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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"> Energetics AHL (HL only) paper 1 questions <a href="#" class="mark-page-favorite pull-right" data-pid="2709" 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">Energetics 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 15.1 and 15.2</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="7-813-2709"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which reaction has the greatest increase in entropy of the system.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> CH<sub>4 </sub>(g) + H<sub>2</sub>O (g) → CO (g) + 3H<sub>2 </sub>(g)</label></p><p><label class="radio"> <input type="radio"> MgCO<sub>3 </sub>(s) → MgO (s) + CO<sub>2 </sub>(g)</label></p><p><label class="radio"> <input type="radio"> <span>Pb(NO<sub>3</sub>) (aq) + 2KI (aq) → 2KNO<sub>3 </sub>(aq) + PbI<sub>2 </sub>(s) </span></label> </p><p><label class="radio"> <input class="c" type="radio"> 2NI<sub>3</sub> (s) → N<sub>2</sub> (g) + 3I<sub>2</sub> (g)</label></p></div><div class="q-explanation"><p>Gases tend to have greater entropy than liquids (or solutions), that tend to have greater entropy than solids. Any change that leads to a greater number of gaseous moles tends to lead to an increase in the entropy of the system.</p><p>Thus the correct answer here is <strong>2NI<sub>3</sub> (s) → N<sub>2</sub> (g) + 3I<sub>2</sub> (g)</strong>, in which 2 moles of solid become 4 moles of gas.</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">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the correct order of increasing (more exothermic) enthalpy change of hydration?</p><p>X<sup>n+</sup> (g) → X<sup>n+</sup> (aq)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> K<sup>+</sup>, Na<sup>+</sup>, Be<sup>2+</sup>, Li<sup>+</sup></label></p><p><label class="radio"> <input class="c" type="radio"> <span>K<sup>+</sup>, Na<sup>+</sup>, Li<sup>+</sup>, Be<sup>2+</sup></span></label></p><p><label class="radio"> <input type="radio"> Be<sup>2+</sup>, K<sup>+</sup>, Na<sup>+</sup>, Li<sup>+</sup></label></p><p><label class="radio"> <input type="radio"> Be<sup>2+</sup><span>, Li<sup>+</sup>, Na<sup>+</sup>, K<sup>+</sup></span></label></p></div><div class="q-explanation"><p>Standard enthalpy of hydration is the enthalpy change when <strong>one mole of</strong> <strong>aqueous solution</strong> is formed from <strong>ions in the gaseous state</strong>.</p><p>Ions with <strong>smaller size</strong> and <strong>higher charge</strong> will have greater enthalpy of hydration values. Therefore in group 1; Li<sup>+</sup> > Na<sup>+</sup> > K<sup>+</sup>. And group 2 ions (like Be<sup>2+</sup>) will have significantly higher enthalpy values.</p><p>The correct answer (increasingly exothermic) is therefore <strong>K<sup>+</sup>, Na<sup>+</sup>, Li<sup>+</sup>, Be<sup>2+</sup></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>Which is correct for this process?</p><p>H<sub>2</sub>O (l) → H<sub>2</sub>O (s)</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> Enthalpy decreases and entropy decreases</label></p><p><label class="radio"> <input type="radio"> Enthalpy increases and entropy decreases</label></p><p><label class="radio"> <input type="radio"> Enthalpy increases and entropy increases</label></p><p><label class="radio"> <input type="radio"> E<span>nthalpy decreases and entropy increases</span></label> </p></div><div class="q-explanation"><p>This phase/state change involves the making of bonds, so the process in exothermic which means that the enthalpy change is negative - so enthalpy decreases.</p><p>Gases tend to have greater entropy than liquids (or solutions), that tend to have greater entropy than solids. Therefore entropy also decreases for this process.</p><p>The correct answer is <strong>Enthalpy decreases and entropy decreases</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">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which equation represents the standard enthalpy of atomisation of astatine, At<sub>2</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> At<span><sub>2(s)</sub> → 2At<sub>(s)</sub></span></label> </p><p><label class="radio"> <input type="radio"> At<span><sub>2(s)</sub> → 2At<sub>(g)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>½At<sub>2(s)</sub> → At<sub>(s)</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>½At<sub>2(s)</sub> → At<sub>(g)</sub></span></label> </p></div><div class="q-explanation"><p>Standard enthalpy of atomisation is a definition that needs to be learned:</p><p><em>Standard enthalpy of atomisation is the enthalpy change when <strong>one mole of</strong> <strong>gaseous atoms</strong> is formed from the element in its <strong>standard state</strong> (at 100kPa and 298K).