{"id":387,"date":"2017-08-21T06:46:53","date_gmt":"2017-08-21T06:46:53","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=387"},"modified":"2017-09-19T19:33:10","modified_gmt":"2017-09-19T19:33:10","slug":"kinetics","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/kinetics\/","title":{"rendered":"Kinetics"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/chemical-calculations-and-yields\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/oxidation-and-reduction\">Next Lesson\u00a0\u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Reactions and Reactivity: Kinetics<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will discuss chemical kinetics, which is the study of rates of reactions.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Balanced equations can be used to find the amount of product made from a given amount of reactant.<\/li>\n<li>The coefficients in a balanced equation represent the stoichiometric ratio of moles of the reactants and the\u00a0products.<\/li>\n<li>Using mole ratios and molar masses, the mass of reactants consumed and products produced can be predicted.<\/li>\n<li>The difference between theoretical and experimental yields is used to\u00a0calculate percent yield, which gives a measure of the efficiency of a\u00a0chemical reaction under given conditions.<\/li>\n<li>Limiting reactants determine the maximum amount of product that can be made during a reaction.<\/li>\n<\/ul>\n<section>\n<h3>Definition of Rate and the Rate Equation<\/h3>\n<p><abbr title=\"Reaction rate: The change in concentration of a substance divided by the change in time\">Rate<\/abbr> is a term used frequently in physical science, especially in physics, where rate of displacement or velocity\u00a0and rate\u00a0of change of velocity or acceleration are well recognized. The term <abbr title=\"The study of rates of chemical reactions\">kinetics<\/abbr> is used to describe the study of rates of chemical reactions.<\/p>\n<p>The\u00a0term \u201crate\u201d implies how one quantity changes with\u00a0respect to time. In velocity, it is a displacement in meters or\u00a0kilometers per\u00a0unit time of seconds or hours. In acceleration, the velocity changes\u00a0per unit\u00a0time, having units of\u00a0 meter\/second\/second. In chemical reactions the\u00a0quantity\u00a0which changes per unit time is the concentration of a reactant or\u00a0product. The\u00a0concentration of a reactant will decrease as it is consumed over time;\u00a0the\u00a0concentration of a product will increase as it is produced over time.\u00a0The units\u00a0of rates of reaction are therefore moles\/liter\/second or M\/s, where M represents moles per liter. Brackets, such as [Cl], are sometimes used to show units of molar concentration.<\/p>\n<p>The diagram below shows the\u00a0decomposition reaction of gaseous\u00a0N<sub>2<\/sub>O<sub>5<\/sub> into its elements as\u00a0represented by the equation<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_001.gif\" width=\"153\" height=\"24\" \/><\/center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/kinetics3.concentration_time_graph.gif\" alt=\"decomposition reaction of gaseous N2O5 into its elements\" width=\"350\" height=\"350\" \/><\/p>\n<p>From\u00a0the slope of the reactant line on the\u00a0diagram, one can see that the rate of change of the reactant is not\u00a0constant.\u00a0The slope is steepest at the beginning of the reaction and\u00a0then\u00a0levels off. The rate is fastest when the concentration of the reactant\u00a0is the\u00a0highest. The rate of the reaction can be defined then as the change in\u00a0molar concentration\u00a0of reactant [<em>R<\/em>] per unit time.\u00a0Or\u00a0<img decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_002.gif\" \/>.<\/p>\n<p>However, the concentration is decreasing and\u00a0rates are always considered positive, so a negative sign is used to show that the concentration of the\u00a0reactant\u00a0has decreased.<\/p>\n<p><center><img decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_003.gif\" \/><\/center>For\u00a0the product, the rate\u00a0can be described by:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_004.gif\" width=\"92\" height=\"41\" \/><\/center>Relationships\u00a0between\u00a0reacting quantities are\u00a0stoichiometric in nature. For example, in the graph above for <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_005.gif\" width=\"153\" height=\"24\" \/>\u00a0the rate of appearance of oxygen would be one-fourth the\u00a0rate of appearance of nitrogen dioxide. The difference in rates is\u00a0clearly shown on the graph, by the differences in the slopes of the\u00a0concentration\u00a0lines.