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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"> Entropy and spontaneity <a href="#" class="mark-page-favorite pull-right" data-pid="891" 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="../358/energetics.html">Energetics</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Entropy and spontaneity</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 60 minutes"><i class="fa fa-clock-o"></i> 60'</span> </ol> <article id="main-article"> <p><img alt="" src="../../images/test-images/entropy.png" style="float: left; width: 160px; height: 92px;">According to the second law of thermodynamics, a process is spontaneous if it causes the total entropy of universe to increase. Entropy is a measure of the number of ways in which the energy in a system can be arranged. <strong>Use the practice questions to help you become more confident in predicting which systems have a positive entropy change</strong>. You should also be able to calculate the entropy change of a system using absolute entropy values.</p> <p>The Gibbs free energy change takes into account both the entropy change of the system and that of the surroundings, the latter a result of the reaction being either exothermic or endothermic. Gibbs free energy calculations are not difficult, but it is easy to trip up by forgetting to convert your entropy energy unit from J to kJ. <strong>Refer to revision card 18 to help you organise your working</strong>. You should also be able to predict the spontaneity of a process knowing the enthalpy change of the system, the entropy change of the system and the reaction temperature. Remember that a spontaneous reaction here refers to a reaction that is thermodynamically feasible and does not give any indication about the rate of the reaction.</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>Ensure you are confident using the terms below and learn the asterisked* definitions</p> <p>Entropy<em>, </em>spontaneous<span calibri="" light="" style="font-size:12.0pt;font-family:"></span>, absolute entropy (<em>S<sup><s>o</s></sup></em>)*, Gibbs free energy change (<em>Δ</em><em>G</em>), Gibbs free energy of formation (<em>ΔG</em><span calibri="" light="" style="font-size:12.0pt;font-family:"><em><sup><s>o</s></sup><sub>f</sub></em>)</span><span calibri="" light="" style="font-size:12.0pt;font-family:">* </span></p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="3" data-mode="" data-topics="637" data-subject-id="7" data-n-flashcards="5" style="text-align:center">Show flashcards</a></div><hr> <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>What is entropy?</p> </div> </div> <div class="panel-body"> <div>An excellent video explanation from TED Ed</div> <div style="text-align: center;"> </div> <div style="text-align: center;"><iframe align="middle" allowfullscreen="" frameborder="0" height="410" scrolling="no" src="//www.youtube.com/embed/YM-uykVfq_E?rel=0" width="700"></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-142" class="dynamic-gallery carousel slide" data-id="142"><div class="carousel-inner" role="listbox"><div class="item active"><a class="fancy" href="../../../std-galleries/7-142/screenshot-2019-09-05-at-212942.png" data-fancybox="gallery-142" title="" data-caption=""><img alt="" src="../../../std-galleries/7-142/screenshot-2019-09-05-at-212942.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-142/screenshot-2019-09-05-at-213000.png" data-fancybox="gallery-142" title="" data-caption=""><img alt="" src="../../../std-galleries/7-142/screenshot-2019-09-05-at-213000.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-142/screenshot-2019-09-05-at-213011.png" data-fancybox="gallery-142" title="" data-caption=""><img 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data-index="8"><img title="Click to view" src="../../../std-galleries/7-142/screenshot-2019-09-05-at-213210-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="7-447-891"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the statements below is the <strong>best</strong> definition of entropy?