</em></p><p>Thus <strong>½At<sub>2(s)</sub> → At<sub>(g) </sub></strong>is the correct answer, since one mole of gaseous atoms<strong> </strong>are formed (from the element in its standard state).</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>Which equation represents lattice enthalpy?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> K<span>Cl (s) → K<sup>+</sup>(g) + Cl<sup>−</sup>(g)</span></label></p><p><label class="radio"> <input type="radio"> <span>KCl (s) → K<sup>+</sup>(s) + Cl<sup>−</sup>(s)</span></label></p><p><label class="radio"> <input type="radio"> <span>KCl (g) → K<sup>+</sup>(g) + Cl<sup>−</sup>(g)</span></label></p><p><label class="radio"> <input type="radio"> <span>KCl (s) → K<sup>+</sup>(aq) + Cl<sup>−</sup>(aq)</span></label></p></div><div class="q-explanation"><p>Lattice enthalpy is a definition that needs to be learned:</p><p>Lattice enthalpy is the enthalpy change when <strong>one mole </strong>of an ionic compound in its <strong>standard state</strong> (solid) forms <b>gaseous ions</b>.</p><p>Thus<strong> KCl (s) → K<sup>+</sup>(g) + Cl<sup>−</sup>(g)<sub> </sub></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">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>What are the signs for ΔH<sup><s>o</s></sup> and ΔS<sup><s>o</s></sup> for this reaction which is spontaneous at low temperature and non-spontaneous at very high temperatures?</p><p>4Fe (s) + 3O<sub>2</sub> (g) → 2Fe<sub>2</sub>O<sub>3 </sub>(s)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> ΔH<sup><s>o</s></sup> is positive ; ΔS<sup><s>o</s></sup> is positive</label></p><p><label class="radio"> <input class="c" type="radio"> ΔH<sup><s>o</s></sup> is negative ; ΔS<sup><s>o</s></sup> is negative</label></p><p><label class="radio"> <input type="radio"> <span>ΔH<sup><s>o</s></sup> is positive ; ΔS<sup><s>o</s></sup> is negative</span></label> </p><p><label class="radio"> <input type="radio"> ΔH<sup><s>o</s></sup> is negative ; ΔS<sup><s>o</s></sup> is positive </label></p></div><div class="q-explanation"><p>For a reaction to be thermodynamically spontaneous the <strong>change in Gibb's free energy</strong> (<strong>ΔG</strong>) must be equal to or less than zero.</p><p>ΔG = ΔH−TΔS (this should be learned)</p><p>(Note for explanation: |x| means <strong>magnitude</strong> of x, that is the numerical value of x without a sign. And remember that temperature,<strong> T</strong> in the expression is given in Kelvin (K) so is always positive.)</p><p>For the reaction to be become spontaneous at lower temperature, ΔH and ΔS would both need to be negative; then ΔG will be negative when |TΔS| < |ΔH|; this will occur as temperature decreases, as |TΔS| becomes smaller as temperature becomes smaller.</p><p>Therefore the correct answer is<b> </b><strong>ΔH<sup><s>o</s></sup> is negative ; ΔS<sup><s>o</s></sup> is negative</strong></p><p><strong>Notes on incorrect answers</strong></p><p>If ΔH is positive and ΔS is positive then ΔG will be negative when |TΔS| > |ΔH|; this will occur as temperature increases, as |TΔS| becomes greater as temperature becomes greater.<span new="" roman="" times=""></span></p><p>For the reaction to always be spontaneous (at any temperature), ΔH would need to be negative and ΔS would need to be positive; then ΔG will always be negative.</p><p>For the reaction to never be spontaneous (at any temperature), ΔH would need to be positive and ΔS would need to be negative; then ΔG will always be positive.</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>What is the enthalpy of solution for magnesium chloride, MgCl<sub>2</sub> (s), in kJ mol<sup>−1</sup>?