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title="Suggested study time: 30 minutes"><i class="fa fa-clock-o"></i> 30'</span> </ol> <article id="main-article"> <p><strong>Topic 17.1</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="7-840-2750"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which is correct for a spontaneous reaction?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>ΔG<sup><s>o</s></sup> will be negative and <em>K<sub>c</sub></em> < 1</span></label> </p><p><label class="radio"> <input type="radio"> <span>ΔG<sup><s>o</s></sup> will be positive and <em>K<sub>c</sub></em> > 1</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>ΔG<sup><s>o</s></sup> will be negative and <em>K<sub>c</sub> </em>> 1</span></label> </p><p><label class="radio"> <input type="radio"> <span>ΔG<sup><s>o</s></sup> will be positive and <em>K<sub>c</sub> </em>< 1</span></label> </p></div><div class="q-explanation"><p>ΔG<sup><s>o</s></sup> = −<em>RT </em>ln<em>K</em></p><p>For a reaction to be spontaneous Gibb's free energy must be negative. This means that the equilibrium constant, <em>K</em>, must be greater than one (Natural log, ln, of a number less than 1 will be negative).</p><p><strong>ΔG<sup><s>o</s></sup> will be negative and K<sub>c</sub> > 1 </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">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which is true when this reaction is at equilibrium?</p><p>2NO<sub>2</sub> (g) ⇌ N<sub>2</sub>O<sub>4</sub> (g)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Entropy is at a minimum and Gibb's free energy is at a minimum </span></label> </p><p><label class="radio"> <input type="radio"> <span>Entropy is at a maximum and Gibb's free energy is at a maximum</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Entropy is at a maximum and Gibb's free energy is at a minimum</span></label> </p><p><label class="radio"> <input type="radio"> <span>Entropy is at a minimum and Gibb's free energy is at a maximum</span></label> </p></div><div class="q-explanation"><p>You should know that Gibb's free energy, ΔG<sup><s>o</s></sup>, is always negative for a spontaneous reaction, so we might expect the position of equilibrium to correspond to the minimum value of Gibb's free energy.</p><p>The total entropy of a system is inversely related to the Gibb's free energy, so when ΔG<sup><s>o</s></sup> is at a minimum, ΔS<sub>total</sub>, is at a maximum.</p><p>You need to learn that for any reaction at equilibrium: The position of equilibrium corresponds to a maximum value of entropy and a minimum value of Gibb's free energy.</p><p>Thus, <strong>Entropy is at a maximum and Gibb's free energy is at a minimum</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">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which is correct given that at temperature, <em>T, ΔG<sup><s>o</s></sup></em> = −12.9 kJmol<sup>−1</sup> at equilibrium.</p><p>N<sub>2</sub>O<sub>4 </sub>(g) ⇌ 2NO<sub>2 </sub>(g)</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><em>K<sub>c</sub> </em>< 1 and the equilibrium favours the reactants</span></label></p><p><label class="radio"> <input type="radio"> <span><em>K<sub>c</sub> </em>< 1 and the equilibrium favours the products</span></label> </p><p><label class="radio"> <input type="radio"> <span><em>K<sub>c</sub></em> > 1 and the equilibrium favours the reactants</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <em>K<sub>c</sub></em> > 1 and the equilibrium favours the products </label></p></div><div class="q-explanation"><p><em>ΔG<sup><s>o</s></sup> = −RTlnK</em></p><p><em>ΔG<sup><s>o</s></sup></em> is negative, so <em>K</em> must be greater than one (Natural log, ln, of a value less than one will give a positive number, which would mean that <em>ΔG<sup><s>o</s></sup></em> would be positive<em>).</em></p><p>And if <em>K </em>> 1, then the products must be favoured over the reactants (since products are on the top of the <em>Kc </em>expression).</p><p>The correct answer is therefore<strong> </strong><strong><em>K<sub>c</sub></em> > 1 and the equilibrium favours the products</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>Iodine and bromine gases were mixed and allowed to reach equilibrium. What is the value of the equilibrium constant?