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paper 1 questions</a></label></li></ul></div> <button id="show-periodic-table" class="btn 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 core (SL and HL) paper 1 questions <a href="#" class="mark-page-favorite pull-right" data-pid="2708" 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 core (SL and HL) 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 5.1, 5.2 and 5.3</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="7-812-2708"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Using the standard enthalpies of combustion given below, what is enthalpy change for the following reaction?</p><p style="text-align: center;">C<sub>7</sub>H<sub>16</sub> (l) → C<sub>2</sub>H<sub>4 </sub>(g) + C<sub>5</sub>H<sub>12</sub> (l)</p><p>ΔHc°(C<sub>7</sub>H<sub>16 (l)</sub>) = −4817 kJ mol<sup>−1</sup></p><p>ΔHc°(C<sub>2</sub>H<sub>4 (g)</sub>) = −1411 kJ mol<sup>−1</sup></p><p>ΔHc°(C<sub>5</sub>H<sub>12</sub> <sub>(l)</sub>) = −3509 kJ mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> + 1411 + 3509 − 4817 </label></p><p><label class="radio"> <input type="radio"> <span>− 1411 − 3509 + 4817</span></label> </p><p><label class="radio"> <input type="radio"> <span>+ 1411 − 3509 − 4817</span></label> </p><p><label class="radio"> <input type="radio"> <span>+ 1411 + 3509 + 4817</span></label> </p></div><div class="q-explanation"><p>We are given ΔH combustion data so we construct the energy cycle by linking the reactants and products through the combustion products (arrows point down to combustion products):</p><p><img alt="" src="../../images/energetics-thermochemistry/hess-generic-comb.png" style="width: 240px; height: 103px;"></p><p>The enthalpy change for the reaction given (A) can then be found: The alternative route going via the combustion products is therefore A = B−C (−C because we are going against the arrow so the sign must be inverted).</p><p>All changes are for <strong>one mole </strong>of combustion, so the calculation is as follows:</p><p>B = −4817</p><p>C = (−1411)+(−3509) = −1411−3509</p><p>A = B−C = −4817−(−1411−3509) = −4817+1411+3509 which rearranged in the order given in the answer gives + 1411 + 3509 − 4817</p><p>The correct answer is therefore <strong>+ 1411 + 3509 − 4817</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">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which equation represents the P−H bond enthalpy in PH<sub>3</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>PH<sub>3(g)</sub> → P<sub>(g)</sub> + 1½H<sub>2(g)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>PH<sub>3(g) </sub>→ P<sub>(g)</sub> + 3H<sub>(g)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>PH<sub>3(g)</sub> → PH<sub>2(g)</sub> + H<sub>(g)</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> ⅓<span>PH<sub>3(g)</sub> → ⅓P<sub>(g)</sub> + H<sub>(g)</sub></span></label> </p></div><div class="q-explanation"><p>Bond enthalpy is defined as the <strong>average</strong> enthalpy required to break one mole of covalent bonds in the gaseous state.</p><p>So the best representation is going to be a chemical change showing the average breaking of one mole of P−H bonds in the gaseous state.</p><p>PH<sub>3(g)</sub> → PH<sub>2(g)</sub> + H<sub>(g)</sub> shows a single P−H bond breaking and by definition (stoichiometric coefficients) we have one mole of bonds breaking. But, when we break the bonds in any molecule one-by-one the bonds will all have different enthalpies as the environment of the bonds is changing.</p><p>Therefore we need to break <strong>all three bonds</strong> in PH<sub>3</sub> and find the <strong>average enthalpy</strong>:</p><p>⅓PH<sub>3(g)</sub> → ⅓P<sub>(g)</sub> + H<sub>(g) </sub>is thus the correct answer.</p><p><strong>Incorrect answers</strong></p><p>PH<sub>3(g) </sub>→ P<sub>(g)</sub> + 3H<sub>(g)</sub> and PH<sub>3(g)</sub> → P<sub>(g)</sub> + 1½H<sub>2(g) </sub>both show <strong>three </strong>moles of P−H bonds breaking, with the hydrogen products shown as atoms or molecules.