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btn-default btn-block" style="margin-bottom: 10px"><i class="fa fa-table"></i> Periodic table</button> <div class="hidden-xs hidden-sm"> <button class="btn btn-default btn-block text-xs-center" data-toggle="modal" data-target="#modal-feedback" style="margin-bottom: 10px"><i class="fa fa-send"></i> Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> Kinetics AHL (HL only) paper 1 questions <a href="#" class="mark-page-favorite pull-right" data-pid="2721" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../2075/paper-1-exam-questions.html">Paper 1 Exam Questions</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Kinetics AHL (HL only) paper 1 questions</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 30 minutes"><i class="fa fa-clock-o"></i> 30'</span> </ol> <article id="main-article"> <p><strong>Topics 16.1 and 16.2</strong></p> <p><strong>Paper 1 style questions </strong>are multiple choice. You are <strong>not permitted to use a calculator or the data book</strong> for these questions, but you should use a periodic table.</p> <p>A <strong>periodic table pop-up</strong> is available on the left hand menu.</p> <div class="greenBg"> <div class="tib-quiz" data-stats="7-815-2721"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>What are the units of the rate constant for a third order reaction?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>mol<sup>–3</sup> dm<sup>9</sup> s<sup>–1</sup></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>mol<sup>–2</sup> dm<sup>6</sup> s<sup>–1</sup></span></label></p><p><label class="radio"> <input type="radio"> mol<sup>–1</sup> dm<sup>3</sup> s<sup>–1</sup><span></span></label> </p><p><label class="radio"> <input type="radio"> <span>s<sup>–1</sup></span></label> </p></div><div class="q-explanation"><p>Units of the rate constant can be worked out by cancelling out in the equation in the same way as calculating the numerical value for the rate constant:</p><p>e.g. Rate = k[Y]<sup>3</sup></p><p>k = Rate / [Y]<sup>3</sup></p><p>units = mol dm<sup>–3</sup> s<sup>–1</sup> / (mol dm<sup>–3</sup>)(mol dm<sup>–3</sup>)(mol dm<sup>–3</sup>)</p><p>So cancel one <em>moles per decimetre cubed</em> from top and bottom:</p><p>units = s<sup>–1</sup> / (mol dm<sup>–3</sup>)(mol dm<sup>–3</sup>)</p><p>Mulitiply out brackets</p><p>units = s<sup>–1</sup> / mol<sup>2</sup> dm<sup>−6</sup></p><p>Move units from bottom to top and therefore invert the sign:</p><p><strong>mol<sup>–2</sup> dm<sup>6</sup> s<sup>–1</sup> </strong>is the correct answer.</p><p>Or they can be learnt:</p><table border="0" cellpadding="0" cellspacing="0" width="650"><colgroup><col><col span="4"></colgroup><tbody><tr height="29"><td height="29" style="height:39px;width:265px;"><p style="text-align: center;">Overall order of reaction</p></td><td style="width:151px;"><p style="text-align: center;">1st</p></td><td style="width:151px;"><p style="text-align: center;">2nd</p></td><td style="width:151px;"><p style="text-align: center;">3rd</p></td><td style="width:151px;"><p style="text-align: center;">4th</p></td></tr><tr height="29"><td height="29" style="height:39px;width:265px;"><p style="text-align: center;">Rate constant units</p></td><td style="width:151px;"><p style="text-align: center;">s<sup>–1</sup></p></td><td style="width:151px;"><p style="text-align: center;">mol<sup>–1</sup> dm<sup>3</sup> s<sup>–1</sup></p></td><td style="width:151px;"><p style="text-align: center;">mol<sup>–2</sup> dm<sup>6</sup> s<sup>–1</sup></p></td><td style="width:151px;"><p style="text-align: center;">mol<sup>–3</sup> dm<sup>9 </sup>s<sup>–1</sup></p></td></tr></tbody></table></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>Which gives the value of <span class="math-tex">\( ln A\)</span> when a graph of <span class="math-tex">\( ln k\)</span> (y-axis) is plotted against <span class="math-tex">\(1 \over T\)</span>(x-axis)?