/sup><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm285627376\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idm27901936\"><strong>Determining Rate Laws from Initial Rates:<\/strong><\/p>\r\nUsing the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:\r\n<div id=\"fs-idm275365520\" style=\"text-align: center\" data-type=\"equation\">2NO(<em>g<\/em>) + Cl<sub>2<\/sub>(<em>g<\/em>) \u27f6 2NOCl<sub>2<\/sub>(<em>g<\/em>)<\/div>\r\n<table id=\"fs-idm285249664\" class=\"medium unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row, and it labels the columns, \u201cTrial,\u201d \u201c[ N O ] ( mol \/ L),\u201d \u201c[ C l subscript 2 ] ( mol \/ L ),\u201d and \u201cnegative capital delta [ N O] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 ).\u201d Under the column \u201cTrial\u201d are the numbers: 1, 2, and 3. Under the column \u201c[ N O ] ( mol \/ L)\u201d are the numbers: 0.10, 0.10, and 0.15. Under the column \u201c[ C l subscript 2 ] ( mol \/ L )\u201d are the numbers: 0.10, 0.15, and 0.10. Under the column \u201cnegative delta [ N O] divided by delta t ( mol L superscript negative 1 s superscript negative 1\u201d are the numbers: 0.00300, 0.00450, 0.00675.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th data-align=\"left\">Trial<\/th>\r\n<th data-align=\"left\">[NO] (mol\/L)<\/th>\r\n<th data-align=\"left\">[Cl<sub>2<\/sub>] (mol\/L)<\/th>\r\n<th data-align=\"left\"><img class=\"alignnone wp-image-1716\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3d-300x68.png\" alt=\"\" width=\"197\" height=\"45\" \/><\/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\">0.10<\/td>\r\n<td data-align=\"left\">0.10<\/td>\r\n<td data-align=\"left\">0.00300<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">2<\/td>\r\n<td data-align=\"left\">0.10<\/td>\r\n<td data-align=\"left\">0.15<\/td>\r\n<td data-align=\"left\">0.00450<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">3<\/td>\r\n<td data-align=\"left\">0.15<\/td>\r\n<td data-align=\"left\">0.10<\/td>\r\n<td data-align=\"left\">0.00675<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idm237007488\"><strong>Solution:<\/strong><\/p>\r\nThe rate law for this reaction will have the form:\r\n<div id=\"fs-idm220277104\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[NO]<em><sup>m<\/sup><\/em>[Cl<sub>2<\/sub>]<sup><em>n<\/em><\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm256345824\">As in <a class=\"autogenerated-content\" href=\"#fs-idm234815200\">(Figure)<\/a>, approach this problem in a stepwise fashion, determining the values of <em data-effect=\"italics\">m<\/em> and <em data-effect=\"italics\">n<\/em> from the experimental data and then using these values to determine the value of <em data-effect=\"italics\">k<\/em>. In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of <em data-effect=\"italics\">m<\/em> and <em data-effect=\"italics\">n<\/em>:<\/p>\r\n\r\n<ol id=\"fs-idm47375696\" class=\"stepwise\" type=\"1\">\r\n \t<li>\r\n<p id=\"fs-idm45955472\"><em data-effect=\"italics\">Determine the value of<\/em> m <em data-effect=\"italics\">from the data in which [NO] varies and [Cl<sub>2<\/sub>] is constant<\/em>. Write the ratios with the subscripts <em data-effect=\"italics\">x<\/em> and <em data-effect=\"italics\">y<\/em> to indicate data from two different trials:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm359376496\" data-type=\"equation\"><img class=\"wp-image-1717 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3e-300x98.png\" alt=\"\" width=\"193\" height=\"63\" \/><\/div>\r\n<p id=\"fs-idm287583408\">Using the third trial and the first trial, in which [Cl<sub>2<\/sub>] does not vary, gives:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idp81408128\" data-type=\"equation\"><img class=\"alignnone size-medium wp-image-1718 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3f-300x48.png\" alt=\"\" width=\"300\" height=\"48\" \/><\/div>\r\n<p id=\"fs-idm272297280\">Canceling equivalent terms in the numerator and denominator leaves:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm185610624\" data-type=\"equation\"><img class=\"wp-image-1719 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3g-300x86.png\" alt=\"\" width=\"175\" height=\"50\" \/><\/div>\r\n<p id=\"fs-idm187825008\">which simplifies to:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm285203088\" style=\"text-align: center\" data-type=\"equation\">2.25 = (1.5)<sup><em>m<\/em><\/sup><\/div>\r\n<p id=\"fs-idm253983824\">Use logarithms to determine the value of the exponent <em data-effect=\"italics\">m<\/em>:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm222931920\" data-type=\"equation\"><img class=\"wp-image-1720 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3h-300x156.png\" alt=\"\" width=\"187\" height=\"97\" \/><\/div>\r\n<p id=\"fs-idm137061360\">Confirm the result<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm10561024\" style=\"text-align: center\" data-type=\"equation\">(1.5)<sup>2<\/sup> = 2.25<\/div><\/li>\r\n \t<li>\r\n<p id=\"fs-idm355457040\"><em data-effect=\"italics\">Determine the value of<\/em> n <em data-effect=\"italics\">from data in which [Cl<sub>2<\/sub>] varies and [NO] is constant.<\/em><span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idp69093344\" data-type=\"equation\"><img class=\"alignnone wp-image-1721 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3i-300x60.png\" alt=\"\" width=\"260\" height=\"52\" \/><\/div>\r\n<p id=\"fs-idm221941584\">Cancelation gives:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idp5205776\" data-type=\"equation\"><img class=\"wp-image-1722 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3j.png\" alt=\"\" width=\"131\" height=\"42\" \/><\/div>\r\n<p id=\"fs-idm265109792\">which simplifies to:<\/p>\r\n\r\n<div id=\"fs-idm97106784\" style=\"text-align: center\" data-type=\"equation\">1.5 = (1.5)<em><sup>n<\/sup><\/em><\/div>\r\n<p id=\"fs-idm256497008\">Thus <em data-effect=\"italics\">n<\/em> must be 1, and the form of the rate law is:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm215496784\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[NO]<sup>2<\/sup>[Cl<sub>2<\/sub>]<\/div><\/li>\r\n \t<li>\r\n<p id=\"fs-idm110493584\"><em data-effect=\"italics\">Determine the numerical value of the rate constant<\/em> k <em data-effect=\"italics\">with appropriate units.