<\/p>\n<p>Thus in the reaction of <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_006.gif\" width=\"129\" height=\"24\" \/>, the rate of disappearance of <em>Y<\/em><sub>2<\/sub> would be one-half the rate of disappearance of\u00a0<em>X<\/em><sub>2<\/sub>.\u00a0The rate of formation of product would be equal to the rate of\u00a0disappearance of <em>X<\/em><sub>2<\/sub>, but twice that of\u00a0the rate of disappearance of <em>Y<\/em><sub>2<\/sub>.\u00a0A\u00a0quantitative\u00a0expression called the rate equation can be written to\u00a0show\u00a0that the rate is proportional to the concentration of the\u00a0reactants and products.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_007.gif\" width=\"86\" height=\"24\" \/><\/center>In this example <em>x<\/em> represents some\u00a0experimentally determined exponent and <em>k<\/em> is the proportionality constant.<\/p>\n<h3>Determining the Rate: the Mathematics of Kinetics<\/h3>\n<p>There are two types of rate\u00a0laws:<\/p>\n<ul>\n<li>The\u00a0differential\u00a0rate law, or simply the rate law, shows how\u00a0the rate\u00a0depends on the concentration of each reactant.<\/li>\n<li>The <abbr title=\"The equation that relates the concentration and time; It is derived from the rate law using integral calculus.\">integrated rate law<\/abbr> shows how the concentration of a substance depends\u00a0upon time.<\/li>\n<\/ul>\n<p>Consider the example\u00a0showing the\u00a0decomposition of N<sub>2<\/sub>O<sub>5<\/sub>. Imagine the following rate data was obtained for different initial\u00a0concentrations of N<sub>2<\/sub>O<sub>5<\/sub>.<\/p>\n<table class=\"gas_law_table\">\n<thead>\n<tr>\n<td valign=\"top\">\n<div align=\"center\"><strong>Initial Concentration N<sub>2<\/sub>O<sub>5<\/sub> (mol\/L) <\/strong><\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\"><strong>Initial Rate of Reaction (mol\/L\/s) <\/strong><\/div>\n<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"bottom\">\n<td>\n<div align=\"center\">0.84 M<\/div>\n<\/td>\n<td>\n<div align=\"center\">3.6 <em>\u00d7<\/em> 10<sup>-4<\/sup><\/div>\n<\/td>\n<\/tr>\n<tr valign=\"bottom\">\n<td>\n<div align=\"center\">0.42 M<\/div>\n<\/td>\n<td>\n<div align=\"center\">1.8 \u00d7 10<sup>-4<\/sup><\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>These results show\u00a0that when the\u00a0concentration of N<sub>2<\/sub>O<sub>5\u00a0<\/sub>is halved, the rate is also halved. This means that the reaction is\u00a0first order\u00a0with respect to the N<sub>2<\/sub>O<sub>5<\/sub> and the rate equation can be\u00a0written as<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_008.gif\" width=\"110\" height=\"25\" \/> or simply <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_009.gif\" width=\"105\" height=\"24\" \/><\/p>\n<p>Let&#8217;s look at an example of how to determine the rate equation when given the experimental data.<\/p>\n<h4>Sample Problem<\/h4>\n<p>The following initial\u00a0rates\u00a0of reaction were measured for\u00a0the concentrations of reactants shown in the table for the reaction of\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_010.gif\" width=\"104\" height=\"18\" \/>. Determine the rate\u00a0equation.<\/p>\n<table class=\"gas_law_table\">\n<thead>\n<tr valign=\"bottom\">\n<td>\n<div align=\"center\"><strong>Experiment number <\/strong><\/div>\n<\/td>\n<td>\n<div align=\"center\"><strong>Initial Conc. <em>A<\/em><\/strong><\/div>\n<\/td>\n<td>\n<div align=\"center\"><strong>Initial Conc. <em>B<\/em><\/strong><\/div>\n<\/td>\n<td>\n<div align=\"center\"><strong>Rate, mol\/L\/s <\/strong><\/div>\n<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"bottom\">\n<td>\n<div align=\"center\">1<\/div>\n<\/td>\n<td>\n<div align=\"center\">0.100 M<\/div>\n<\/td>\n<td>\n<div align=\"center\">0.005 M<\/div>\n<\/td>\n<td>\n<div align=\"center\">1.35 \u00d7 10<sup>-7<\/sup><\/div>\n<\/td>\n<\/tr>\n<tr valign=\"bottom\">\n<td>\n<div align=\"center\">2<\/div>\n<\/td>\n<td>\n<div align=\"center\">0.100 M<\/div>\n<\/td>\n<td>\n<div align=\"center\">0.010 M<\/div>\n<\/td>\n<td>\n<div align=\"center\">2.70 \u00d7 10<sup>-7<\/sup><\/div>\n<\/td>\n<\/tr>\n<tr valign=\"bottom\">\n<td>\n<div align=\"center\">3<\/div>\n<\/td>\n<td>\n<div align=\"center\">0.200 M<\/div>\n<\/td>\n<td>\n<div align=\"center\">0.010 M<\/div>\n<\/td>\n<td>\n<div align=\"center\">1.08 \u00d7 10<sup>-6<\/sup><\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The general form of the equation\u00a0would be <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_011.gif\" width=\"117\" height=\"24\" \/>. The data will be used to\u00a0determine\u00a0the value of the\u00a0exponents. Considering experiments 1 and 2, you can see that keeping\u00a0the concentration of A constant, and doubling the concentration of B,\u00a0causes the rate to increase by a factor of two (the same as the\u00a0increase in the concentration of B). Thus, the order of the reaction\u00a0with respect to B is first order and the value of <em>m<\/em> would be 1.