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>The average kinetic energy of a system</span></label> </p><p><label class="radio"> <input type="radio"> <span>A measure of disorder or chaos in a system</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>A measure of the number of ways of arranging energy within particles</span></label> </p><p><label class="radio"> <input type="radio"> <span>The change in heat energy at constant pressure</span></label> </p></div><div class="q-explanation"><p>The best definition is <strong>A measure of the number of ways of arranging energy within particles</strong> and this is the correct answer.</p><p>Entropy is sometimes described as a 'measure of disorder', but this may be misleading and should be avoided.</p><p>'The change in heat energy at constant pressure' is a definition of Enthalpy change (ΔH). 'The average kinetic energy of a system' is a definition of temperature, T.</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>Which reaction has the greatest positive entropy change, ΔS (of the system).</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>CaCO<sub>3(s) </sub>→ CaO<sub>(s)</sub> + CO<sub>2(g)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>2H<sub>2(g)</sub> + O<sub>2(g) </sub>→ 2H<sub>2</sub>O<sub>(l)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>3H<sub>2(g)</sub> + N<sub>2(g)</sub> → 2NH<sub>3(g)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>Pb(NO<sub>3</sub>)<sub>(aq)</sub> + 2KI<sub>(aq)</sub> → 2KNO<sub>3(aq)</sub> + PbI<sub>2(s)</sub> </span></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 CaCO<sub>3(s) </sub>→ CaO<sub>(s)</sub> + CO<sub>2(g)</sub></p><p>The other reactions involve the creation of liquids or solids from gases, or a decrease in the number of gaseous moles.</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 reaction has the greatest positive entropy change, ΔS (of the system).</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>CH<sub>4(g) </sub>+ 1½O<sub>2(g)</sub> → CO<sub>(g)</sub> + 2H<sub>2</sub>O<sub>(l)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>4(g) </sub>+ 2O<sub>2(g)</sub> → CO<sub>2(g)</sub> + 2H<sub>2</sub>O<sub>(l)</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>CH<sub>4(g) </sub>+ 1½O<sub>2(g)</sub> → CO<sub>(g)</sub> + 2H<sub>2</sub>O<sub>(g)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>4(g) </sub>+ 2O<sub>2(g)</sub> → CO<sub>2(g)</sub> + 2H<sub>2</sub>O<sub>(g)</sub></span></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 CH<sub>4(g) </sub>+ 1½O<sub>2(g)</sub> → CO<sub>(g)</sub> + 2H<sub>2</sub>O<sub>(g)</sub> since 2½ moles of gas react to give 3 moles of gas.</p><p>The other reactions involve the creation of liquid water rather than gaseous water, or no increase in the number of gaseous moles.</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 reaction has an entropy change, ΔS (of the system) that is closest to zero?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>CuCO<sub>3(s)</sub> → CuO<sub>(s)</sub> + CO<sub>2(g)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>2H<sub>2(g) </sub>+ O<sub>2(g)</sub> → 2H<sub>2</sub>O<sub>(g)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>H<sub>2</sub>O<sub>(g)</sub> → H<sub>2</sub>O<sub>(l)</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>H<sub>2(g) </sub>+ Cl<sub>2(g)</sub> → 2HCl<sub>(g)</sub></span></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 is H<sub>2(g) </sub>+ Cl<sub>2(g)</sub> → 2HCl<sub>(g)</sub> since 2 moles of gas react to give 2 moles of gas. So the entropy change of the system is likely to be very close to zero.</p><p>The other reactions involve the creation of liquid water from gaseous water (negative ΔS), a decrease in the number of gaseous moles (negative ΔS), or the creation of a gaseous product from a solid reactant (positive ΔS).</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><em>Calculator question:</em> Using the standard entropy values from the data book (values given below), what is the standard entropy change ΔS<sup><s>o</s></sup> (in JK<sup>–1</sup>mol<sup>–1</sup>) of the system for<span style="font-family:arial,helvetica,sans-serif;"> the</span> combustion of methane?