<\/em> The units for the rate of a reaction are mol<sup>.<\/sup>L<sup>-1.<\/sup>s<sup>-1<\/sup>. The units for <em data-effect=\"italics\">k<\/em> are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol<sup>3<\/sup>\/L<sup>3<\/sup>. The units for <em data-effect=\"italics\">k<\/em> should be mol<sup>\u22122.<\/sup>L<sup>2.<\/sup>s<sup>-1<\/sup> so that the rate is in terms of mol<sup>.<\/sup>L<sup>-1.<\/sup>s<sup>-1<\/sup>.<\/p>\r\n<p id=\"fs-idm53387488\">To determine the value of <em data-effect=\"italics\">k<\/em> once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for <em data-effect=\"italics\">k<\/em>:<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-idm263345248\" style=\"text-align: center\" data-type=\"equation\">0.00300 mol<sup>.<\/sup>L<sup>-1.<\/sup>s<sup>-1<\/sup> = <em>k<\/em>(0.10 mol<sup>.<\/sup>L<sup>-1<\/sup>)<sup>2<\/sup>(0.10 mol<sup>.<\/sup>L<sup>-1<\/sup>)<\/div><\/li>\r\n<\/ol>\r\n<div id=\"fs-idm263345248\" style=\"text-align: center\" data-type=\"equation\"><span style=\"font-size: 1em\"><em>k<\/em> = 3.0 <\/span><span style=\"font-size: 1em\">mol<\/span><sup>-2.<\/sup><span style=\"font-size: 1em\">L<\/span><sup>2.<\/sup><span style=\"font-size: 1em\">s<\/span><sup>-1<\/sup><\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<p id=\"fs-idm209905104\"><strong>Check Your Learning:<\/strong><\/p>\r\nUse the provided initial rate data to derive the rate law for the reaction whose equation is:\r\n<div id=\"fs-idm205740176\" style=\"text-align: center\" data-type=\"equation\">OCl<sup>-<\/sup>(<em>aq<\/em>) + I<sup>-<\/sup>(<em>aq<\/em>) \u27f6 OI<sup>-<\/sup>(<em>aq<\/em>) + Cl<sup>-<\/sup>(<em>aq<\/em>)<\/div>\r\n<table id=\"fs-idm275492048\" class=\"medium unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row and it labels each column, \u201cTrial,\u201d \u201c[ O C l superscript negative sign ] ( mol \/ L),\u201d \u201c[ I superscript negative sign ] ( mol \/ l ),\u201d and \u201cInitial Rate ( mol \/ L \/ s).\u201d Under the column \u201cTrial\u201d are the numbers: 1, 2, and 3. Under the column \u201c[ O C l superscript negative sign ] ( mol \/ L)\u201d are the numbers 0.0040, 0.0020, and 0.0020. Under the column \u201c[ I superscript negative sign ] ( mol \/ l )\u201d are the numbers: 0.0020, 0.0040, and 0.0020. Under the column \u201cInitial Rate ( mol \/ L \/ s)\u201d are the numbers: 0.00184, 0.00092, and 0.00046.\" data-label=\"\">\r\n<thead>\r\n<tr valign=\"middle\">\r\n<th data-align=\"left\">Trial<\/th>\r\n<th data-align=\"left\">[OCl<sup>\u2212<\/sup>] (mol\/L)<\/th>\r\n<th data-align=\"left\">[I<sup>\u2212<\/sup>] (mol\/L)<\/th>\r\n<th data-align=\"left\">Initial Rate (mol<sup>.<\/sup>L<sup>-1.<\/sup>s<sup>-1<\/sup>)<\/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\">0.0040<\/td>\r\n<td data-align=\"left\">0.0020<\/td>\r\n<td data-align=\"left\">0.00184<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">2<\/td>\r\n<td data-align=\"left\">0.0020<\/td>\r\n<td data-align=\"left\">0.0040<\/td>\r\n<td data-align=\"left\">0.00092<\/td>\r\n<\/tr>\r\n<tr valign=\"middle\">\r\n<td data-align=\"left\">3<\/td>\r\n<td data-align=\"left\">0.0020<\/td>\r\n<td data-align=\"left\">0.0020<\/td>\r\n<td data-align=\"left\">0.00046<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idm296296016\">Determine the rate law expression and the value of the rate constant <em data-effect=\"italics\">k<\/em> with appropriate units for this reaction.<\/p>\r\n\r\n<div id=\"fs-idm214856768\" data-type=\"note\">\r\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\r\n<p id=\"fs-idm19062704\"><img class=\"alignnone wp-image-1725\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3k-300x206.png\" alt=\"\" width=\"487\" height=\"334\" \/><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm197875360\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Reaction Order and Rate Constant Units<\/strong><\/h3>\r\n<p id=\"fs-idp9200128\">In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.<\/p>\r\n<p id=\"fs-idm178764304\">Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:<\/p>\r\n\r\n<div id=\"fs-idm213140144\" data-type=\"equation\">NO<sub>2<\/sub> + CO \u27f6 NO + CO<sub>2\u00a0<\/sub> \u00a0 \u00a0 \u00a0 \u00a0rate = <em>k<\/em>[NO<sub>2<\/sub>]<sup>2 <\/sup><\/div>\r\n<div data-type=\"equation\">CH<sub>3<\/sub>CHO \u27f6 CH<sub>4<\/sub> + CO\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0rate = <em>k<\/em>[CH<sub>3<\/sub>CHO]<sup>2<\/sup><\/div>\r\n<div data-type=\"equation\">2N<sub>2<\/sub>O<sub>5<\/sub> \u27f6 NO<sub>2<\/sub> + O<sub>2<\/sub>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 rate = <em>k<\/em>[N<sub>2<\/sub>O<sub>5<\/sub>]<\/div>\r\n<div data-type=\"equation\">2NO<sub>2<\/sub> + F<sub>2<\/sub> \u27f6 2NO<sub>2<\/sub>F\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 rate = <em>k<\/em>[NO<sub>2<\/sub>][F<sub>2<\/sub>]<\/div>\r\n<div data-type=\"equation\">2NO<sub>2<\/sub>Cl \u27f62NO<sub>2<\/sub> + Cl<sub>2\u00a0<\/sub>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0rate = <em>k<\/em>[NO<sub>2<\/sub>Cl]<\/div>\r\n<p id=\"fs-idm284144576\">It is important to note that <em data-effect=\"italics\">rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm168364064\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\r\n<p id=\"fs-idm262547408\">Rate laws (<em data-effect=\"italics\">differential rate laws<\/em>) provide a mathematical description of how changes in the concentration of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the concentration of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm273385984\" class=\"exercises\" data-depth=\"1\">\r\n<div id=\"fs-idm237267424\" data-type=\"exercise\">\r\n<div id=\"fs-idm220253760\" data-type=\"solution\">\r\n<p id=\"fs-idm220253504\"><\/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-idm176826832\">\r\n \t<dt>method of initial rates<\/dt>\r\n \t<dd id=\"fs-idm176826448\">common experimental approach to determining rate laws that involves measuring reaction rates at varying initial reactant concentrations<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm276164832\">\r\n \t<dt>overall