<\/p>\n<p>If\u00a0you consider the experiments 2 and 3 where the concentration of B is\u00a0held constant, and the concentration of A is doubled, the rate is\u00a0increased by a factor of 4;<img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_012.gif\" width=\"106\" height=\"44\" \/>. This means that the rate depends on the square of the concentration of A, and the exponent n has a value of 2.<\/p>\n<p>The rate equation can be written as <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_013.gif\" width=\"106\" height=\"24\" \/>.\u00a0The\u00a0reaction is second order with\u00a0respect to reactant A and first order with respect to reactant B; the\u00a0overall order of the reaction is third order which is the sum of the\u00a0exponents.<\/p>\n<p>We can also use this data to determine the value of the rate constant <em>k<\/em>. We can take the\u00a0experimental data for any of the experiments and substitute it into the\u00a0rate equation. In this case:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_014.gif\" width=\"336\" height=\"24\" align=\"absmiddle\" \/><\/center><\/p>\n<p style=\"text-align: center;\">k = 0.0027 L<sup>2<\/sup>\/mol<sup>2<\/sup>s<\/p>\n<h3>Use of Equations<\/h3>\n<p>To use the integrated form of the rate equation, one must\u00a0plot the concentration of the reactant of interest as a function of time.\u00a0Unless the reaction is zero-order, the graph will be a curve. The rate of the\u00a0reaction can be determined by the slope of the line, but to determine the rate\u00a0equation more calculations must be done. Using calculus, it can be shown that\u00a0the equation relating the concentration as a function of time for a first order\u00a0reaction is:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_016.gif\" width=\"132\" height=\"24\" \/><\/center>[<em>A<\/em>] is\u00a0the concentration at time <em>t<\/em>, and [<em>A<\/em>]<sub>o<\/sub> is the initial\u00a0concentration of reactant. A plot of such data will be a straight line of the\u00a0form <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_017.gif\" width=\"78\" height=\"21\" \/> with ln[<em>A<\/em>]\u00a0corresponding to <em>y<\/em>; ln[<em>A<\/em>]<sub>o<\/sub> is\u00a0the <em>y<\/em>-intercept and \u2013<em>k<\/em> the slope of the line. All first order\u00a0kinetics use this law. If experimental data is plotted using the natural log\u00a0of concentration as a function of time and a straight line is obtained, the\u00a0reaction is first order with respect to that reaction. Such a plot is shown in\u00a0the graph.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/kinetics5.nat_log_concentration_time_graph.gif\" alt=\"graph of a first order plot\" width=\"350\" height=\"350\" \/><\/center><\/p>\n<h3>Calculating Half-life<\/h3>\n<p>The <abbr title=\"The time required for the concentration of a reactant to decrease to one-half its initial value\">half-life<\/abbr> of any reaction is the time it takes\u00a0for the concentration of the reactant to become 50% of the original\u00a0amount. In\u00a0other words, the ratio of [<em>A<\/em>] to [<em>A<\/em>]<sub>0<\/sub> is equal to\u00a00.50.<\/p>\n<p>Substituting the values of [<em>A<\/em>] = 0.5[<em>A<\/em>]<sub>0<\/sub> in the integrated rate equation,\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_018.gif\" width=\"132\" height=\"24\" \/>,\u00a0and rearranging gives:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_019.gif\" width=\"89\" height=\"18\" \/> where\u00a0<em>t\u00a0<\/em>= the half-life, <em>t<\/em><sub>\u00bd<\/sub>.<\/p>\n<p>Because ln0.50 = 0.693, for any first order reaction&lt;\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_020.gif\" width=\"80\" height=\"41\" \/>;\u00a0the half-life is independent of the\u00a0concentration of starting material.<\/p>\n<p>For\u00a0second order reactions, the integrated rate\u00a0law takes the form of <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_021.gif\" width=\"104\" height=\"45\" \/>. A\u00a0plot of <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_022.gif\" width=\"26\" height=\"44\" \/> as a function of time\u00a0will produce a straight-line with a slope equal to <em>k<\/em>.