</p><p>CH<sub>4(g) </sub>+ 2O<sub>2(g)</sub> → CO<sub>2(g)</sub> + 2H<sub>2</sub>O<sub>(l)</sub></p><p>Standard entropy values (ΔS<sup><s>o</s></sup>) in JK<sup>–1</sup>mol<sup>–1</sup><span style="font-family:arial,helvetica,sans-serif;"><span style="font-size: 10pt;"></span></span></p><p>CH<sub>4(g)</sub> +186.0 | <span new="" roman="" style="font-size:12.0pt;font-family:" times=""></span>O<sub>2(g) </sub>+205.0 | CO<sub>2(g)</sub> +213.8 | H<sub>2</sub>O<sub>(l) </sub>+70.0</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>–242.2</span></label> </p><p><label class="radio"> <input type="radio"> –107.2<span></span></label> </p><p><label class="radio"> <input type="radio"> <span>+242.2</span></label> </p><p><label class="radio"> <input type="radio"> <span>+107.2</span></label> </p></div><div class="q-explanation"><p>The standard entropy change can be calculated by finding the difference between the standard entropy of the reactants and the standard entropy of the products; ΔS = S(products) – S(reactants).<span new="" roman="" style="font-size:10.0pt;font-family:Times;mso-fareast-font-family:
" times=""></span></p><p>The standard entropy of the reactants is 186 + 2×205.0 = 596.0</p><p>The standard entropy of the products is 213.8 + 2×70.0 = 353.8</p><p>Thus the standard entropy change is 353.8 – 596.0 = <strong>–242.2 </strong>(in JK<sup>–1</sup>mol<sup>–1</sup>) which is the correct answer.</p><p><strong>Incorrect answers</strong></p><p>+242.2 is obtained if ΔS is found by S(reactants) – S(products) instead of ΔS = S(products) – S(reactants).</p><p>–107.2 is obtained if <strong>one mole</strong> is used for every species in the calculation instead of using two moles of oxygen and two moles of water.</p><p>+107.2 is obtained if <strong>one mole</strong> is used for every species in the calculation instead of using two moles of oxygen and two moles of water <strong>and </strong>ΔS is found by S(reactants) – S(products) instead of ΔS = S(products) – S(reactants).</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><em>Calculator question:</em> Using the standard entropy values from the data book (some values given below), what is the entropy change ΔS (in JK<sup>–1</sup>mol<sup>–1</sup>) of the system for<span style="font-family:arial,helvetica,sans-serif;"> the</span> reaction as written below?</p><p>2H<sub>2(g) </sub>+ O<sub>2(g)</sub> → 2H<sub>2</sub>O<sub>(g)</sub></p><p>Standard entropy values (ΔS<sup><s>o</s></sup>) in JK<sup>–1</sup>mol<sup>–1</sup><span style="font-family:arial,helvetica,sans-serif;"><span style="font-size: 10pt;"></span></span></p><p>H<sub>2(g)</sub> +131.0 | <span new="" roman="" style="font-size:12.0pt;font-family:" times=""></span>O<sub>2(g) </sub>+205.0 | H<sub>2</sub>O<sub>(g)</sub> +188.8 | H<sub>2</sub>O<sub>(l) </sub>+70.0</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> –147.2<span></span></label> </p><p><label class="radio"> <input type="radio"> <span>–278.2</span></label> </p><p><label class="radio"> <input type="radio"> <span>–327</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>–89.4</span></label> </p></div><div class="q-explanation"><p>The entropy change can be calculated by finding the difference between the standard entropy of the reactants and the standard entropy of the products; ΔS = S(products) – S(reactants).<span new="" roman="" style="font-size:10.0pt;font-family:Times;mso-fareast-font-family:
" times=""></span></p><p>The standard entropy of the reactants is 2×131.0 + 205.0 = 467.0</p><p>The standard entropy of the products is 2×188.8 = 377.6</p><p>Thus the standard entropy change is 377.6 – 467.0 = <strong>–89.4 </strong>(in JK<sup>–1</sup>mol<sup>–1</sup>) which is the correct answer.</p><p><strong>Incorrect answers</strong></p><p>–278.2 is obtained if <strong>one mole</strong> of water is used (products) instead of two moles of water.</p><p>–147.2 is obtained if <strong>one mole</strong> is used for every species in the calculation instead of using two moles of hydrogen and two moles of water.</p><p>–327 is obtained if the value used for water is <strong>liquid water </strong>(+70.0) instead of gaseous water (+188.8) as written in the equation.</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>For a chemical reaction to be thermodynamically spontaneous, which of the following <strong>must</strong> <strong>always</strong> be true?