</p><p><em><span class="math-tex">\(ln k = {-E_a \over RT} + ln A\)</span></em></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> The p<span>oint on the graph where the curve flattens</span></label></p><p><label class="radio"> <input class="c" type="radio"> Intercept on the y-axis</label></p><p><label class="radio"> <input type="radio"> The gradient of the line</label></p><p><label class="radio"> <input type="radio"> Intercept on the x-axis<span> </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 plot will generate a straight line, and 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 <strong>Intercept on the y-axis.</strong></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">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the rate equation for the reaction mechanism below?</p><p>Step 1 (fast step): 2A<sub>(g)</sub> → B<sub>(g) </sub>+ C<sub>(g)</sub></p><p>Step 2 (slow step): B<sub>(g)</sub> + D<sub>(g)</sub> → E<sub>(g)</sub> + F<sub>(g)</sub></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Rate = k[A]<sup>2</sup></span></label></p><p><label class="radio"> <input type="radio"> <span>Rate = k[A]<sup>2</sup>[B][D]</span></label></p><p><label class="radio"> <input class="c" type="radio"> <span>Rate = k[A]<sup>2</sup>[D]</span></label></p><p><label class="radio"> <input type="radio"> <span>Rate = k[B][D]</span></label></p></div><div class="q-explanation"><p>Only reactants that are <strong>in </strong>the rate determining step – slowest step – (or contribute to <strong>an intermediate that is in</strong> the rate determining step) can appear in the rate equation and therefore have an effect upon rate. One mole of B and D appear in the slow/rate determining step and thus [D] has an order of 1. B, however, is an <strong>intermediate </strong>(since it is made in step 1 and used up in step 2). Two moles of A are used to make the intermediate B, and therefore [A] has an order of 2.</p><p>In a sense, the rate equation that looks like it should be <em>Rate = k[B][D]</em> becomes Rate = k[A]<sup>2</sup>[D], since intermediates cannot appear in the rate equation.</p><p><strong>Rate = k[A]<sup>2</sup>[D]</strong> is the correct answer.</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>The following data was obtained for the chemical reaction: X + Y → <em>Products</em></p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td style="text-align: center;">[X] (mol dm<sup>–3</sup>)</td><td style="text-align: center;">[Y] (mol dm<sup>–3 </sup>)</td><td style="text-align: center;">Initial Rate (mol dm<sup>–3 </sup>s<sup>–1</sup>)</td></tr><tr><td style="text-align: center;">0.20</td><td style="text-align: center;">0.10</td><td style="text-align: center;">2.0 × 10<sup>–3</sup></td></tr><tr><td style="text-align: center;">0.20</td><td style="text-align: center;">0.30</td><td style="text-align: center;">1.8 × 10<sup>–2</sup></td></tr><tr><td style="text-align: center;">0.40</td><td style="text-align: center;">0.10</td><td style="text-align: center;">4.0 × 10<sup>–3</sup></td></tr></tbody></table><p>What is the order of reaction with respect to [X] and the order of reaction with respect to [Y]?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>1st with respect to [X] and 2nd with respect to [Y]</span></label></p><p><label class="radio"> <input type="radio"> <span>1st with respect to [X] and zero with respect to [Y]</span></label> </p><p><label class="radio"> <input type="radio"> <span>2nd with respect to [X] and 2nd with respect to [Y]</span></label> </p><p><label class="radio"> <input type="radio"> 2nd with respect to [X] and 1st with respect to [Y]<span></span></label> </p></div><div class="q-explanation"><p>The correct answer is <strong>1st with respect to [X] and 2nd with respect to [Y].</strong></p><p>When the concentration of Y is tripled (experiments 1 to 2) the rate increases nine fold (0.002 to 0.018 – make sure you understand <em>standard form</em>; to a positive power of ten moves the decimal place that many places to the right, negative power to the left). [Y] × 3 causes rate × 9, and 3<sup>2</sup>=9, so [Y] is second order.</p><p>When the concentration of X is doubled (experiments 1 to 3) the rate also doubles (0.002 to 0.004). [X] × 2 causes rate × 2, and 2<sup>1</sup>=2, so [X] is first order.