\u00a0Calculations show that the equation for the half-life is\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_023.gif\" width=\"77\" height=\"45\" \/>.<\/p>\n<p>The table\u00a0below shows the\u00a0characteristics of three\u00a0different orders of reactions for the integrated form.<\/p>\n<table class=\"gas_law_table\">\n<thead>\n<tr>\n<td valign=\"top\">\n<div align=\"center\"><strong>Order <\/strong><\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\"><strong>Rate expression <\/strong><\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\"><strong>Equation <\/strong><\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\"><strong>Half-life <\/strong><\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\"><strong>Linear plot <\/strong><\/div>\n<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"middle\">\n<td height=\"51\">\n<div align=\"center\">0<\/div>\n<\/td>\n<td>\n<div align=\"center\">Rate = <em>k<\/em><\/div>\n<\/td>\n<td>\n<div align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_024.gif\" width=\"108\" height=\"24\" \/><\/div>\n<\/td>\n<td>\n<div align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_025.gif\" width=\"36\" height=\"42\" \/><\/div>\n<\/td>\n<td>\n<div align=\"center\">[<em>A<\/em>] vs. <em>t<\/em><\/div>\n<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td height=\"49\">\n<div align=\"center\">1<\/div>\n<\/td>\n<td>\n<div align=\"center\">Rate =<em>k<\/em>[<em>A<\/em>]<\/div>\n<\/td>\n<td>\n<div align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_027.gif\" width=\"132\" height=\"24\" \/><\/div>\n<\/td>\n<td>\n<div align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_028.gif\" width=\"48\" height=\"41\" \/><\/div>\n<\/td>\n<td>\n<div align=\"center\">ln[<em>A<\/em>] vs <em>t<\/em><\/div>\n<\/td>\n<\/tr>\n<tr valign=\"middle\">\n<td>\n<div align=\"center\">2<\/div>\n<\/td>\n<td>\n<div align=\"center\">Rate = <em>k<\/em>[<em>A<\/em>]<sup>2<\/sup><\/div>\n<\/td>\n<td>\n<div align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_029.gif\" width=\"104\" height=\"45\" \/><\/div>\n<\/td>\n<td>\n<div align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_030.gif\" width=\"45\" height=\"45\" \/><\/div>\n<\/td>\n<td>\n<div align=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"non_block_image no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_031.gif\" width=\"26\" height=\"44\" align=\"absmiddle\" \/>vs.<br \/>\n<em>t<\/em><\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<h4>Sample Problem<\/h4>\n<p>The following data were\u00a0obtained for the decomposition of\u00a0gaseous HI. Determine the order of the reaction, the rate equation, the\u00a0value\u00a0of <em>k<\/em>, and the half-life of the reaction.<\/p>\n<table class=\"gas_law_table\">\n<tbody>\n<tr>\n<td valign=\"top\">\n<div align=\"left\"><strong>Time (hr) <\/strong><\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">0<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">2<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">4<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">6<\/div>\n<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" height=\"26\">\n<div align=\"left\"><strong>[HI] <\/strong><\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">1.00<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">0.50<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">0.33<\/div>\n<\/td>\n<td valign=\"top\">\n<div align=\"center\">0.25<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Three curves must be plotted\u00a0according to the information in the summary chart above. The plot that\u00a0produces a straight line will show the order of the reaction.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/kinetics6.three_curves.gif\" alt=\"Three curves plotted according to the summary chart\" width=\"600\" height=\"300\" \/><\/center>The third plot\u00a0is the straight line, so this must be a second order reaction, and the\u00a0equation would be <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_032.gif\" width=\"90\" height=\"24\" \/><\/p>\n<p>The value of k is equal to the slope of the line:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_033.gif\" width=\"354\" height=\"69\" \/><\/center>Thus, the\u00a0half life of the reaction is<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_035.gif\" width=\"305\" height=\"45\" \/><\/center>Looking at the original graph of\u00a0concentration vs. time will confirm that it takes about 2 hours for the\u00a0concentration to decrease to half the initial concentration.