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>The reaction must be exothermic </span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>ΔH−TΔS ≤ 0</span></label> </p><p><label class="radio"> <input type="radio"> <span>ΔH < 0</span></label> </p><p><label class="radio"> <input type="radio"> <span>The entropy change of the system (ΔS) must be positive</span></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 (there must be an overall gain in entropy in the universe). Since ΔG = ΔH−TΔS (this should be learned), the correct answer is <strong>ΔH−TΔS ≤ 0</strong> since this is stating that ΔG ≤ 0.</p><p><strong>Incorrect answers (</strong>refer to ΔG = ΔH−TΔS)</p><p>The reaction does not always have to be exothermic. (ΔH < 0 shows an exothermic reaction.) Exothermic reactions will often be spontaneous, but ΔG may not be negative (or zero) if ΔS is negative. Endothermic reactions (positive ΔH) can be spontaneous; ΔG can be negative (or zero) if ΔS is positive.</p><p>If the entropy change of the system ΔS is positive this may lead to a spontaneous reaction; ΔG will be negative (or zero) if the reaction is also exothermic (negative ΔH). However, if the reaction is endothermic (positive ΔH) then ΔG may not be less than or equal to zero so the reaction will not be spontaneous.</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>In the following reaction the signs for the energy changes are as follows: ΔH is positive and ΔS is positive. When will the reaction be thermodynamically spontaneous?</p><p>CaCO<sub>3(s)</sub> → CaO<sub>(s)</sub> + CO<sub>2(g)</sub></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> The reaction will become spontaneous at lower temperature<span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>The reaction will become spontaneous at higher temperature</span></label> </p><p><label class="radio"> <input type="radio"> The reaction will always be spontaneous<span></span></label> </p><p><label class="radio"> <input type="radio"> <span>The reaction will never be spontaneous</span></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 (there must be an overall gain in entropy in the universe). Since ΔG = ΔH−TΔS (this should be learned), the correct answer is <strong>the reaction will become spontaneous at higher temperature</strong> since at some point as temperature increases ΔG ≤ 0.</p><p>Looking carefully at the Gibb's free energy equation, ΔG = ΔH−TΔS, 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="" style="font-size:12.0pt;font-family:" times=""></span></p><p>Note: |x| means <strong>magnitude</strong> of x, that is the numerical value of x without a sign.</p><p>Remember: Temperature,<strong> T</strong> in the expression is given in Kelvin (K) so is always positive.</p><p><strong>Incorrect answers (</strong>refer to ΔG = ΔH−TΔS)</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>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">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which units are entropy changes (ΔS) and enthalpy changes (ΔH) usually quoted in?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>ΔH in kJ mol<sup>−1</sup> ΔS in kJ mol<sup>−1</sup> </span></label> </p><p><label class="radio"> <input type="radio"> <span>ΔH in J K<sup>−1</sup> mol<sup>−1</sup> ΔS in J K<sup>−1</sup> mol<sup>−1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>ΔH in J K<sup>−1</sup> mol<sup>−1</sup> ΔS in kJ mol<sup>−1</sup></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>ΔH in kJ mol<sup>−1</sup> ΔS in J K<sup>−1</sup> mol<sup>−1</sup></span></label> </p></div><div class="q-explanation">This is an important thing to remember and check! ΔH values are given in kJ mol<sup>−1</sup> and ΔS values are given in J K<sup>−1</sup> mol<sup>−1</sup>. So it is important to convert kJ into J by multiplying by 1000 (or vica versa) when calculating ΔG using ΔG = ΔH−TΔS. Thus, <strong>ΔH in kJ mol<sup>−1</sup> ΔS in J K<sup>−1</sup> mol</strong><sup>−1</sup> is the correct answer.