reaction order<\/dt>\r\n \t<dd id=\"fs-idm263526528\">sum of the reaction orders for each substance represented in the rate law<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm185628880\">\r\n \t<dt>rate constant (<em data-effect=\"italics\">k<\/em>)<\/dt>\r\n \t<dd id=\"fs-idm213521696\">proportionality constant in a rate law<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm215968704\">\r\n \t<dt>rate law<\/dt>\r\n \t<dd id=\"fs-idm215968320\">(also, rate equation) (also, differential rate laws) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-idm333854336\">\r\n \t<dt>reaction order<\/dt>\r\n \t<dd id=\"fs-idm47186480\">value of an exponent in a rate law (for example, zero order for 0, first order for 1, second order for 2, and so on)<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox textbox--learning-objectives\">\n<h3><strong>Learning Objectives<\/strong><\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Explain the form and function of a rate law<\/li>\n<li>Use rate laws to calculate reaction rates<\/li>\n<li>Use rate and concentration data to identify reaction orders and derive rate laws<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm238441168\">As described in the previous module, the rate of a reaction is often affected by the concentrations of reactants. <strong>Rate laws <\/strong>(sometimes called <em data-effect=\"italics\">differential rate laws<\/em>) are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation<\/p>\n<div id=\"fs-idm248316432\" style=\"text-align: center\" data-type=\"equation\"><em>aA<\/em> + <em>bB<\/em> \u27f6 <em>products<\/em><\/div>\n<p id=\"fs-idm656673904\">where <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">b<\/em> are stoichiometric coefficients. The rate law for this reaction is written as:<\/p>\n<div id=\"fs-idm276589968\" style=\"text-align: center\" data-type=\"equation\">rate =<em>k<\/em>[A]<em><sup>m<\/sup><\/em>[B]<em><sup>n<\/sup><\/em><\/div>\n<p id=\"fs-idp18257184\">in which [<em data-effect=\"italics\">A<\/em>] and [<em data-effect=\"italics\">B<\/em>] represent the molar concentrations of reactants, and <em data-effect=\"italics\">k<\/em> is the <strong>rate constant<\/strong>, which is specific for a particular reaction at a particular temperature. The exponents <em data-effect=\"italics\">m<\/em> and <em data-effect=\"italics\">n<\/em> are the <strong>reaction orders<\/strong> and are typically positive integers, though they can be fractions, negative, or zero. The rate constant <em data-effect=\"italics\">k<\/em> and the reaction orders <em data-effect=\"italics\">m<\/em> and <em data-effect=\"italics\">n<\/em> must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant <em data-effect=\"italics\">k<\/em> is independent of the reactant concentrations, but it does vary with temperature.<\/p>\n<p id=\"fs-idm115903376\">The reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is <em data-effect=\"italics\">m<\/em> order with respect to <em data-effect=\"italics\">A<\/em> and <em data-effect=\"italics\">n<\/em> order with respect to <em data-effect=\"italics\">B<\/em>. For example, if <em data-effect=\"italics\">m<\/em> = 1 and <em data-effect=\"italics\">n<\/em> = 2, the reaction is first order in <em data-effect=\"italics\">A<\/em> and second order in <em data-effect=\"italics\">B<\/em>. The <strong>overall reaction order<\/strong> is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall (1 + 2 = 3). A few specific examples are shown below to further illustrate this concept.<\/p>\n<p id=\"fs-idm217470464\">The rate law:<\/p>\n<div id=\"fs-idm204886480\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[H<sub>2<\/sub>O<sub>2<\/sub>]<\/div>\n<p id=\"fs-idm388320\">describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:<\/p>\n<div id=\"fs-idm185553664\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[C<sub>4<\/sub>H<sub>6<\/sub>]<sup>2<\/sup><\/div>\n<p id=\"fs-idm105991808\">describes a reaction that is second order in C<sub>4<\/sub>H<sub>6<\/sub> and second order overall. The rate law:<\/p>\n<div id=\"fs-idm41933536\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[H<sup>+<\/sup>][OH<sup>&#8211;<\/sup>]<\/div>\n<p id=\"fs-idm221631328\">describes a reaction that is first order in H<sup>+<\/sup>, first order in OH<sup>\u2212<\/sup>, and second order overall.<\/p>\n<div id=\"fs-idm235878704\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm114205984\"><strong>Writing Rate Laws from Reaction Orders:<\/strong><\/p>\n<p>An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:<\/p>\n<div id=\"fs-idm189808080\" style=\"text-align: center\" data-type=\"equation\">NO<sub>2<\/sub>(<em>g<\/em>) + CO(<em>g<\/em>) \u27f6 NO(<em>g<\/em>) + CO<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<p id=\"fs-idm213018368\">is second order in NO<sub>2<\/sub> and zero order in CO at 100 \u00b0C. What is the rate law for the reaction?<\/p>\n<p id=\"fs-idm222185424\"><strong>Solution:<\/strong><\/p>\n<p>The reaction will have the form:<\/p>\n<div id=\"fs-idm212887136\" style=\"text-align: center\" data-type=\"equation\">rate =<em>k<\/em>[NO<sub>2<\/sub>]<em><sup>m<\/sup><\/em>[CO]<em><sup>n<\/sup><\/em><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm221521296\">The reaction is second order in NO<sub>2<\/sub>; thus <em data-effect=\"italics\">m<\/em> = 2. The reaction is zero order in CO; thus <em data-effect=\"italics\">n<\/em> = 0. The rate law is:<\/p>\n<div id=\"fs-idm49710800\" style=\"text-align: center\" data-type=\"equation\">rate =<em>k<\/em>[NO<sub>2<\/sub>]<sup>2<\/sup>[CO]<sup>0<\/sup> = <em>k<\/em>[NO<sub>2<\/sub>]<sup>2<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm250199824\">Remember that a number raised to the zero power is equal to 1, thus [CO]<sup>0<\/sup> = 1, which is why the CO concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of NO<sub>2<\/sub>. A later chapter section on reaction mechanisms will explain how a reactant\u2019s concentration can have no effect on a reaction rate despite being involved in the reaction.