<\/p>\n<h3>Collision Theory<\/h3>\n<p>In order for a reaction to\u00a0occur, the reactant particles\u00a0must collide with the proper energy and proper orientation for the\u00a0formation of\u00a0the <abbr title=\"The chemical species present at the maximum in energy in a reaction coordinate diagram. It is a high energy state and consists of atoms in some intermediate state of bond-breaking and bond-forming between reactants and products.\">transition\u00a0state<\/abbr> complex. The\u00a0minimum energy\u00a0required is called the <abbr title=\"The minimum energy that must be added to the reactants for a reaction to occur. On a reaction coordinate diagram, it is the energy required to go from reactants to the highest point on the energy coordinate.\">activation\u00a0energy<\/abbr>, <em>E<\/em><sub>a<\/sub>. There are three factors which affect the\u00a0rate\u00a0constant:<\/p>\n<ul>\n<li>A <abbr title=\"The term in the Arrhenius equation which accounts for only certain orientations producing effective collisions\">steric\u00a0factor<\/abbr>, <em>p<\/em>, takes into\u00a0account that only certain\u00a0orientations of the species involved will result in effective\u00a0collisions.<\/li>\n<li>The <abbr title=\"The term in the Arrhenius equation which gives the number of molecular collisions in a unit time at unit concentration\">frequency\u00a0factor<\/abbr>, <em>Z<\/em>, gives the\u00a0number of\u00a0collisions which occur in a given period of time.<\/li>\n<li>An energy factor, <em>f<\/em>, accounts for the number of particles\u00a0with the minimum energy necessary for reaction, where <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_037.gif\" width=\"72\" height=\"26\" \/>.<\/li>\n<\/ul>\n<p>The Arrhenius equation relates the factors to the rate constant <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_038.gif\" width=\"84\" height=\"26\" \/>. The steric\u00a0factor and frequency factor are combined in the variable <em>A<\/em>.\u00a0A useful form of the\u00a0equation is obtained by taking the\u00a0natural logarithm of both sides, which yields an equation in the form <em>y = mx + b<\/em>, which can be\u00a0graphed as a\u00a0straight line:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_038.gif\" width=\"84\" height=\"26\" \/><\/center><em>R<\/em> is the\u00a0ideal gas constant and <em>T<\/em> is the\u00a0Kelvin temperature. Because the rate constant is\u00a0temperature dependent, if ln<em>k<\/em> is plotted as a function of 1\/<em>T<\/em>, the slope of the\u00a0line will be <em>E<\/em><sub>a<\/sub>\/<em>R<\/em>.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/kinetics9.frequency_1nk.gif\" alt=\"placeholder.png\" width=\"350\" height=\"350\" \/><\/center>Increasing the temperature increases both the number of collisions and the energy of collisions.<\/p>\n<h3>Factors Affecting Rates<\/h3>\n<p>The\u00a0Collision\u00a0Theory assumes that in order for a reaction to\u00a0occur the reacting\u00a0particles must collide with the proper orientation and proper energy.\u00a0Any\u00a0factor that enhances the ability of the particles to interact will\u00a0necessarily\u00a0affect the rate. The five factors normally considered are:<\/p>\n<ul>\n<li>Temperature<\/li>\n<li>Concentration\u00a0of reactants<\/li>\n<li>Nature\u00a0of reactants<\/li>\n<li>Surface\u00a0area<\/li>\n<li>Presence\u00a0of a catalyst<\/li>\n<\/ul>\n<h3>Temperature<\/h3>\n<p>As we have seen, temperature affects the\u00a0rate of a reaction. According to the kinetic molecular theory, molecules are in constant\u00a0motion.\u00a0The higher the temperature, the higher the kinetic energy due to an\u00a0increase in\u00a0velocity of the molecules. In order for a reaction to occur, the\u00a0reactant\u00a0particles must collide with\u00a0sufficient\u00a0energy and the proper orientation\u00a0for the\u00a0formation of the transition\u00a0state\u00a0complex. Increasing the temperature increases both the number of\u00a0collisions and\u00a0the energy of collisions. An approximate\u00a0rule of\u00a0thumb\u00a0(which is derived using the\u00a0Arrhenius\u00a0equation) is that an increase of 10<sup>o <\/sup>C doubles\u00a0the rate of a\u00a0reaction. For example, if the rate of a reaction at 0<sup>o <\/sup>C\u00a0is 5.0\u00a0mol\/L-min,\u00a0we can expect that\u00a0increasing the temperature to 10<sup>o <\/sup>C\u00a0would make the rate\u00a010 mol\/L-min.<\/p>\n<h3>Concentration<\/h3>\n<p>The concentration of a\u00a0reactant is the relative amount of\u00a0reactant per unit volume. Remember, the more collisions that occur, the greater the chance that a reaction will have\u00a0the proper orientation.