</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><em>Calculator question: </em>The hydrogenation of propyne is shown below:</p><p>C<sub>3</sub>H<sub>4(g) </sub>+ 2H<sub>2(g) </sub> → C<sub>3</sub>H<sub>8(g)</sub></p><p>The energy changes for the reaction are as follows:</p><p><span style="font-family:arial,helvetica,sans-serif;">ΔH<sup><s>o</s></sup> = −287 kJ mol<sup>–1</sup></span></p><p><span style="font-family:arial,helvetica,sans-serif;">ΔS<sup><s>o</s></sup> = −240 J K<sup>–1</sup> mol<sup>–1</sup></span></p><p>What is the standard free energy change (ΔG<sup><s>o</s></sup>) at 298K (in kJ mol<sup>–1</sup>) to three sig figs for<span style="font-family:arial,helvetica,sans-serif;"> the</span> reaction as written above, and is the reaction spontaneous at this temperature?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>–215; yes, it is spontaneous</span></label> </p><p><label class="radio"> <input type="radio"> +71200; yes, it is spontaneous<span></span></label> </p><p><label class="radio"> <input type="radio"> <span>+71200; no, it is not spontaneous</span></label> </p><p><label class="radio"> <input type="radio"> <span>–215; no, it is not spontaneous</span></label> </p></div><div class="q-explanation"><p>The standard free energy change (ΔG<sup><s>o</s></sup>) can be calculated by using ΔG<sup><s>o</s></sup> = ΔH<sup><s>o</s></sup>–TΔS<sup><s>o</s></sup>.<span new="" roman="" style="font-size:10.0pt;font-family:Times;mso-fareast-font-family:
" times=""></span></p><p>Temperature must be in K (as given in the question).</p><p>Units for enthalpy include <strong>kilojoules</strong> and units for entropy include <strong>joules</strong>. So it is best to convert entropy into kilojoules by dividing by 1000; −240 <strong>J</strong> K<sup>–1</sup> mol<sup>–1</sup> becomes −0.240 <strong>kJ </strong>K<sup>–1</sup> mol<sup>–1</sup>.</p><p>Therefore ΔG<sup><s>o</s></sup> = −287−(298×−0.240)</p><p>= −215.48 = –215 kJ mol<sup>–1</sup> to three sig figs</p><p>The standard free energy value is negative, so the reaction is spontaneous. Reactions are spontaneous (meaning 'allowed'; that they will take place if given sufficient activation energy) if the ΔG value is less than or equal to zero.</p><p>Thus the correct answer is <strong>–215; yes, it is spontaneous</strong>.</p><p><strong>Incorrect answers</strong></p><p>+71200 (+71233) is obtained if entropy value (−240) is <strong>not converted into kilojoules</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">11</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Calculator question: </em>The reduction of iron oxide is shown below:</p><p>Fe<sub>2</sub>O<sub>3(s) </sub>+ 3H<sub>2(g) </sub> → 2Fe(s) + 3H<sub>2</sub>O<sub>(l)</sub></p><p>The energy changes for the reaction are as follows:</p><p><span style="font-family:arial,helvetica,sans-serif;">ΔH<sup><s>o</s></sup> = +96.0 kJ mol<sup>–1</sup></span></p><p><span style="font-family:arial,helvetica,sans-serif;">ΔS<sup><s>o</s></sup> = +138 J K<sup>–1</sup> mol<sup>–1</sup></span></p><p>What is the standard free energy change (ΔG<sup><s>o</s></sup>) at 20°C (in kJ mol<sup>–1</sup>) to three sig figs for<span style="font-family:arial,helvetica,sans-serif;"> the</span> reaction as written above?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>−2660</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>+55.6</span></label> </p><p><label class="radio"> <input type="radio"> −40300<span></span></label> </p><p><label class="radio"> <input type="radio"> <span>+93.2</span></label> </p></div><div class="q-explanation"><p>The standard free energy change (ΔG<sup><s>o</s></sup>) can be calculated by using ΔG<sup><s>o</s></sup> = ΔH<sup><s>o</s></sup>–TΔS<sup><s>o</s></sup>.<span new="" roman="" style="font-size:10.0pt;font-family:Times;mso-fareast-font-family:
" times=""></span></p><p>Temperature must be in K (Add 273 to °C to get temperature in K); so 20+273= 293K.</p><p>Units for enthalpy include <strong>kilojoules</strong> and units for entropy include <strong>joules</strong>. So it is best to convert entropy into kilojoules by dividing by 1000; +138 <strong>J</strong> K<sup>–1</sup> mol<sup>–1</sup> becomes +0.138 <strong>kJ </strong>K<sup>–1</sup> mol<sup>–1</sup>.</p><p>Therefore ΔG<sup><s>o</s></sup> = +96−(293×+0.138) = 96−40.434</p><p>= +55.566 = +55.6 kJ mol<sup>–1</sup> to three sig figs</p><p>Thus the correct answer is <strong>+55.6</strong></p><p>(The standard free energy value is positive, so the reaction is not spontaneous. Reactions are spontaneous (meaning 'allowed'; that they will take place if given sufficient activation energy) if the ΔG value is less than or equal to zero.)</p><p><strong>Incorrect answers</strong></p><p>−40300 (−40338) is obtained if entropy value (+138) is <strong>not converted into kilojoules</strong>.</p><p>+93.2 (+93.24) is obtained if the <strong>temperature is not converted into Kelvin</strong>; 20 is used as a value instead of 293.</p><p>−2660 (−2664) is obtained if entropy value (+138) is <strong>not converted into kilojoules </strong><u>and </u>the <strong>temperature is not converted into Kelvin</strong>; 20 is used as a value instead of 293.</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><em>Calculator question: </em>The reduction of iron oxide is shown below:</p><p>Fe<sub>2</sub>O<sub>3(s) </sub>+ 3H<sub>2(g) </sub> → 2Fe(s) + 3H<sub>2</sub>O<sub>(l)</sub></p><p>The energy changes for the reaction are as follows:</p><p><span style="font-family:arial,helvetica,sans-serif;">ΔH<sup><s>o</s></sup> = +96.0 kJ mol<sup>–1</sup></span></p><p><span style="font-family:arial,helvetica,sans-serif;">ΔS<sup><s>o</s></sup> = +138 J K<sup>–1</sup> mol<sup>–1</sup></span></p><p>The reaction is not spontaneous at 298K (25°C) but does becomes spontaneous at higher temperature. To three sig figs, at what temperature (in Kelvin, K) will the reaction become spontaneous?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> 54.9<span></span></label> </p><p><label class="radio"> <input type="radio"> <span>0.696</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>696</span></label> </p><p><label class="radio"> <input type="radio"> 13.2<span></span></label> </p></div><div class="q-explanation"><p>The standard free energy change (ΔG<sup><s>o</s></sup>) can be calculated by using ΔG<sup><s>o</s></sup> = ΔH<sup><s>o</s></sup>–TΔS<sup><s>o</s></sup>.<span new="" roman="" style="font-size:10.0pt;font-family:Times;mso-fareast-font-family:
" times=""></span></p><p>Reactions are spontaneous (meaning 'allowed'; that they will take place if given sufficient activation energy) if the ΔG value is less than or equal to zero. So to find the point at which the reaction becomes spontaneous we must calculate the temperature at which ΔG<sup><s>o</s></sup> = 0.</p><p>Units for enthalpy include <strong>kilojoules</strong> and units for entropy include <strong>joules</strong>. So it is best to convert entropy into kilojoules by dividing by 1000; +138 <strong>J</strong> K<sup>–1</sup> mol<sup>–1</sup> becomes +0.138 <strong>kJ </strong>K<sup>–1</sup> mol<sup>–1</sup>.</p><p>Therefore ΔG<sup><s>o</s></sup> = 0 = +96−(T×+0.138)</p><p>Rearranging:</p><p>−(T×+0.138) = −96</p><p>(T×+0.138) = 96</p><p>T = 96/0.138</p><p>= 695.652174 = 696K to three sig figs</p><p>Thus the correct answer is <strong>696</strong></p><p><strong>Incorrect answers</strong></p><p>0.696 is obtained if entropy value (+138) is <strong>not converted into kilojoules</strong>.</p><p>54.9 (54.876) is the<strong> ΔG<sup><s>o</s></sup> </strong>value (in kJ mol<sup>–1</sup>) obtained if the <strong>values given in the question are plugged into </strong>ΔG<sup><s>o</s></sup> = ΔH<sup><s>o</s></sup>–TΔS<sup><s>o</s></sup> (at 298K).</p><p>13.2 (13.248) is obtained if T = 96×0.138 instead of 96÷0.138 (rearranging error).</p></div><div class="actions"><span class="score" data-score="0"></span><button 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questions</a></h5> <h4>Paper 2</h4> <h5>Core (SL&HL): <a href="../2710/energetics-core-sl-hl-paper-2-questions.html" title="Energetics core (SL & HL) paper 2 questions">Energetics core (SL & HL) paper 2 questions</a></h5> <h5>AHL (HL only): <a href="../2711/energetics-ahl-hl-only-paper-2-questions.html" title="Energetics AHL (HL only) paper 2 questions">Energetics AHL (HL only) paper 2 questions</a></h5> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="page-container panel-self-assessment" data-id="891"> <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>Entropy and spontaneity</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" 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