</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>When equal masses of X and Z absorb the same amount of energy their temperatures rise by 10°C and 20°C respectively.</p><p>Which is correct?</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> The specific heat capacity of X is half that of Z</label></p><p><label class="radio"><input type="radio"> The specific heat capacity of X is ten times that of Z</label></p><p><label class="radio"><input class="c" type="radio"> The specific heat capacity of X is twice that of Z</label></p><p><label class="radio"><input type="radio"> The specific heat capacity of X is the same as Z</label></p></div><div class="q-explanation"><p>Specific heat capacity is the amount of energy needed (J) to raise one unit of mass (g) by one unit of temperature (K or °C - the interval/unit size is the same so either if fine). E.g. the specific heat capacity of water is 4.18 JK<sup>−1</sup>g<sup>−1</sup></p><p>In this question the same amount of energy raises X by 10°C and Z by 20°C. So it takes more energy to raise the temperature of X than Z. In fact, as the relationship between mass and energy needed to raise the temperature is linear, it must take twice as much energy to raise X by 20°C. So X must have twice the specific heat capacity of Z.</p><p><strong>The specific heat capacity of X is twice that of Y</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">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which is correct for the reaction?</p><p>2NO<sub>2</sub> → N<sub>2</sub>O<sub>4</sub> ΔH = −57.6 kJ mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>The products are less stable than the reactants and the reaction is endothermic.</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>The products are more stable than the reactants and the reaction is exothermic.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The products are less stable than the reactants and the reaction is exothermic.</span></label> </p><p><label class="radio"> <input type="radio"> <span>The products are more stable than the reactants and the reaction is endothermic.</span></label> </p></div><div class="q-explanation"><p>A negative ΔH indicates that the reaction is exothermic.</p><p>If a reaction is exothermic, then heat energy (enthaply) is transferred from the system to the surroundings. Thus the system loses energy; ΔH is negative, and the energy of the products is lower than the energy of the reactants; so the products are <strong>more stable </strong>than the reactants.</p><p>Thus<strong>, the products are more stable than the reactants and the reaction is exothermic</strong> is the correct answer.</p><p>On the left; exothermic - on the right; endothermic.</p><p><img alt="" src="../../images/energetics-thermochemistry/exo-endo.png" style="width: 486px; height: 270px;"></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>What is the enthalpy change for the incomplete combustion of ethene using average bond enthalpies given below.</p><p><span style="font-size: 15px;">C<sub>2</sub></span><span style="font-size: 15px;">H<sub>4 </sub></span>(g) + 2O<sub>2 </sub>(g) → 2CO (g) + 2H<sub>2</sub>O (g)</p><p>Bond enthalpies (in kJ mol<sup>−1</sup>) that you may require: C−H 414 / C=C 614 / O=O 498 / C≡O 1077 / O−H 463</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> [2(1077) + 2(463)] − [614 + 4(414) + 2(498)]</label></p><p><label class="radio"> <input type="radio"> [614 + 4(414) + 2(498)] − [2(1077) + 2(463)]</label></p><p><label class="radio"> <input class="c" type="radio"> [614 + 4(414) + 2(498)] − [2(1077) + 4(463)]</label></p><p><label class="radio"> <input type="radio"> [2(1077) + 4(463)] − [614 + 4(414) + 2(498)]</label></p></div><div class="q-explanation"><p>Total energy change = bonds broken − bonds made</p><p>Bonds present in reactants: 1×C=C, 4×C−H, 2×O=O</p><p>Bonds present in products: 2×C≡O, 4×O−H</p><p>Assuming all the bonds in reactants are broken and all bonds in products are made:</p><p>Enthalpy for bonds broken: 614 + 4(414) + 2(498)</p><p>Enthalpy for bonds made: 2(1077) + 4(463)</p><p>Thus the correct answer is<strong> [614 + 4(414) + 2(498)] − [2(1077) + 4(463)]</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>Using the standard enthalpies given below, what is the standard enthalpy of combustion of ethane in <strong>kJ mol<sup>−1</sup></strong>?