\u00a0Thus, higher concentrations of reactant molecules result in a greater rate of collisions between reactant molecules,\u00a0and a faster rate of reaction. Just\u00a0how\u00a0much the rate is increased for a particular reaction must be determined\u00a0experimentally.<\/p>\n<h3>Nature of Reactants<\/h3>\n<p>The nature of the substances\u00a0which are reacting does\u00a0affect the rate. In general, ionic reactions occur faster than\u00a0molecular ones\u00a0in solution. The latter require the breaking of covalent bonds and the\u00a0formation of new ones. For example, a weak acid such as acetic\u00a0acid\u00a0reacts much more slowly with zinc metal than does the strong acid,\u00a0like hydrochloric\u00a0acid, of the same concentration. The difference is in the ability of the\u00a0hydrochloric\u00a0acid to ionize more completely than the acetic acid. Another example\u00a0would be\u00a0the different reaction rates of metals when they react with dilute hydrochloric acid.\u00a0Magnesium\u00a0metal reacts much more rapidly with acids than does lead.<\/p>\n<h3>Surface Area<\/h3>\n<p>The ability of reactants to\u00a0meet affects the rate of a\u00a0reaction. One way to enhance the opportunity for the reactants to come\u00a0into\u00a0contact with each other is to increase their surface area. Consider\u00a0attempting to\u00a0burn a charcoal briquette using the flame from a candle. The rate of\u00a0combustion\u00a0is very slow, because only the charcoal particles on the outside\u00a0surface of the\u00a0briquette are in contact with the oxygen in the air. If the briquette\u00a0is ground\u00a0up and blown into the candle flame the reaction rate is very rapid, in\u00a0fact it\u00a0is explosive. Grain elevator explosions are another impressive example of such surface area phenomena. These\u00a0explosions\u00a0occur, not in full grain elevators, but\u00a0in ones\u00a0that are almost empty of grain, but full of grain dust. A\u00a0spark\u00a0from a light switch can provide enough energy to start a reaction which\u00a0can\u00a0bring down the elevator.<\/p>\n<h3>Presence of a Catalyst<\/h3>\n<p>A <abbr title=\"A substance that speeds up a reaction without being permanently changed in the process\">catalyst<\/abbr> is a substance which alters the rate of\u00a0a chemical reaction without itself being consumed. Since the activation energy is the minimum\u00a0amount of energy required for a reaction to occur, if the particles do\u00a0not have\u00a0that minimum energy, no reaction occurs. Catalysts provide a different reaction pathway with a lower\u00a0activation\u00a0energy. Frequently the reaction mechanism for catalyzed reactions has\u00a0multiple\u00a0steps, each with a lower activation energy than the uncatalyzed\u00a0reaction.<br \/>\nCatalysts may be <abbr title=\"A catalyst that is in the same phase as the reacting substances \">homogeneous <\/abbr>(same\u00a0phase) or\u00a0<abbr title=\"A catalyst that is in a different phase from the reacting substances\">heterogeneous<\/abbr>(different phase) with the reactants. Enzymes are examples of\u00a0biological\u00a0catalysts which allow reactions in the body to occur at much lower\u00a0temperatures\u00a0than would be required in a test tube.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/kinetics8.Catalyst_effect.png\" alt=\"Effect of a catalyst on a reaction \" width=\"292\" height=\"243\" \/><\/center><\/p>\n<p class=\"figcaption\">Effect of a catalyst on a reaction<\/p>\n<p>Other substances called inhibitors also alter the rate of\u00a0reactions by slowing them. For example, some preservatives that are found in\u00a0commercial food\u00a0products are inhibitors. Anti-depressant medications selectively\u00a0inhibit\u00a0chemical reactions in the body. The action of inhibitors is complex, but inhibitors work by reducing the\u00a0effectiveness of a catalyst. It is\u00a0generally thought that they offer alternate reaction pathways that result in other\u00a0products.<\/p>\n<h3>Reaction Mechanisms<\/h3>\n<p>A <abbr title=\"Sequence of bond forming and bond breaking steps that occurs during the conversion of reactants to products in a chemical reaction\">reaction\u00a0mechanism<\/abbr> is a proposed series of steps by which a reaction\u00a0is thought to\u00a0proceed. Each step in a mechanism is considered as an elementary step\u00a0which may\u00a0involve one particle\u00a0(a unimolecular step), two\u00a0particles (a bimolecular step),\u00a0or in rare cases, three particles\u00a0(a termolecular step). These steps have\u00a0corresponding\u00a0rate equations as shown below, in which the rate equation is equal to\u00a0the\u00a0constant multiplied by the molar concentration of each reactant.