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href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../359/kinetics.html">Kinetics</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Activation energy</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 45 minutes"><i class="fa fa-clock-o"></i> 45'</span> </ol> <article id="main-article"> <p> <img alt="" src="../../images/test-images/match.png" style="width: 160px; height: 179px; float: left;">This is a short topic. The Arrhenius equation is given in the data book in all its relevant forms so the important thing to do is to become familiar with the equation and to ensure that you know how to plot a graph using the Arrhenius equation in order to determine activation energy − this occasionally appears on examination papers.</p> <hr class="hidden-separator"> <div class="panel panel-has-colored-body panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <p>There are no additional key terms required. Ensure that you are confident using the terms from the other chemical kinetics topics and familiar with the Arrhenius equation.</p> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Using the Arrhenius equation to find activation energy</p> </div> </div> <div class="panel-body"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/390579864"></iframe></div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <p> The revision cards contain all of the essential content:</p> <div id="carousel-171" class="dynamic-gallery carousel slide" data-id="171"><div class="carousel-inner" role="listbox"><div class="item active"><a class="fancy" href="../../../std-galleries/7-171/screenshot-2020-03-06-at-151156.png" data-fancybox="gallery-171" title="" data-caption=""><img alt="" src="../../../std-galleries/7-171/screenshot-2020-03-06-at-151156.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-171/screenshot-2020-03-06-at-151222.png" data-fancybox="gallery-171" title="" data-caption=""><img alt="" src="../../../std-galleries/7-171/screenshot-2020-03-06-at-151222.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-171/screenshot-2020-03-06-at-151822.png" data-fancybox="gallery-171" title="" data-caption=""><img alt="" src="../../../std-galleries/7-171/screenshot-2020-03-06-at-151822.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-171/screenshot-2020-03-06-at-151348.png" data-fancybox="gallery-171" title="" data-caption=""><img alt="" src="../../../std-galleries/7-171/screenshot-2020-03-06-at-151348.png"></a></div></div><a class="left carousel-control" href="#carousel-171" role="button" data-slide="prev"><i class="fa fa-fw fa-chevron-left"></i></a><a class="right carousel-control" href="#carousel-171" role="button" data-slide="next"><i class="fa fa-fw fa-chevron-right"></i></a></div><ol class="std-carousel-indicators"><li data-index="0"><img title="Click to view" src="../../../std-galleries/7-171/screenshot-2020-03-06-at-151156-thumb128.jpg"><li><li data-index="1"><img title="Click to view" src="../../../std-galleries/7-171/screenshot-2020-03-06-at-151222-thumb128.jpg"><li><li data-index="2"><img title="Click to view" src="../../../std-galleries/7-171/screenshot-2020-03-06-at-151822-thumb128.jpg"><li><li data-index="3"><img title="Click to view" src="../../../std-galleries/7-171/screenshot-2020-03-06-at-151348-thumb128.jpg"><li></li></ol> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test yourself</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <div class="tib-quiz" data-stats="7-417-894"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>The Arrhenius equation is given in the data book:</p><p><span class="math-tex">\(k = Ae^{-E_a \over RT}\)</span></p>What happens to the value of the rate constant (<em>k</em>) if the temperature increases?</div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>The rate constant increases </span></label> </p><p><label class="radio"> <input type="radio"> <span>The rate constant decreases</span></label> </p><p><label class="radio"> <input type="radio"> <span>Unknown</span></label> </p><p><label class="radio"> <input type="radio"> <span>The rate constant stays the same</span></label> </p></div><div class="q-explanation"><p>An increase in temperature will always increase the value of the rate constant (k is only constant at constant temperature). This can be shown by substituting values into the expression:</p><p>Using 40000J mol<sup>−1</sup> for E<sub>a</sub> and temperature of 298K.</p><p><span class="math-tex">\(e^{-E_a \over RT}\)</span> = e<sup>−40000/(8.31×298) </sup>= 9.6 × 10<sup>−8</sup></p><p>Using 40000J mol<sup>−1</sup> for E<sub>a</sub> and temperature of 308K.