<\/p>\n<p id=\"fs-idm273206544\"><strong>Check Your Learning:<\/strong><\/p>\n<p>The rate law for the reaction:<\/p>\n<div id=\"fs-idm122197664\" style=\"text-align: center\" data-type=\"equation\">H<sub>2<\/sub>(<em>g<\/em>) + 2NO(<em>g<\/em>) \u27f6N<sub>2<\/sub>O(<em>g<\/em>) + H<sub>2<\/sub>O(<em>g<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm122425904\">has been determined to be rate = <em data-effect=\"italics\">k<\/em>[NO]<sup>2<\/sup>[H<sub>2<\/sub>]. What are the orders with respect to each reactant, and what is the overall order of the reaction?<\/p>\n<div id=\"fs-idm168583904\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm141295376\">order in NO = 2; order in H<sub>2<\/sub> = 1; overall order = 3<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-idm117052240\"><strong>Check Your Learning:<\/strong><\/p>\n<p>In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH<sub>3<\/sub>OH) and ethyl acetate (CH<sub>3<\/sub>CH<sub>2<\/sub>OCOCH<sub>3<\/sub>) as a sample reaction before studying the chemical reactions that produce biodiesel:<\/p>\n<div id=\"fs-idm155639824\" style=\"text-align: center\" data-type=\"equation\">CH<sub>3<\/sub>OH + CH<sub>3<\/sub>CH<sub>2<\/sub>OCOCH<sub>3<\/sub> \u27f6 CH<sub>3<\/sub>OCOCH<sub>3<\/sub> + CH<sub>3<\/sub>CH<sub>2<\/sub>OH<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm213233440\">The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be:<\/p>\n<div id=\"fs-idm236204576\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[CH<sub>3<\/sub>OH]<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm219218784\">What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?<\/p>\n<div id=\"fs-idm253536096\" data-type=\"note\">\n<div data-type=\"title\">Answer:<\/div>\n<p id=\"fs-idm215326320\">order in CH<sub>3<\/sub>OH = 1; order in CH<sub>3<\/sub>CH<sub>2<\/sub>OCOCH<sub>3<\/sub> = 0; overall order = 1<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm147402384\">A common experimental approach to the determination of rate laws is the <strong>method of initial rates<\/strong>. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.<\/p>\n<div id=\"fs-idm234815200\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm162746544\"><strong>Determining a Rate Law from Initial Rates:<\/strong><\/p>\n<p>Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (<a class=\"autogenerated-content\" href=\"#CNX_Chem_12_03_OzoneHole\">(Figure)<\/a>). One such reaction is the combination of nitric oxide, NO, with ozone, O<sub>3<\/sub>:<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Chem_12_03_OzoneHole\" class=\"scaled-down\">\n<div class=\"bc-figcaption figcaption\">A contour map showing stratospheric ozone concentration and the \u201cozone hole\u201d that occurs over Antarctica during its spring months. (credit: modification of work by NASA)<\/div>\n<p><span id=\"fs-idm204213280\" data-type=\"media\" data-alt=\"A view of Earth\u2019s southern hemisphere is shown. A nearly circular region of approximately half the diameter of the image is shown in shades of purple, with Antarctica appearing in a slightly lighter color than the surrounding ocean areas. Immediately outside this region is a narrow bright blue zone followed by a bright green zone. In the top half of the figure, the purple region extends slightly outward from the circle and the blue zone extends more outward to the right of the center as compared to the lower half of the image. In the upper half of the image, the majority of the space outside the purple region is shaded green, with a few small strips of interspersed blue regions. The lower half however shows the majority of the space outside the central purple zone in yellow, orange, and red. The red zones appear in the lower central and left regions outside the purple zone. To the lower right of this image is a color scale that is labeled \u201cTotal Ozone (Dobsone units).\u201d This scale begins at 0 and increases by 100\u2019s up to 700. At the left end of the scale, the value 0 shows a very deep purple color, 100 is indigo, 200 is blue, 300 is green, 400 is a yellow-orange, 500 is red, 600 is pink, and 700 is white.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/CNX_Chem_12_03_OzoneHole-2.jpg\" alt=\"A view of Earth\u2019s southern hemisphere is shown. A nearly circular region of approximately half the diameter of the image is shown in shades of purple, with Antarctica appearing in a slightly lighter color than the surrounding ocean areas. Immediately outside this region is a narrow bright blue zone followed by a bright green zone. In the top half of the figure, the purple region extends slightly outward from the circle and the blue zone extends more outward to the right of the center as compared to the lower half of the image. In the upper half of the image, the majority of the space outside the purple region is shaded green, with a few small strips of interspersed blue regions. The lower half however shows the majority of the space outside the central purple zone in yellow, orange, and red. The red zones appear in the lower central and left regions outside the purple zone. To the lower right of this image is a color scale that is labeled \u201cTotal Ozone (Dobsone units).\u201d This scale begins at 0 and increases by 100\u2019s up to 700. At the left end of the scale, the value 0 shows a very deep purple color, 100 is indigo, 200 is blue, 300 is green, 400 is a yellow-orange, 500 is red, 600 is pink, and 700 is white.\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<\/div>\n<div data-type=\"equation\"><\/div>\n<div id=\"fs-idm215850976\" style=\"text-align: center\" data-type=\"equation\">NO(<em>g<\/em>) + O<sub>3<\/sub>(<em>g<\/em>) \u27f6 NO<sub>2<\/sub>(<em>g<\/em>) + O<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm116899456\">This reaction has been studied in the laboratory, and the following rate data were determined at 25 \u00b0C.