</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>The following data was obtained for the chemical reaction: Y + Z → <em>Products</em></p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td style="text-align: center;">[Y] (mol dm<sup>–3</sup>)</td><td style="text-align: center;">[Z] (mol dm<sup>–3 </sup>)</td><td style="text-align: center;">Initial Rate (mol dm<sup>–3 </sup>s<sup>–1</sup> ×10<sup>−2</sup>)</td></tr><tr><td style="text-align: center;">0.20</td><td style="text-align: center;">0.20</td><td style="text-align: center;">0.22</td></tr><tr><td style="text-align: center;">0.40</td><td style="text-align: center;">0.20</td><td style="text-align: center;">0.88</td></tr><tr><td style="text-align: center;">0.40</td><td style="text-align: center;">0.40</td><td style="text-align: center;">0.88</td></tr></tbody></table><p>What is the overall order of reaction?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> Zero</label></p><p><label class="radio"> <input type="radio"> 1st</label></p><p><label class="radio"> <input class="c" type="radio"> <span>2nd</span></label></p><p><label class="radio"> <input type="radio"> 3rd</label></p></div><div class="q-explanation"><p>When the concentration of only Y is doubled (experiments 1 to 2) the rate is quadrupled. [Y] × 2 causes rate × 4, so [Y] is second order. When the concentration of only Z is doubled (experiments 2 to 3) the rate doesn't change, so [Z] is zero order. Thus [Y] is 2nd order and [Z] is zero order. Adding the orders gives 2nd order (or order 2) overall.</p><p>So the correct answer is<strong> 2nd.</strong></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>What can be deduced from this graph?</p><p><img alt="" src="../../images/chemical-kinetics/zeroorder.png" style="width: 160px; height: 124px;"></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> The reactant has a constant half-life</label></p><p><label class="radio"> <input class="c" type="radio"> The reaction is zero order with respect to the reactant</label></p><p><label class="radio"> <input type="radio"> The reaction is first order with respect to the reactant</label></p><p><label class="radio"> <input type="radio"> The reaction is second order with respect to the reactant<span></span></label> </p></div><div class="q-explanation"><p>The relationships between reactant concentration and time are shown below:</p><p><img alt="" src="../../images/chemical-kinetics/correctgraphs1.png" style="width: 480px; height: 133px;"></p><p>A straight line (decreasing concentration does not impact rate) indicates zero order.</p><p>A curved line with a constant half-life indicates first order.</p><p>A curved line with a half-life that is not constant indicates second order.</p><p><strong>The reaction is zero order with respect to the reactant</strong> is therefore the correct answer.</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">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>The rate expression for a reaction is Rate = k[A]<sup>2</sup>[B]</p><p>By which factor will the rate of reaction increase if [A] and [B] are both increased by a factor of 3?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 6</label></p><p><label class="radio"> <input type="radio"> 9</label></p><p><label class="radio"> <input class="c" type="radio"> 27</label></p><p><label class="radio"> <input type="radio"> 12</label></p></div><div class="q-explanation"><p>When the concentration of A is tripled the rate increases nine fold: [A] × 3 causes rate × 9, 3<sup>2</sup> = 9.</p><p>When the concentration of B is tripled the rate is also tripled: [B] × 3 causes rate × 3, and 3<sup>1</sup> = 3.</p><p>Rate × 9 and rate × 3 gives a total (9×3) of × 27.</p><p>So <strong>27</strong> is the correct answer.</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">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>The mechanism below represents the S<sub>N</sub>1 reaction between 2-bromo-2-methyl propane and a hydroxide ion:</p><p>Step 1 (slow step): (CH<sub>3</sub>)<sub>3</sub>CBr → (CH<sub>3</sub>)<sub>3</sub>C<sup>+</sup> + Br<sup>–</sup></p><p>Step 2 (fast step): (CH<sub>3</sub>)<sub>3</sub>C<sup>+</sup> + OH<sup>–</sup> → (CH<sub>3</sub>)<sub>3</sub>COH</p><p>The hydroxide ion does not take part in step 1 and the bromide ion does not take part in step 2.