</p><p style="text-align: center;">2C<sub>2</sub>H<sub>6</sub> (g) + 7O<sub>2</sub> (g) → 4CO<sub>2</sub> (g) + 6H<sub>2</sub>O (l)</p><p>C (s) + O<sub>2</sub> (g) → CO<sub>2</sub> (g) = <strong><em>X</em></strong> kJ mol<sup>−1</sup></p><p>H<sub>2</sub> (g) + ½O<sub>2</sub> (g) → H<sub>2</sub>O (l) = <strong><em>Y</em></strong> kJ mol<sup>−1</sup></p><p>2C (s) + 3H<sub>2</sub> (g) → C<sub>2</sub>H<sub>6</sub> (g) = <strong><em>Z</em></strong> kJ mol<sup>−1</sup></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> 2<em><strong>X</strong></em> + 3<em><strong>Y</strong></em> + <em><strong>Z</strong></em></label></p><p><label class="radio"><input type="radio"> 4<em><strong>X</strong></em> + 6<em><strong>Y</strong></em> + 2<em><strong>Z</strong></em></label></p><p><label class="radio"><input type="radio"> 4<em><strong>X</strong></em> + 6<em><strong>Y</strong></em> − 2<em><strong>Z</strong></em></label></p><p><label class="radio"><input class="c" type="radio"> 2<em><strong>X</strong></em> + 3<em><strong>Y</strong></em> − <em><strong>Z</strong></em></label></p></div><div class="q-explanation"><p>We are given ΔH formation data so we construct the energy cycle by linking the reactants and products through the elements (arrows point up from elements):</p><p><img alt="" src="../../images/energetics-thermochemistry/hess-generic-form.png" style="width:240px;height:100px;"></p><p>The enthalpy change for the reaction given (A) can then be found: The alternative route going via the elements is therefore A = −B+C (−B because we are going against the arrow so the sign must be inverted).</p><p>As <strong>written in the question</strong>, the changes are:</p><p>B = 2<strong><em>Z</em></strong> (O<sub>2</sub> is already an element so there is no change)</p><p>C = 4<em><strong>X</strong></em> + 6<em><strong>Y</strong></em></p><p>A = −B+C = −2<em><strong>Z</strong></em> + 4<strong><em>X</em></strong> + 6<em><strong>Y</strong></em></p><p>or rearranged: 4<em><strong>X</strong></em> + 6<em><strong>Y</strong></em> − 2<em><strong>Z</strong></em></p><p>But this is for <strong>two moles</strong> of ethane! The standard enthalpy of combustion is always <strong>per mole</strong> (hence units <strong>kJ mol<sup>−1</sup></strong>).</p><p>The correct answer is therefore 2<em><strong>X</strong></em> + 3<em><strong>Y</strong></em> − <em><strong>Z</strong></em></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>Two 50.0 cm<sup>3</sup> aqueous solutions containing 0.020 mol of hydrochloric acid and 0.020 mol of sodium hydroxide respectively are both at 298K.</p><p>When the two solutions are mixed the temperature rises by <i>y</i> °C.</p><p>Assume that the density of the final solution is 1.00 g cm<sup>−3</sup> and the specific heat capacity is 4.18 J K<sup>−1</sup> g<sup>−1.</sup></p><p>What is the enthalpy change of neutralisation in <strong>kJ mol<sup>−1</sup></strong>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({100 \times 4.18 \times y \over 0.020}\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({100 \times 4.18 \times (y+273) \over 1000 \times 0.040}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({100 \times 4.18 \times y \over 0.040}\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span class="math-tex">\({100 \times 4.18 \times y \over 1000 \times 0.020}\)</span></label></p></div><div class="q-explanation"><p>Using Q=mcΔT.</p><p>Q (heat energy) = mass × specific heat capacity × temperature change (these three values are for whatever substance is absorbing the heat energy; in this case, the final solution). The final solution is assumed to have a density of 1.00 g cm<sup>−3</sup> (so 100cm<sup>3</sup> is 100g; that is the two solutions poured together).