<\/p>\n<table id=\"fs-idm205685856\" class=\"medium unnumbered\" summary=\"This table has four columns and six rows. The first row is a header row, and it labels each column: \u201cTrial,\u201d \u201c[ N O ] ( mol \/ L),\u201d \u201c[ O subscript 3 ] ( mol \/ L ),\u201d and \u201ccapital delta [ N O subscript 2 ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, 3, 4, and 5. Under the \u201c[ N O ] ( mol \/ L)\u201d column are the numbers: 1.00 times ten to the negative six power; 1.00 times ten to the negative six power; 1.00 times ten to the negative six power; 2.00 times ten to the negative six power; and 3.00 times ten to the negative six power. Under the \u201c[ O subscript 3 ] ( mol \/ L )\u201d column are the numbers: 3.00 times ten to the negative six; 6.00 times ten to the negative six; 9.00 times ten to the negative six; 9.00 times ten to the negative six; and 9.00 times ten to the negative six. Under the column \u201ccapital delta [ N O subscript 2 ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 )\u201d are the numbers: 6.60 times ten to the negative five; 1.32 times ten to the negative four; 1.98 times ten to the negative four; 3.96 times ten to the negative four; and 5.94 times ten to the negative four.\" data-label=\"\">\n<thead>\n<tr valign=\"middle\">\n<th data-align=\"left\">Trial<\/th>\n<th data-align=\"left\">[NO] (mol\/L)<\/th>\n<th data-align=\"left\">[O<sub>3<\/sub>] (mol\/L)<\/th>\n<th data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1711\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3a.png\" alt=\"\" width=\"154\" height=\"34\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3a.png 235w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3a-65x14.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3a-225x50.png 225w\" sizes=\"auto, (max-width: 154px) 100vw, 154px\" \/><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">1<\/td>\n<td data-align=\"left\">1.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">3.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">6.60 \u00d7 10<sup>\u22125<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">2<\/td>\n<td data-align=\"left\">1.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">6.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">1.32 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">3<\/td>\n<td data-align=\"left\">1.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">9.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">1.98 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">4<\/td>\n<td data-align=\"left\">2.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">9.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">3.96 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">5<\/td>\n<td data-align=\"left\">3.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">9.00 \u00d7 10<sup>\u22126<\/sup><\/td>\n<td data-align=\"left\">5.94 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idm234408416\">Determine the rate law and the rate constant for the reaction at 25 \u00b0C.<\/p>\n<p id=\"fs-idm236104112\"><strong>Solution:<\/strong><\/p>\n<p>The rate law will have the form:<\/p>\n<div id=\"fs-idm14888352\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[NO]<em><sup>m<\/sup><\/em>[O<sub>3<\/sub>]<sup><em>n<\/em><\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm187393984\">Determine the values of <em data-effect=\"italics\">m<\/em>, <em data-effect=\"italics\">n<\/em>, and <em data-effect=\"italics\">k<\/em> from the experimental data using the following three-part process:<\/p>\n<ol id=\"fs-idp46707232\" class=\"stepwise\" type=\"1\">\n<li>\n<p id=\"fs-idp221247360\"><em data-effect=\"italics\">Determine the value of<\/em> m <em data-effect=\"italics\">from the data in which [NO] varies and [O<sub>3<\/sub>] is constant.<\/em> In the last three experiments, [NO] varies while [O<sub>3<\/sub>] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and <em data-effect=\"italics\">m<\/em> in the rate law is equal to 1.<\/p>\n<\/li>\n<li>\n<p id=\"fs-idm59663552\"><em data-effect=\"italics\">Determine the value of<\/em> n <em data-effect=\"italics\">from data in which [O<sub>3<\/sub>] varies and [NO] is constant.<\/em> In the first three experiments, [NO] is constant and [O<sub>3<\/sub>] varies. The reaction rate changes in direct proportion to the change in [O<sub>3<\/sub>]. When [O<sub>3<\/sub>] doubles from trial 1 to 2, the rate doubles; when [O<sub>3<\/sub>] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O<sub>3<\/sub>], and <em data-effect=\"italics\">n<\/em> is equal to 1.The rate law is thus:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm236326672\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[NO]<em><sup>1<\/sup><\/em>[O<sub>3<\/sub>]<sup><em>1<\/em><\/sup><em> = k[NO][O<sub>3<\/sub>]<br \/>\n<\/em><\/div>\n<\/li>\n<li>\n<p id=\"fs-idm10220816\"><em data-effect=\"italics\">Determine the value of<\/em> k <em data-effect=\"italics\">from one set of concentrations and the corresponding rate<\/em>. The data from trial 1 are used below:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm208899344\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1712 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3b-300x114.png\" alt=\"\" width=\"316\" height=\"120\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3b-300x114.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3b-65x25.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3b-225x86.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3b-350x133.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3b.png 446w\" sizes=\"auto, (max-width: 316px) 100vw, 316px\" \/><\/div>\n<\/li>\n<\/ol>\n<p id=\"fs-idm352713280\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:<\/p>\n<div id=\"fs-idm196768496\" style=\"text-align: center\" data-type=\"equation\">CH<sub>3<\/sub>CHO(<em>g<\/em>) \u27f6 CH<sub>4<\/sub>(<em>g<\/em>) + CO(<em>g<\/em>)<\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm167718688\">Determine the rate law and the rate constant for the reaction from the following experimental data:<\/p>\n<table id=\"fs-idm276791216\" class=\"medium unnumbered\" summary=\"This table has three columns and four rows. The first row is a header row, and it labels each column, \u201cTrial,\u201d \u201c[ C H subscript 3 C H O ] ( mol \/ L),\u201d and \u201cnegative capital delta [ C H subscript 3 C H O ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 ).\u201d Under the \u201cTrial\u201d column are the numbers: 1, 2, and 3. Under the \u201c[ C H subscript 3 C H O ] ( mol \/ L)\u201d are the numbers: 1.75 times ten to the negative three; 3.50 times ten to the negative three; and 7.00 times ten to the negative three. Under the column \u201cnegative capital delta [ C H subscript 3 C H O ] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 )\u201d are the numbers: 2.06 times ten to the negative 11; 8.24 times ten to the negative 11; and 3.30 times ten to the negative ten.