</p><p>Which of the following is correct?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> The reaction is first order with respect to the hydroxide ion </label></p><p><label class="radio"> <input type="radio"> The reaction is second order overall</label></p><p><label class="radio"> <input type="radio"> The rate expression is rate=k[(CH<sub>3</sub>)<sub>3</sub>CBr][OH<sup>–</sup>]</label></p><p><label class="radio"> <input class="c" type="radio"> The molecularity of the rate determining step is 1</label></p></div><div class="q-explanation"><p>This is a two step reaction with the first (slowest) step being the rate determining step (RDS).</p><p>The RDS has only the halogenoalkane species as a reactant (it is unimolecular) so the rate expression will be rate=k[(CH<sub>3</sub>)<sub>3</sub>CBr]. That is first order with respect to (CH<sub>3</sub>)<sub>3</sub>CBr, and first order overall.</p><p>Molecularity is the number of particles that react together in a given reaction/step. In step 1 the only reactant is (CH<sub>3</sub>)<sub>3</sub>CBr so the molecularity is 1 (a unimolecular step). (In step 2 the reactants are (CH<sub>3</sub>)<sub>3</sub>C<sup>+</sup> and OH<sup>–</sup> so the molecularity is 2 (a bimolecular step).</p><p><strong>The molecularity of the rate determining step is 1</strong> is the correct answer.</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">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the equation for the over<span style="color:#FFFFFF;"></span>all reaction given the reaction mechanism below?</p><p>Step 1 (fast step): A + B → C<sub> </sub>+ 2D</p><p>Step 2 (fast step): D + B → E + F</p><p>Step 3 (slow step): F + D + C → H</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> A + 2B → E + H<span></span></label></p><p><label class="radio"> <input type="radio"> <span>A + 2B + C + 2D + F → E +H</span></label></p><p><label class="radio"> <input type="radio"> A + 2B + D → E + F + H <span></span></label></p><p><label class="radio"> <input type="radio"> A + B → H<span></span></label></p></div><div class="q-explanation"><p>Intermediates need to be cancelled. That is, if a species appears as a product in one step and then as a reactant in another step, it should be cancelled from the equation:</p><p>Step 1 (fast step): A + B → <span style="color:#FFFFFF;"><span style="background-color:#008000;">C</span></span> + <span style="color:#FFFFFF;"><span style="background-color:#FF0000;">2D</span></span></p><p>Step 2 (fast step): <span style="color:#FFFFFF;"><span style="background-color:#FF0000;">D</span></span> + B → E + <span style="color:#FFFFFF;"><span style="background-color:#0000FF;">F</span></span></p><p>Step 3 (slow step): <span style="color:#FFFFFF;"><span style="background-color:#0000FF;">F</span></span> + <span style="color:#FFFFFF;"><span style="background-color:#FF0000;">D</span></span> + <span style="color:#FFFFFF;"><span style="background-color:#008000;">C</span></span> → H</p><p>This leaves only the non-highlighted species, so <strong>A + 2B → E + H</strong><b> </b>is the correct answer.</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">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following represents a concentration vs rate graph for a zero order reactant?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <img alt="" src="../../images/chemical-kinetics/second.png" style="width: 120px; height: 107px;"></label></p><p><label class="radio"> <input type="radio"> <img alt="" src="../../images/chemical-kinetics/zeroorder.png" style="width: 120px; height: 93px;"></label></p><p><label class="radio"> <input class="c" type="radio"> <img alt="" src="../../images/chemical-kinetics/zero.png" style="width: 120px; height: 107px;"></label></p><p><label class="radio"> <input type="radio"> <img alt="" src="../../images/chemical-kinetics/first.png" style="width: 120px; height: 112px;"></label></p></div><div class="q-explanation"><p>The relationships between reactant concentration and rate are shown below:</p><p><img alt="" src="../../images/chemical-kinetics/correctgraphs2.png" style="width: 600px; height: 180px;"></p><p>A straight horizontal line indicates zero order.