<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>A \u2192 B + C <\/em>with rate = <em>k<\/em>[<em>A<\/em>]<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>A + B \u2192 C + D <\/em> with rate =\u00a0<em>k<\/em>[<em>A<\/em>][<em>B<\/em>]<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>A + B + C \u2192 D + E <\/em> with rate = <em>k<\/em>[<em>A<\/em>][<em>B<\/em>][<em>C<\/em>]<\/p>\n<p>When a mechanism is\u00a0proposed, two requirements must be\u00a0satisfied:<\/p>\n<ul>\n<li>The\u00a0sum of the elementary steps must give the overall balanced equation for\u00a0the\u00a0reaction.<\/li>\n<li>The\u00a0mechanism must agree with the experimentally determined rate law.<\/li>\n<\/ul>\n<p>The slowest step in a\u00a0mechanism is called the <abbr title=\"The slowest step in a reaction mechanism\">rate-determining\u00a0step<\/abbr>. The overall reaction can only proceed as fast as the\u00a0slowest step. As\u00a0an example, consider the formation of a compound by the reaction of\u00a0nitrogen\u00a0dioxide with fluorine.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_046.gif\" width=\"209\" height=\"24\" \/> with rate = <em>k<\/em>[NO<sub>2<\/sub>][F<sub>2<\/sub>] determined by experiment.<\/p>\n<p class=\"center\">A proposed mechanism is <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_048.gif\" width=\"178\" height=\"25\" \/> <em>slow<\/em><\/p>\n<p class=\"center\">With a second step of <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_049.gif\" width=\"150\" height=\"25\" \/> <em>fast<\/em><\/p>\n<p>The first requirement is met\u00a0because the algebraic sum of\u00a0the two steps is the net reaction. The second requirement is also met\u00a0because\u00a0the rate of the overall reaction is that of the rate-determining step\u00a0which is\u00a0bimolecular with rate = <em>k<\/em><sub>1<\/sub>[NO<sub>2<\/sub>][F<sub>2<\/sub>].\u00a0The proposed mechanism is acceptable because it meets both\u00a0requirements, but it is not necessarily the <em>correct\u00a0<\/em>mechanism. Chemists\u00a0need to perform many additional experiments to decide the most probable\u00a0mechanisms.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>The mechanism for the reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is shown below. What is the overall balanced equation and what is the expected rate law?<\/p>\n<p class=\"center\">NO<sub>2 <\/sub>+ NO<sub>2<\/sub> \u2192 NO<sub>3<\/sub>+ NO <em>slow<br \/>\n<\/em>NO<sub>3<\/sub> + CO \u2192 NO<sub>2<\/sub>+ CO<sub>2<\/sub> <em>fast<\/em><\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The overall balanced equation will be the algebraic sum of the two elementary\u00a0steps which is<br \/>\n<span class=\"center\">NO<sub>3<\/sub> + CO \u2192 NO<sub>2<\/sub>+ CO<sub>2<\/sub><\/span><br \/>\nThe rate law would be for the slowest step which is second order in NO<sub>2<\/sub> and zero order in CO, so it would be rate = <em>k<\/em><sub>1<\/sub>[NO<sub>2<\/sub>]<sup>2<\/sup><\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>A certain compound decomposes in air. The rate constant for the first order reaction is 0.423 h<sup>-1<\/sup>.<\/p>\n<p>What is the half-life of the compound?<\/p>\n<p>If a 0.050 M sample is opened, what will be the concentration after 8.0 hours?<\/p>\n<p>How long will it take for the concentration to drop to 0.0030M?<\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">For first\u00a0order kinetics the integrated rate equation takes the form<img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_077.gif\" width=\"132\" height=\"24\" \/>. At the\u00a0half life the concentration [A] = 0.50[Ao]. Substituting in the\u00a0equation, a general form for all first order reactions is <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_078.gif\" width=\"82\" height=\"41\" \/>. In this\u00a0case, the half-life is 1.6 hours. Using the same integrated rate\u00a0equation, one can substitute in the values for initial concentration of\u00a00.050 M, a time of 8.0 hours for <em>t<\/em>, and determine the concentration of\u00a0the substance to be 0.0017 M after 8.0 hours. To determine how long it\u00a0takes for the concentration to drop to 0.0030 M, the equation is solved\u00a0for time and the values of <em>k<\/em> and the initial concentration are\u00a0substituted. It requires 6.6 hours for the concentration to get to\u00a00.0030 M.