\" data-label=\"\">\n<thead>\n<tr valign=\"middle\">\n<th data-align=\"left\">Trial<\/th>\n<th data-align=\"left\">[CH<sub>3<\/sub>CHO] (mol\/L)<\/th>\n<th data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1713\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3c.png\" alt=\"\" width=\"206\" height=\"40\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3c.png 289w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3c-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3c-225x44.png 225w\" sizes=\"auto, (max-width: 206px) 100vw, 206px\" \/><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">1<\/td>\n<td data-align=\"left\">1.75 \u00d7 10<sup>\u22123<\/sup><\/td>\n<td data-align=\"left\">2.06 \u00d7 10<sup>\u221211<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">2<\/td>\n<td data-align=\"left\">3.50 \u00d7 10<sup>\u22123<\/sup><\/td>\n<td data-align=\"left\">8.24 \u00d7 10<sup>\u221211<\/sup><\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">3<\/td>\n<td data-align=\"left\">7.00 \u00d7 10<sup>\u22123<\/sup><\/td>\n<td data-align=\"left\">3.30 \u00d7 10<sup>\u221210<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-idm217041008\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm234756944\">rate = <em>k<\/em>[CH<sub>3<\/sub>CHO]<sup>2<\/sup> with <em>k<\/em> = 6.73 x 10<sup>-6<\/sup> L<sup>.<\/sup>mol<sup>-1.<\/sup>s<sup>-1<\/sup><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm285627376\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idm27901936\"><strong>Determining Rate Laws from Initial Rates:<\/strong><\/p>\n<p>Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:<\/p>\n<div id=\"fs-idm275365520\" style=\"text-align: center\" data-type=\"equation\">2NO(<em>g<\/em>) + Cl<sub>2<\/sub>(<em>g<\/em>) \u27f6 2NOCl<sub>2<\/sub>(<em>g<\/em>)<\/div>\n<table id=\"fs-idm285249664\" class=\"medium unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row, and it labels the columns, \u201cTrial,\u201d \u201c[ N O ] ( mol \/ L),\u201d \u201c[ C l subscript 2 ] ( mol \/ L ),\u201d and \u201cnegative capital delta [ N O] divided by capital delta t ( mol L superscript negative 1 s superscript negative 1 ).\u201d Under the column \u201cTrial\u201d are the numbers: 1, 2, and 3. Under the column \u201c[ N O ] ( mol \/ L)\u201d are the numbers: 0.10, 0.10, and 0.15. Under the column \u201c[ C l subscript 2 ] ( mol \/ L )\u201d are the numbers: 0.10, 0.15, and 0.10. Under the column \u201cnegative delta [ N O] divided by delta t ( mol L superscript negative 1 s superscript negative 1\u201d are the numbers: 0.00300, 0.00450, 0.00675.\" data-label=\"\">\n<thead>\n<tr valign=\"middle\">\n<th data-align=\"left\">Trial<\/th>\n<th data-align=\"left\">[NO] (mol\/L)<\/th>\n<th data-align=\"left\">[Cl<sub>2<\/sub>] (mol\/L)<\/th>\n<th data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1716\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3d-300x68.png\" alt=\"\" width=\"197\" height=\"45\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3d-300x68.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3d-65x15.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3d-225x51.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3d-350x79.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3d.png 359w\" sizes=\"auto, (max-width: 197px) 100vw, 197px\" \/><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">1<\/td>\n<td data-align=\"left\">0.10<\/td>\n<td data-align=\"left\">0.10<\/td>\n<td data-align=\"left\">0.00300<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">2<\/td>\n<td data-align=\"left\">0.10<\/td>\n<td data-align=\"left\">0.15<\/td>\n<td data-align=\"left\">0.00450<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">3<\/td>\n<td data-align=\"left\">0.15<\/td>\n<td data-align=\"left\">0.10<\/td>\n<td data-align=\"left\">0.00675<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idm237007488\"><strong>Solution:<\/strong><\/p>\n<p>The rate law for this reaction will have the form:<\/p>\n<div id=\"fs-idm220277104\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[NO]<em><sup>m<\/sup><\/em>[Cl<sub>2<\/sub>]<sup><em>n<\/em><\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm256345824\">As in <a class=\"autogenerated-content\" href=\"#fs-idm234815200\">(Figure)<\/a>, approach this problem in a stepwise fashion, determining the values of <em data-effect=\"italics\">m<\/em> and <em data-effect=\"italics\">n<\/em> from the experimental data and then using these values to determine the value of <em data-effect=\"italics\">k<\/em>. In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of <em data-effect=\"italics\">m<\/em> and <em data-effect=\"italics\">n<\/em>:<\/p>\n<ol id=\"fs-idm47375696\" class=\"stepwise\" type=\"1\">\n<li>\n<p id=\"fs-idm45955472\"><em data-effect=\"italics\">Determine the value of<\/em> m <em data-effect=\"italics\">from the data in which [NO] varies and [Cl<sub>2<\/sub>] is constant<\/em>. Write the ratios with the subscripts <em data-effect=\"italics\">x<\/em> and <em data-effect=\"italics\">y<\/em> to indicate data from two different trials:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm359376496\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1717 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3e-300x98.png\" alt=\"\" width=\"193\" height=\"63\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3e-300x98.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3e-65x21.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3e-225x73.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3e-350x114.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3e.png 369w\" sizes=\"auto, (max-width: 193px) 100vw, 193px\" \/><\/div>\n<p id=\"fs-idm287583408\">Using the third trial and the first trial, in which [Cl<sub>2<\/sub>] does not vary, gives:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idp81408128\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-1718 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3f-300x48.png\" alt=\"\" width=\"300\" height=\"48\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3f-300x48.