</p><p>A straight line indicates first order.</p><p>A curved line indicates second order.</p><p>The fourth graph with a straight line sloping down indicates the relationship between<strong> time and reactant concentration</strong>, seen here:</p><p><img alt="" src="../../images/chemical-kinetics/correctgraphs1.png" style="width: 600px; height: 166px;"></p><p>Therefore the correct answer is</p><p><img alt="" src="../../images/chemical-kinetics/zero.png" style="width: 120px; height: 107px;"></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 class="page-container panel-self-assessment" data-id="2721"> <div class="panel-heading">MY PROGRESS</div> <div class="panel-body understanding-rate"> <div class="msg"></div> <label class="label-lg">Self-assessment</label><p>How much of <strong>Kinetics AHL (HL only) paper 1 questions</strong> have you understood?</p><div class="slider-container text-center"><div id="self-assessment-slider" class="sib-slider self-assessment " data-value="1" data-percentage=""></div></div> <label class="label-lg">My notes</label> <textarea name="page-notes" class="form-control" rows="3" placeholder="Write your notes here..."></textarea> </div> <div class="panel-footer text-xs-center"> <span id="last-edited" class="mb-xs-3"> </span> <div class="actions mt-xs-3"> <button id="save-my-progress" type="button" class="btn btn-sm btn-primary text-center btn-xs-block"> <i class="fa fa-fw fa-floppy-o"></i> Save </button> </div> </div></div> <div id="modal-feedback" class="modal fade" tabindex="-1" role="dialog"> <div class="modal-dialog" role="document"> <div class="modal-content"> <div class="modal-header"> <h4 class="modal-title">Feedback</h4> <button type="button" class="close hidden-xs hidden-sm" data-dismiss="modal" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div> <div class="modal-body"> <div class="errors"></div> <p><strong>Which of the following best describes your feedback?</strong></p> <form method="post" style="overflow: hidden"> <div class="form-group"> <div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Recommendation"> Recommend</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Problem"> Report a problem</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Improvement"> Suggest an improvement</label></div><div class="radio"><label style="color: #121212;"><input type="radio" name="feedback-type" value="Other"> Other</label></div> </div> <hr> <div class="row"> <div class="col-md-6"> <div class="form-group"> <label for="feedback-name">Name</label> <input type="text" class="form-control" name="feedback-name" placeholder="Name" value=" "> </div> </div> <div class="col-md-6"> <div class="form-group"> <label for="feedback-email">Email address</label> <input type="email" class="form-control" name="feedback-email" placeholder="Email" value="@airmail.cc"> </div> </div> </div> <div class="form-group"> <label for="feedback-comments">Comments</label> <textarea class="form-control" name="feedback-comments" style="resize: vertical;"></textarea> </div> <input type="hidden" name="feedback-ticket" value="082b9c9c4ae3624d"> <input type="hidden" name="feedback-url" value="https://studyib.net/chemistry/page/2721/kinetics-ahl-hl-only-paper-1-questions"> <input type="hidden" name="feedback-subject" value="7"> <input type="hidden" name="feedback-subject-name" value="Chemistry"> <div class="pull-left"> </div> </form> </div> <div class="modal-footer"> <button type="button" class="btn btn-primary btn-xs-block feedback-submit mb-xs-3 pull-right"> <i class="fa fa-send"></i> Send </button> <button type="button" class="btn btn-default btn-xs-block m-xs-0 pull-left" data-dismiss="modal"> Close </button> </div> </div> </div></div> <style type="text/css" media="screen">/* Important part */
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