</p><p><span class="math-tex">\(e^{-E_a \over RT}\)</span> = e<sup>−40000/(8.31×308)</sup>= 1.63 × 10<sup>−</sup><sup>7</sup></p><p>Assuming that E<sub>a </sub>and A remain constant the increase in value of the 'e' expression must lead to an increase in the rate constant, k.</p>Thus the correct answer is 'The rate constant increases'.</div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>The Arrhenius equation is given in the data book:</p><p><span class="math-tex">\(k = Ae^{-E_a \over RT}\)</span></p>What does the pre-exponential factor<em>, A</em>, represent?</div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>The activation energy.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The gas constant.</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>The frequency of collisions that have the correct orientation.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The rate constant.</span></label> </p></div><div class="q-explanation">The correct answer is 'The frequency of collisions that have the correct orientation', and needs to be learned. The activation energy is given by E<sub>a</sub>, the gas constant by R and the rate constant by k.</div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>Taking natural logs of the Arrhenius equation<em> </em>gives the expression below:</p><p><em><span class="math-tex">\(ln k = {-E_a \over RT} + ln A\)</span></em></p><p>This can be used to determine the activation energy for a reaction because:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Plotting a graph of 1/T against ln k gives the intercept on the y-axis equal to E<sub>a</sub>. </span></label> </p><p><label class="radio"> <input type="radio"> <span>Plotting a graph of 1/T against ln k is a linear plot with gradient equal to E<sub>a</sub>/R.</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Plotting a graph of 1/T against ln k is a linear plot with gradient equal to –E<sub>a</sub>/R.</span></label> </p><p><label class="radio"> <input type="radio"> <span>Plotting a graph of temperature against rate is a linear plot with gradient equal to –E<sub>a</sub>.</span></label> </p></div><div class="q-explanation"><p>The equation given in the question can be compared to the equation for a straight line <span class="math-tex">\(y = mx + c\)</span>:</p><p><span class="math-tex">\(ln k = {-E_a \over R}.{1 \over T} + ln A\)</span></p><p>The temperature term has been separated from the activation energy term, and this makes it easier to see that if <em>y = lnk </em>and <em>x = 1/T</em> then the gradient, <em>m, </em>will be <em>–E<sub>a</sub>/R </em>and the intercept, <em>c</em>, will be <em>lnA.</em></p><p>So the correct answer is <em>Plotting a graph of 1/T against ln k is a linear plot with gradient equal to –E<sub>a</sub>/R</em>. The other answers are not correct as the expression has been incorrectly interpreted.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>A student did an experiment to find the activation energy for a reaction that produced a precipitate. The student varied the temperature and measured how long it took for the precipitate to obsure a black cross written on a piece of white paper. The student collected results for 6 different temperatures, and drew a graph to calculate activation energy.</p><p>What should the student plot to find E<sub>a</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span><strong>1/temperature </strong>against <strong>ln 1/time</strong> </span></label> </p><p><label class="radio"> <input type="radio"> <span><strong>temperature </strong>against <strong>ln time</strong></span></label> </p><p><label class="radio"> <input type="radio"> <span><strong>temperature </strong>against <strong>ln 1/time</strong></span></label> </p><p><label class="radio"> <input type="radio"> <span><strong>1/temperature </strong>against <strong>ln time</strong></span></label> </p></div><div class="q-explanation"><p>The Arrhenius equation is given in the data book:</p><p><span class="math-tex">\(lnk = {-E_a \over RT} + lnA\)</span></p><p>and can be compared to the equation for a straight line <span class="math-tex">\(y = mx + c\)</span>:</p><p><span class="math-tex">\(ln k = {-E_a \over R}.{1 \over T} + ln A\)</span></p><p>The temperature term has been separated from the activation energy term, and this makes it easier to see that if <em>y = lnk </em>and <em>x = 1/T</em> then the gradient, <em>m, </em>will be <em>–E<sub>a</sub>/R </em>and the intercept, <em>c</em>, will be <em>lnA.</em></p><p>It is important to remember that the rate constant is proportional to rate, and that rate is propotional to 1/time.</p><p>The expression <em>lnk</em> is the natural log of the rate constant, but because the activation energy is calculated from a <strong>gradient </strong>any value <strong>proportional to</strong> <em>lnk </em>can be plotted.