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3f-65x10.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3f-225x36.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3f-350x56.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3f.png 583w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<p id=\"fs-idm272297280\">Canceling equivalent terms in the numerator and denominator leaves:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm185610624\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1719 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3g-300x86.png\" alt=\"\" width=\"175\" height=\"50\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3g-300x86.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3g-65x19.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3g-225x65.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3g.png 316w\" sizes=\"auto, (max-width: 175px) 100vw, 175px\" \/><\/div>\n<p id=\"fs-idm187825008\">which simplifies to:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm285203088\" style=\"text-align: center\" data-type=\"equation\">2.25 = (1.5)<sup><em>m<\/em><\/sup><\/div>\n<p id=\"fs-idm253983824\">Use logarithms to determine the value of the exponent <em data-effect=\"italics\">m<\/em>:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm222931920\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1720 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3h-300x156.png\" alt=\"\" width=\"187\" height=\"97\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3h-300x156.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3h-65x34.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3h-225x117.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3h-350x182.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3h.png 362w\" sizes=\"auto, (max-width: 187px) 100vw, 187px\" \/><\/div>\n<p id=\"fs-idm137061360\">Confirm the result<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm10561024\" style=\"text-align: center\" data-type=\"equation\">(1.5)<sup>2<\/sup> = 2.25<\/div>\n<\/li>\n<li>\n<p id=\"fs-idm355457040\"><em data-effect=\"italics\">Determine the value of<\/em> n <em data-effect=\"italics\">from data in which [Cl<sub>2<\/sub>] varies and [NO] is constant.<\/em><span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idp69093344\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1721 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3i-300x60.png\" alt=\"\" width=\"260\" height=\"52\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3i-300x60.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3i-65x13.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3i-225x45.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3i-350x70.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3i.png 589w\" sizes=\"auto, (max-width: 260px) 100vw, 260px\" \/><\/div>\n<p id=\"fs-idm221941584\">Cancelation gives:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idp5205776\" data-type=\"equation\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1722 aligncenter\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3j.png\" alt=\"\" width=\"131\" height=\"42\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3j.png 296w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3j-65x21.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3j-225x72.png 225w\" sizes=\"auto, (max-width: 131px) 100vw, 131px\" \/><\/div>\n<p id=\"fs-idm265109792\">which simplifies to:<\/p>\n<div id=\"fs-idm97106784\" style=\"text-align: center\" data-type=\"equation\">1.5 = (1.5)<em><sup>n<\/sup><\/em><\/div>\n<p id=\"fs-idm256497008\">Thus <em data-effect=\"italics\">n<\/em> must be 1, and the form of the rate law is:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm215496784\" style=\"text-align: center\" data-type=\"equation\">rate = <em>k<\/em>[NO]<sup>2<\/sup>[Cl<sub>2<\/sub>]<\/div>\n<\/li>\n<li>\n<p id=\"fs-idm110493584\"><em data-effect=\"italics\">Determine the numerical value of the rate constant<\/em> k <em data-effect=\"italics\">with appropriate units.<\/em> The units for the rate of a reaction are mol<sup>.<\/sup>L<sup>-1.<\/sup>s<sup>-1<\/sup>. The units for <em data-effect=\"italics\">k<\/em> are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol<sup>3<\/sup>\/L<sup>3<\/sup>. The units for <em data-effect=\"italics\">k<\/em> should be mol<sup>\u22122.<\/sup>L<sup>2.<\/sup>s<sup>-1<\/sup> so that the rate is in terms of mol<sup>.<\/sup>L<sup>-1.<\/sup>s<sup>-1<\/sup>.<\/p>\n<p id=\"fs-idm53387488\">To determine the value of <em data-effect=\"italics\">k<\/em> once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for <em data-effect=\"italics\">k<\/em>:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm263345248\" style=\"text-align: center\" data-type=\"equation\">0.00300 mol<sup>.<\/sup>L<sup>-1.<\/sup>s<sup>-1<\/sup> = <em>k<\/em>(0.10 mol<sup>.<\/sup>L<sup>-1<\/sup>)<sup>2<\/sup>(0.10 mol<sup>.<\/sup>L<sup>-1<\/sup>)<\/div>\n<\/li>\n<\/ol>\n<div id=\"fs-idm263345248\" style=\"text-align: center\" data-type=\"equation\"><span style=\"font-size: 1em\"><em>k<\/em> = 3.0 <\/span><span style=\"font-size: 1em\">mol<\/span><sup>-2.<\/sup><span style=\"font-size: 1em\">L<\/span><sup>2.<\/sup><span style=\"font-size: 1em\">s<\/span><sup>-1<\/sup><\/div>\n<div data-type=\"equation\"><\/div>\n<p id=\"fs-idm209905104\"><strong>Check Your Learning:<\/strong><\/p>\n<p>Use the provided initial rate data to derive the rate law for the reaction whose equation is:<\/p>\n<div id=\"fs-idm205740176\" style=\"text-align: center\" data-type=\"equation\">OCl<sup>&#8211;<\/sup>(<em>aq<\/em>) + I<sup>&#8211;<\/sup>(<em>aq<\/em>) \u27f6 OI<sup>&#8211;<\/sup>(<em>aq<\/em>) + Cl<sup>&#8211;<\/sup>(<em>aq<\/em>)<\/div>\n<table id=\"fs-idm275492048\" class=\"medium unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row and it labels each column, \u201cTrial,\u201d \u201c[ O C l superscript negative sign ] ( mol \/ L),\u201d \u201c[ I superscript negative sign ] ( mol \/ l ),\u201d and \u201cInitial Rate ( mol \/ L \/ s).