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>The decomposition of HX has an activation energy of 192 kJ\/mol. The rate constant at 900\u00b0C is 0.0250\u00a0L\/mol\/h.<\/p>\n<p>What is the rate constant at 800\u00b0C?<\/p>\n<p>And at what temperature will the rate constant be 0.00625 L\/mol\/h?<\/p>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The\u00a0Arrhenius equation can be expressed in the form of the\u00a0Clausius-Clapeyron equation as<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_084.gif\" width=\"268\" height=\"46\" \/>Subtracting the equations to eliminate\u00a0the value of A, gives<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_085.gif\" width=\"141\" height=\"50\" \/>The value for R is 8.31 J\/mol K. The Celsius\u00a0temperatures must be converted to kelvins, and the activation energy\u00a0converted to joules from kilojoules. Substituting in the equation<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_086.gif\" width=\"252\" height=\"45\" \/>Solving for k<sup>2<\/sup>, gives a value of 3.99 \u00d7 10<sup>-3<\/sup> L\/mol\/h at 800\u00b0C. To determine the temperature at which the rate constant is\u00a00.00625 L\/mol\/h, the same equation is used and solved for T<sub>2<\/sub>.<br \/>\n<img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_087.gif\" \/>The temperature is 1262 K or 989\u00b0C.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>The following data were obtained for the reaction shown below at various temperatures. Plot the\u00a0data (ln<em>k<\/em> vs 1\/<em>T<\/em>) and calculate the activation energy.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_088.gif\" width=\"177\" height=\"24\" \/><\/center><\/p>\n<table class=\"gas_law_table\">\n<tbody>\n<tr valign=\"bottom\">\n<td>\n<div align=\"center\"><strong><em>k<\/em> (s<sup>-1<\/sup>)<\/strong><\/div>\n<\/td>\n<td>\n<div align=\"center\">0.048<\/div>\n<\/td>\n<td>\n<div align=\"center\">2.3<\/div>\n<\/td>\n<td>\n<div align=\"center\">49<\/div>\n<\/td>\n<td>\n<div align=\"center\">590<\/div>\n<\/td>\n<\/tr>\n<tr valign=\"bottom\">\n<td>\n<div align=\"center\"><strong><em>T<\/em> (<sup>o<\/sup>C) <\/strong><\/div>\n<\/td>\n<td>\n<div align=\"center\">500<\/div>\n<\/td>\n<td>\n<div align=\"center\">600<\/div>\n<\/td>\n<td>\n<div align=\"center\">700<\/div>\n<\/td>\n<td>\n<div align=\"center\">800<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">Plotting the ln<em>k<\/em> as a function\u00a0of 1\/<em>T<\/em> gives a straight line. The activation energy is equal to\u00a0\u2013R<strong>\u00b7<\/strong>(slope of the line). In this case the slope is -2.6 \u00d7 10<sup>4<\/sup>.\u00a0So calculating the activation energy is\u00a0<span class=\"center\">(8.31 J\/mol K)(2.6 10<sup>4<\/sup>) = 216 kJ\/mol<\/span><\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>The reaction <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s5_061.gif\" width=\"205\" height=\"24\" \/> &gt;is second order in NO\u00a0and first order in bromine gas. The rate of reaction is 1.6\u00d710<sup>-8\u00a0<\/sup>mol\/L\/min when [NO] = 0.030 M and [Br<sub>2<\/sub>] = 0.020 M.<br \/>\nWhat is the value of\u00a0<em>k<\/em>?<\/p>\n<ol>\n<li>1.8\u00d710<sup>-5<\/sup> L<sup>2<\/sup>\/mol<sup>2<\/sup>min<\/li>\n<li>1.2\u00d710<sup>-5<\/sup> mol\/L\/min<\/li>\n<li>2.7\u00d710<sup>-5<\/sup> L<sup>2<\/sup>\/mol<sup>2<\/sup>min<\/li>\n<li>8.9\u00d710<sup>-4 <\/sup>L<sup>2<\/sup>\/mol<sup>2<\/sup>min<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is D. It is correct since the form of the equation is rate = <em>k<\/em>[NO]<sup>2<\/sup>[Br<sub>2<\/sub>].<\/p>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/chemical-calculations-and-yields\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/oxidation-and-reduction\">Next Lesson\u00a0\u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/section>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson\u00a0\u27a1 Reactions and Reactivity: Kinetics Objective In this lesson we will discuss chemical kinetics, which is the study of rates of reactions. Previously we covered&#8230; Balanced equations can be used to find the amount of product made from a given amount of reactant. The coefficients in a balanced equation represent the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-387","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/387","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=387"}],"version-history":[{"count":25,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/387\/revisions"}],"predecessor-version":[{"id":906,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/387\/revisions\/906"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}