</p><table border="0" cellpadding="0" cellspacing="0" style="width:100%;"><tbody><tr><td><strong>Enthalpy change</strong></td><td style="text-align: center;"> </td><td style="text-align: center;"><strong>Value (kJ mol<sup>−1</sup>)</strong></td></tr><tr><td>Lattice enthalpy MgCl<sub>2 </sub>(s)</td><td style="text-align: center;"> </td><td style="text-align: center;">+2540</td></tr><tr><td>Enthalpy of hydration, Mg<sup>2+ </sup>(g)</td><td> </td><td style="text-align: center;">−1963</td></tr><tr><td>Enthalpy of hydration, Cl<sup>−</sup> (g)</td><td> </td><td style="text-align: center;">−359</td></tr></tbody></table></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>−2540 + 1963 + 359</span></label></p><p><label class="radio"> <input class="c" type="radio"> 2540 − 1963 + 2(−359)</label></p><p><label class="radio"> <input type="radio"> <span>2540 − 1963 − 359</span></label> </p><p><label class="radio"> <input type="radio"> <span>−2540 + 1963 − 2(−359)</span></label></p></div><div class="q-explanation"><p>It is worth learning the energy cycle (shown with a generic metal, M, and non-metal, X):</p><p><img alt="" src="../../images/energetics-thermochemistry/hydrationetc.png" style="width: 360px; height: 143px;"></p><p>Following the alternative route, ΔH solution is therefore equal to lattice enthalpy <strong>plus</strong> ΔH hydration for the ions.</p><p>In MgCl<sub>2</sub>, there is one Mg<sup>2+</sup> ion and <strong>two</strong> Cl<sup>−</sup> ions, so:</p><p>ΔH solution<sub> </sub>(MgCl<sub>2</sub>) = +2540 + (−1963) + 2(−359)</p><p>Which gives the correct answer <strong>2540 − 1963 + 2(−359)</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">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which value represents the lattice enthalpy (in kJ mol<sup>−1</sup>) for magnesium fluoride, MgF<sub>2 </sub>as shown on the diagram?</p><p><img alt="" src="../../images/energetics-thermochemistry/bh-mgf2-full.png" style="width: 480px; height: 317px;"></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> −(−<span>1124) + 148 + 2(79) + 738 +1450 + 2(−328)</span></label></p><p><label class="radio"> <input type="radio"> <span>1124 + 148 + 2(79) + 738 +1450 + 2(328)</span></label> </p><p><label class="radio"> <input type="radio"> −1124 +148 + 2(79) + 738 +1450 + 2(−328)</label></p><p><label class="radio"> <input type="radio"> −2(−328) − 1450 − 738 − 2(79) − 148 + 1124</label></p></div><div class="q-explanation"><p>Lattice enthalpy as shown on the diagram begins at <strong>MgF<sub>2</sub> (s) </strong>and ends at <strong>Mg<sup>2+</sup> (g) + 2F<sup>−</sup> (g)</strong>, therefore we can take the alternative route beginning with going against the −1124 arrow and following around the energy cycle, inverting the signs of the energy changes when going against the arrows:</p><p>Thus, ΔH<sub>LE</sub>(MgF<sub>2</sub>) = <strong>−(−1124) + 148 + 2(79) + 738 + 1450 + 2(−328)</strong>, which 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">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which statements are correct for ionic compounds?</p><div><p><strong>1: </strong>Lattice enthalpy increases as ionic radii decrease.</p><p><strong>2: </strong>Lattice enthalpy increases as the charge on the cation decreases.</p><p><strong>3: </strong>Solubility in water depends on the relative magnitude of the lattice enthalpy compared to the hydration enthalpy </p></div></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p></div><div class="q-explanation"><p>Lattice enthalpy increases with <strong>increasing charge </strong>of the ions and with <strong>decreasing size</strong> of the ions.</p><p>Solubility in water does depend on the relative magnitude of the lattice enthalpy and hydration enthalpy, as shown on the diagram below:</p><p><img alt="" src="../../images/energetics-thermochemistry/hydrationetc.png" style="width: 360px; height: 143px;"></p><p>Therefore <strong>1 and 3 only</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">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>The combustion of ethanol is exothermic. Which is correct for this reaction?</p><p>C<sub>2</sub>H<sub>5</sub>OH (l) + 3O<sub>2</sub> (g) → 2CO<sub>2</sub> (g) + 3H<sub>2</sub>O<span style="font-size: 11.25px;"> </span>(g)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> ΔH<sup><s>o</s></sup> is negative ; ΔS<sup><s>o</s></sup> is positive ; the reaction is non-spontaneous </label></p><p><label class="radio"> <input type="radio"> <span>ΔH<sup><s>o</s></sup> is positive ; ΔS<sup><s>o</s></sup> is positive ; the reaction is non-spontaneous</span></label></p><p><label class="radio"> <input class="c" type="radio"> ΔH<sup><s>o</s></sup> is negative ; ΔS<sup><s>o</s></sup> is positive ; the reaction is spontaneous</label></p><p><label class="radio"> <input type="radio"> ΔH<sup><s>o</s></sup> is positive ; ΔS<sup><s>o</s></sup> is negative ; the reaction is spontaneous</label></p></div><div class="q-explanation"><p>The reaction is exothermic, so ΔH must be negative.