</p><p>Q = 100 × 4.18 × <em>y<strong> </strong></em><strong>Joules</strong></p><p>The enthalpy change of neutralisation must be given <strong>per mole </strong>(of HCl or NaOH) <strong>and in kJ </strong>(not in Joules).</p><p>Dividing by 0.020 will give the value <strong>per mole</strong> of HCl (or NaOH), and dividing by 1000 will give the value in<strong> kJ</strong> <strong>mol<sup>−1</sup></strong>.</p><p>Thus the answer is <span class="math-tex">\({100 \times 4.18 \times y \over 1000 \times 0.020}\)</span> </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 expression gives the mass, in g, of propan-1-ol required to produce 790 kJ of heat energy upon complete combustion?</p><p>(M<sub>r</sub> for propan-1-ol = 60.1, ΔH<sup><s>o</s></sup>c = −2021 kJ mol<sup>−1</sup>)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({2021 \times 60.1 \over 790}\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({2021 \over 790 \times 60.1}\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({790 \over 2021 \times 60.1}\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span class="math-tex">\({790 \times 60.1 \over 2021}\)</span></label></p></div><div class="q-explanation"><p>One mole of propan-1-ol produces 2021 kJ of heat energy and has a molar mass of 60.1g.</p><p>The fraction of one mole needed to produce 790 kJ of energy will therefore be <span class="math-tex">\({790 \over 2021}\)</span></p><p>Therefore the mass of propan-1-ol needed will be the mole fraction × molar mass, which is <span class="math-tex">\({790 \over 2021}\)</span> × 60.1</p><p>Thus the answer is <span class="math-tex">\({790 \times 60.1 \over 2021}\)</span> </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>Why is the enthalpy change for this reaction calculated from bond enthalpy data less accurate than that calculated using standard enthalpies of formation?</p><p>C<sub>3</sub>H<sub>6</sub> (g) + 4½O<sub>2</sub> (g) → 3CO<sub>2</sub> (g) + 3H<sub>2</sub>O (g)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> Elements do not have standard enthalpies of formation</label></p><p><label class="radio"> <input type="radio"> All of the reactants and products are gases</label></p><p><label class="radio"> <input class="c" type="radio"> Bond enthalpies are average values of many compounds</label></p><p><label class="radio"> <input type="radio"> The equation is incorrectly balanced</label></p></div><div class="q-explanation"><p>Bond enthalpy is defined as the <strong>average</strong> enthalpy required to break one mole of covalent bonds in the <strong>gaseous </strong>state.</p><p>All of the species in this equation are gaseous, so that is not a problem for a bond enthalpy calculation. However, since bond enthalpies are average values, the enthalpy change calculated using bond enthalpies will not be accurate to this reaction.</p><p>Therefore the correct answer is <strong>Bond enthalpies are average values of many compounds</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">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>What can be deduced from the knowledge that ozone absorbs ultraviolet radiation of a greater wavelength than oxygen?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> The bonds between atoms in ozone are stronger than those in oxygen </label></p><p><label class="radio"> <input type="radio"> Only ozone can form free radicals </label></p><p><label class="radio"> <input type="radio"> Ozone absorbs radiation of a higher energy than oxygen </label></p><p><label class="radio"> <input class="c" type="radio"> The bonds between atoms in oxygen require more energy to break than those in ozone</label></p></div><div class="q-explanation"><p>Both oxygen and ozone will absorb ultraviolet radiation in the atmosphere and the bonds of both molecules may break to form oxygen free radicals. Ozone absorbs energy of a greater wavelength. Wavelength is inversely proportional to frequency and hence energy, so ozone absorbs radiation of a lower energy. </p><p>The double bond in oxygen (O<sub>2</sub>) is therefore stronger and requires more energy to break than the '1½' bonds in ozone (O<sub>3</sub>).