</p><p>So the correct answer is <em><strong>1/temperature</strong> against <strong>ln 1/time</strong></em>. The other answers are not correct as the expression has been incorrectly interpreted, or rate constant has been incorrectly correlated with time (rather than 1/time).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>A student did an experiment to find the activation energy for a reaction that produced a precipitate. The student varied the temperature and measured how long it took for the precipitate to obsure a black cross written on a piece of white paper. The student collected results for 6 different temperatures, and drew a graph of <strong>1/temperature (x axis) </strong>against <strong>ln 1/time (y axis) </strong>to calculate activation energy.</p><p>One result was that the reaction took 46 seconds (to obsure the cross) at a temperature of 40°C.</p><p>What values (2 sig figs) should the student use to plot this point on the graph?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2.5 × 10<sup>–2</sup> (x axis) and –3.8 (y axis)</span></label> </p><p><label class="radio"> <input type="radio"> <span>3.2 × 10<sup>–3</sup> (x axis) and 2.2 × 10<sup>–2</sup> (y axis)</span></label> </p><p><label class="radio"> <input type="radio"> <span>2.5 × 10<sup>–2</sup> (x axis) and 2.2 × 10<sup>–2</sup> (y axis)</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>3.2 × 10<sup>–3</sup> (x axis) and –3.8 (y axis) </span></label> </p></div><div class="q-explanation"><p>For 1/T, remember that temperature must be in <strong>Kelvin</strong>:</p><p>40°C is 40+273=313K</p><p>1/313 = 0.003194888, which is <strong>3.2 × 10<sup>–3</sup></strong><sup> </sup> in standard form (2 sig figs)</p><p>For 1/t, don't forget to take the natural log (ln):</p><p>1/46 = 0.02173913</p><p>and ln 0.02173913 = <strong>–3.8</strong></p><p>Thus <strong>3.2 × 10<sup>–3</sup> (x axis) and –3.8 (y axis)</strong> is the correct answer.</p><p>The incorrect answers: 1/40 = 0.025, which is <strong>2.5 × 10<sup>–3</sup></strong><sup> </sup> in standard form (not converted to <strong>Kelvin</strong>)</p><p>And 1/46 = 0.02173913 which is <strong>2.2 × 10</strong><sup>–2</sup><sup> </sup> in standard form (no<strong> natural log </strong>taken)</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>The Arrhenius equation, in the natural logarithmic form, is given in the data book:</p><p><span class="math-tex">\(lnk = {-E_a \over RT} + lnA\)</span></p><p>In an experiment studying the gas phase reaction of ethene and bromine at a temperature of 299K, the rate constant (<em>k</em>) was found to be 0.128 (mol<sup>–1</sup> dm<sup>3</sup> s<sup>–1</sup>) and the activation energy was found to be 40kJ mol<sup>−1</sup>.</p><p>What is the numerical value of the pre-exponential factor<em>, A</em><em>?</em></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2.04</span></label> </p><p><label class="radio"> <input type="radio"> <span>7.69</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1.26 × 10<sup>6</sup> </span></label> </p><p><label class="radio"> <input type="radio"> <span>14.0</span></label> </p></div><div class="q-explanation"><p>The activation energy must be converted to J (rather than kJ) since the gas constant R (given in the data book) has a value of 8.31 J K<sup>−1</sup> mol<sup>−1</sup>. Ea = 40000 J mol<sup>−1</sup>.</p><p>ln0.128 = −40000/(8.31×299) + ln<em>A</em></p><p>−2.055725015 = −16.09858775 + ln<em>A</em></p><p>lnA = 14.04286274</p><p>A = e<sup>14.04286274</sup> = 1255271.874</p><p>So the correct answer is 1.26 × 10<sup>6</sup> (3 dp)</p><p>Incorrect answers:</p><p>7.69 uses 40 rather than 40000</p><p>14.0 is the natural log of the answer (not converted to A)</p><p>2.04 is the natural log of the answer (not converted to A) using 40 rather than 40000</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Exam-style questions</p> </div> </div> <div class="panel-body"> <div> <h4>Paper 1</h4> <h5>Core (SL&HL): <a href="../2712/kinetics-core-sl-and-hl-paper-1-questions.html" title="Kinetics core (SL and HL) paper 1 questions">Kinetics core (SL and HL) paper 1 questions</a></h5> <h5>AHL (HL only): <a href="../2721/kinetics-ahl-hl-only-paper-1-questions.html" title="Kinetics AHL (HL only) paper 1 questions">Kinetics AHL (HL only) paper 1 questions</a></h5> <h4>Paper 2</h4> <h5>Core (SL&HL): <a href="../2719/kinetics-core-sl-hl-paper-2-questions.html" title="Kinetics core (SL & HL) paper 2 questions">Kinetics core (SL & HL) paper 2 questions</a></h5> <h5>AHL (HL only): <a href="../2720/kinetics-ahl-hl-only-paper-2-questions.html" title="Kinetics AHL (HL only) paper 2 questions">Kinetics AHL (HL only) paper 2 questions</a></h5> </div> </div> <div class="panel-footer"> <div> 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