</p><p>The reaction involves 1 mole of liquid and 3 moles of gas producing 5 moles of gas as written, so there must be a gain in entropy, since entropy increases gas>liquid>solid. Therefore ΔS must be positive.</p><p>For a reaction to be thermodynamically spontaneous the <strong>change in Gibb's free energy</strong> (<strong>ΔG</strong>) must be equal to or less than zero.</p><p>ΔG = ΔH−TΔS (this should be learned)</p><p>ΔH is negative and ΔS is positive, so ΔG will always be negative, since −TΔS will always be negative. Therefore the reaction will always be spontaneous (at any temperature).</p><p><strong>ΔH<sup><s>o</s></sup> is negative ; ΔS<sup><s>o</s></sup> is positive ; the reaction is spontaneous</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">11</div><div class="exercise shadow-bottom"><div class="q-question"><p>Consider the values of ΔH<sup><s>o</s></sup> and ΔS<sup><s>o</s></sup> for this reaction. Which statement is correct?</p><p style="text-align: center;">N<sub>2</sub> (g) + 3H<sub>2</sub> (g) → 2NH<sub>3 </sub>(g)</p><p>ΔH<sup><s>o</s></sup> = −92 kJ mol<sup>−1</sup></p><p>ΔS<sup><s>o</s></sup> = −198 J K<sup>−1</sup> mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> ΔG<sup><s>o</s></sup> is negative at low temperatures</label></p><p><label class="radio"> <input type="radio"> ΔG<sup><s>o</s></sup> is negative at all temperatures</label></p><p><label class="radio"> <input type="radio"> <span>ΔG<sup><s>o</s></sup> is positive at low temperatures</span></label></p><p><label class="radio"> <input type="radio"> ΔG<sup><s>o</s></sup> is positive at all temperatures</label></p></div><div class="q-explanation"><p>ΔG = ΔH−TΔS (this should be learned)</p><p>(Note for explanation: |x| means <strong>magnitude</strong> of x, that is the numerical value of x without a sign. And remember that temperature,<strong> T</strong> in the expression is given in Kelvin (K) so is always positive.)</p><p>If ΔH and ΔS are both negative; then ΔG will be negative when |TΔS| < |ΔH|; this will occur as temperature decreases, as |TΔS| becomes smaller as temperature becomes smaller.</p><p>Therefore the correct answer is<b> </b><strong>ΔG<sup><s>o</s></sup> is negative at low temperatures</strong></p><p><strong>Notes</strong></p><p>For a reaction to be thermodynamically spontaneous the change in Gibb's free energy (ΔG) must be equal to or less than zero.</p><p>If ΔH is positive and ΔS is positive then ΔG will be negative when |TΔS| > |ΔH|; this will occur as temperature increases, as |TΔS| becomes greater as temperature becomes greater.<span new="" roman="" times=""></span></p><p>If ΔH is negative and ΔS is positive; then ΔG will always be negative.</p><p>If ΔH is positive and ΔS is negative; then ΔG will always be positive.</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">12</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which equation represents the second electron affinity of sulfur?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> S<span>(s) + 2e<sup>−</sup> → S<sup>2</sup><sup>−</sup>(s)</span></label> </p><p><label class="radio"> <input class="c" type="radio"> S<span><sup>−</sup>(g) + e<sup>−</sup> → S<sup>2−</sup>(g)</span></label> </p><p><label class="radio"> <input type="radio"> S<sup>−</sup>(s) + e<sup>−</sup> → S<sup>2</sup><sup>−</sup>(s) </label></p><p><label class="radio"> <input type="radio"> S<span>(g) + 2e<sup>−</sup> → S<sup>2−</sup>(g)</span></label> </p></div><div class="q-explanation"><p>First electron affinity and second electron affinity are definitions that need to be learned:</p><p>First electron affinity is the enthalpy change when <strong>one electron is gained </strong>by <strong>each atom</strong> <strong>in one mole</strong> of <strong>gaseous atoms</strong>.</p><p>Second electron affinity is the enthalpy change when <strong>one electron is gained </strong>by <strong>each 1− ion</strong> <strong>in one mole</strong> of <strong>gaseous ions</strong>.</p><p>Thus <strong>S<sup>−</sup><sub>(g)</sub> + e<sup>−</sup> → S<sup>2−</sup><sub>(g)</sub></strong><strong><sub> </sub></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="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" 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