</p><p>The correct answer is therefore <strong>The bonds between atoms in oxygen require more energy to break than those in ozone</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>What is the enthalpy of formation of carbon monoxide, in kJ mol<sup>−1</sup>, as represented by ΔH<sub>3</sub>?</p><p><img alt="" src="../../images/energetics-thermochemistry/hesseleveldiagram.png" style="width: 480px; height: 339px;"></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> 282−393</label></p><p><label class="radio"> <input type="radio"> 393+282</label></p><p><label class="radio"> <input type="radio"> 393−282 </label></p><p><label class="radio"> <input type="radio"> −393−282</label></p></div><div class="q-explanation"><p class="MsoNormal"><span calibri="" style="font-size:14.0pt;font-family:"></span></p><div><p>ΔH<sub>3 </sub>is the enthalpy change from elements to carbon monoxide. An alternative route is to go from elements to carbon dioxide and then back up to carbon dioxide. This alternative route 'goes with' the first arrow of the enthalpy change ΔH<sub>1</sub><span style="font-size: 11.25px;"> </span>so the sign (indicating the exothermic or endothermic nature) of the enthalpy change is unchanged. And this alternative route then 'goes against' the second arrow of the enthalpy change ΔH<sub>2</sub> so the sign (indicating the exothermic or endothermic nature) of the enthalpy change is inverted (minus-minus is a plus).</p><p>Therefore the enthalpy change is −393−(−282) = 282−393</p><p>Thus, the correct answer is <strong>282−393</strong></p></div><p class="MsoNormal"><span calibri="" style="font-size:14.0pt;font-family:"></span></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>5.85g of solid sodium chloride, NaCl(s), was added to water to form 25.0g of solution. The maximum decrease in temperature was 3.7°C.</p><p>What is the enthalpy change, in kJ mol<sup>−1</sup> for this process?</p><p>(Assume that the specific heat capacity of the solution is 4.18 J K<sup>−1</sup> g−1 and the molar mass of NaCl is 58.5 g mol<sup>−1</sup>.)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(\Delta H=- {25\times 4.18 \times 3.7 \over 1000 \times 0.1}\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(\Delta H=- {25\times 4.18 \times (3.7+273) \over 0.1}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(\Delta H=+ {25\times 4.18 \times 3.7 \over 0.1}\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span class="math-tex">\(\Delta H=+ {25\times 4.18 \times 3.7 \over 1000 \times 0.1}\)</span></label></p></div><div class="q-explanation"><p>Using Q=mcΔT.</p><p>Q (heat energy) = mass × specific heat capacity × temperature change (these three values are for whatever substance is absorbing the heat energy; in this case, the resulting solution).</p><p>Q = 25.0 × 4.18 × 3.7<strong> Joules</strong></p><p>The enthalpy change must be given <strong>per mole </strong><strong>and in kJ </strong>(not in Joules).</p><p>Moles of sodium chloride = <span class="math-tex">\({5.85 \over 58.5}\)</span>= 0.100 (or 0.1 for simplicity)</p><p>Dividing by 0.1 will give the value <strong>per mole</strong> of NaCl, and dividing by 1000 will give the value in<strong> kJ</strong> <strong>mol<sup>−1</sup></strong>.</p><p>The temperature decreases so the reaction is endothermic, and therefore ΔH is positive.</p><p>Thus the answer is <span class="math-tex">\(\Delta H=+ {25\times 4.18 \times 3.7 \over 1000 \times 0.1}\)</span> </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="2708"> <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>Energetics core (SL and HL) 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/2708/energetics-core-sl-and-hl-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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