\u201d Under the column \u201cTrial\u201d are the numbers: 1, 2, and 3. Under the column \u201c[ O C l superscript negative sign ] ( mol \/ L)\u201d are the numbers 0.0040, 0.0020, and 0.0020. Under the column \u201c[ I superscript negative sign ] ( mol \/ l )\u201d are the numbers: 0.0020, 0.0040, and 0.0020. Under the column \u201cInitial Rate ( mol \/ L \/ s)\u201d are the numbers: 0.00184, 0.00092, and 0.00046.\" data-label=\"\">\n<thead>\n<tr valign=\"middle\">\n<th data-align=\"left\">Trial<\/th>\n<th data-align=\"left\">[OCl<sup>\u2212<\/sup>] (mol\/L)<\/th>\n<th data-align=\"left\">[I<sup>\u2212<\/sup>] (mol\/L)<\/th>\n<th data-align=\"left\">Initial Rate (mol<sup>.<\/sup>L<sup>-1.<\/sup>s<sup>-1<\/sup>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td data-align=\"left\">1<\/td>\n<td data-align=\"left\">0.0040<\/td>\n<td data-align=\"left\">0.0020<\/td>\n<td data-align=\"left\">0.00184<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">2<\/td>\n<td data-align=\"left\">0.0020<\/td>\n<td data-align=\"left\">0.0040<\/td>\n<td data-align=\"left\">0.00092<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td data-align=\"left\">3<\/td>\n<td data-align=\"left\">0.0020<\/td>\n<td data-align=\"left\">0.0020<\/td>\n<td data-align=\"left\">0.00046<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idm296296016\">Determine the rate law expression and the value of the rate constant <em data-effect=\"italics\">k<\/em> with appropriate units for this reaction.<\/p>\n<div id=\"fs-idm214856768\" data-type=\"note\">\n<div data-type=\"title\"><strong>Answer:<\/strong><\/div>\n<p id=\"fs-idm19062704\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1725\" src=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3k-300x206.png\" alt=\"\" width=\"487\" height=\"334\" srcset=\"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3k-300x206.png 300w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3k-768x528.png 768w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3k-65x45.png 65w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3k-225x155.png 225w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3k-350x240.png 350w, https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-content\/uploads\/sites\/1463\/2021\/07\/12.3k.png 933w\" sizes=\"auto, (max-width: 487px) 100vw, 487px\" \/><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm197875360\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Reaction Order and Rate Constant Units<\/strong><\/h3>\n<p id=\"fs-idp9200128\">In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.<\/p>\n<p id=\"fs-idm178764304\">Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:<\/p>\n<div id=\"fs-idm213140144\" data-type=\"equation\">NO<sub>2<\/sub> + CO \u27f6 NO + CO<sub>2\u00a0<\/sub> \u00a0 \u00a0 \u00a0 \u00a0rate = <em>k<\/em>[NO<sub>2<\/sub>]<sup>2 <\/sup><\/div>\n<div data-type=\"equation\">CH<sub>3<\/sub>CHO \u27f6 CH<sub>4<\/sub> + CO\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0rate = <em>k<\/em>[CH<sub>3<\/sub>CHO]<sup>2<\/sup><\/div>\n<div data-type=\"equation\">2N<sub>2<\/sub>O<sub>5<\/sub> \u27f6 NO<sub>2<\/sub> + O<sub>2<\/sub>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 rate = <em>k<\/em>[N<sub>2<\/sub>O<sub>5<\/sub>]<\/div>\n<div data-type=\"equation\">2NO<sub>2<\/sub> + F<sub>2<\/sub> \u27f6 2NO<sub>2<\/sub>F\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 rate = <em>k<\/em>[NO<sub>2<\/sub>][F<sub>2<\/sub>]<\/div>\n<div data-type=\"equation\">2NO<sub>2<\/sub>Cl \u27f62NO<sub>2<\/sub> + Cl<sub>2\u00a0<\/sub>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0rate = <em>k<\/em>[NO<sub>2<\/sub>Cl]<\/div>\n<p id=\"fs-idm284144576\">It is important to note that <em data-effect=\"italics\">rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.<\/em><\/p>\n<\/div>\n<div id=\"fs-idm168364064\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\"><strong>Key Concepts and Summary<\/strong><\/h3>\n<p id=\"fs-idm262547408\">Rate laws (<em data-effect=\"italics\">differential rate laws<\/em>) provide a mathematical description of how changes in the concentration of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the concentration of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.<\/p>\n<\/div>\n<div id=\"fs-idm273385984\" class=\"exercises\" data-depth=\"1\">\n<div id=\"fs-idm237267424\" data-type=\"exercise\">\n<div id=\"fs-idm220253760\" data-type=\"solution\">\n<p id=\"fs-idm220253504\">\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-idm176826832\">\n<dt>method of initial rates<\/dt>\n<dd id=\"fs-idm176826448\">common experimental approach to determining rate laws that involves measuring reaction rates at varying initial reactant concentrations<\/dd>\n<\/dl>\n<dl id=\"fs-idm276164832\">\n<dt>overall reaction order<\/dt>\n<dd id=\"fs-idm263526528\">sum of the reaction orders for each substance represented in the rate law<\/dd>\n<\/dl>\n<dl id=\"fs-idm185628880\">\n<dt>rate constant (<em data-effect=\"italics\">k<\/em>)<\/dt>\n<dd id=\"fs-idm213521696\">proportionality constant in a rate law<\/dd>\n<\/dl>\n<dl id=\"fs-idm215968704\">\n<dt>rate law<\/dt>\n<dd id=\"fs-idm215968320\">(also, rate equation) (also, differential rate laws) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants<\/dd>\n<\/dl>\n<dl id=\"fs-idm333854336\">\n<dt>reaction order<\/dt>\n<dd id=\"fs-idm47186480\">value of an exponent in a rate law (for example, zero order for 0, first order for 1, second order for 2, and so on)<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":1392,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-707","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":695,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/707","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":10,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/707\/revisions"}],"predecessor-version":[{"id":2161,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapters\/707\/revisions\/2161"}],"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\/707\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/media?parent=707"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/pressbooks\/v2\/chapter-type?post=707"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/contributor?post=707"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/aperrott\/wp-json\/wp\/v2\/license?post=707"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}