</p><p style="text-align: center;">I<sub>2 </sub>(g) + Br<sub>2 </sub>(g)<sub> </sub>⇌ 2IBr<sub> </sub>(g)</p><table border="0" cellpadding="0" cellspacing="0" style="width:100%;"><tbody><tr><td style="text-align: center;"> </td><td style="text-align: center;">[I<sub>2</sub>]</td><td style="text-align: center;"> </td><td style="text-align: center;">[Br<sub>2</sub>]</td><td style="text-align: center;"> </td><td style="text-align: center;">[IBr]</td></tr><tr><td style="text-align: center;">Initial concentration</td><td style="text-align: center;">0.40</td><td style="text-align: center;"> </td><td style="text-align: center;">0.40</td><td style="text-align: center;"> </td><td style="text-align: center;">0</td></tr><tr><td style="text-align: center;">Equilibrium concentration</td><td style="text-align: center;">0.20</td><td style="text-align: center;"> </td><td style="text-align: center;">0.20</td><td style="text-align: center;"> </td><td style="text-align: center;"><em>x</em></td></tr></tbody></table></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>4.0</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.40</span></label> </p><p><label class="radio"> <input type="radio"> 1.0 </label></p><p><label class="radio"> <input type="radio"> 10 </label></p></div><div class="q-explanation"><p>Using the initial-change-equilibrum (ICE) moles table (or RICE if you include the mole <strong>ratio</strong>):</p><table border="0" cellpadding="0" cellspacing="0" style="width:100%;"><tbody><tr><td style="text-align: center;"> </td><td style="text-align: center;">Br<sub>2(g)</sub></td><td style="text-align: center;">+</td><td style="text-align: center;">I<sub>2(g)</sub></td><td style="text-align: center;">⇌</td><td style="text-align: center;">2IBr<sub>(g)</sub></td></tr><tr><td style="text-align: center;">R</td><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">2</td></tr><tr><td style="text-align: center;">I</td><td style="text-align: center;">0.40</td><td style="text-align: center;"> </td><td style="text-align: center;">0.40</td><td style="text-align: center;"> </td><td style="text-align: center;">0</td></tr><tr><td style="text-align: center;">C</td><td style="text-align: center;"><strong>−0.20</strong></td><td style="text-align: center;"> </td><td style="text-align: center;"><strong>−0.20</strong></td><td style="text-align: center;"> </td><td style="text-align: center;"><strong>+0.40</strong></td></tr><tr><td style="text-align: center;">E</td><td style="text-align: center;">0.20</td><td style="text-align: center;"> </td><td style="text-align: center;">0.20</td><td style="text-align: center;"> </td><td style="text-align: center;"><strong>0.40</strong></td></tr></tbody></table><p>The non-bold information is given in the question.</p><p>The change in concentration (which is the same as the change in number of moles in this example, since the volume terms will all cancel) for I<sub>2</sub> and Br<sub>2</sub> is <strong>−0.20.</strong></p><p>The mole ratio for I<sub>2</sub> or Br<sub>2</sub> to IBr is 1:2, so 0.20 moles used up will produced <strong>+0.40</strong> moles of IBr.</p><p><span class="math-tex">\(K_c = {[IBr]^2 \over [I_2][Br_2]}\)</span></p><p><span class="math-tex">\(K_c = {0.40^2 \over 0.20 \times0.20}\)</span></p><p><span class="math-tex">\(K_c = {0.16 \over 0.04}\)</span>= 4</p><p>The correct answer is<strong> 4.0</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">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>1.0 mol of N<sub>2</sub> (g), 1.0 mol of H<sub>2 </sub>(g) and 1.0 mol of NH<sub>3</sub> (g) are sealed in a 1.0 dm<sup>3</sup> cylinder and left to reach equilibrium.</p><p>N<sub>2</sub> (g) + 3H<sub>2 </sub>(g) ⇌ 2NH<sub>3</sub> (g)</p><p>If the equilibrium concentration of N<sub>2</sub> (g) is 0.9 mol dm<sup>−3</sup>, what are the equilibrium concentrations of H<sub>2</sub> (g) and NH<sub>3</sub> (g) in mol dm<sup>−3</sup>?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> [H<sub>2</sub>] = 0.7 and [NH<sub>3</sub>] = 1.2 </label></p><p><label class="radio"> <input type="radio"> [H<sub>2</sub>] = 2.7 and [NH<sub>3</sub>] = 1.8 </label></p><p><label class="radio"> <input type="radio"> [H<sub>2</sub>] = 0.9 and [NH<sub>3</sub>] = 1.1<span></span></label> </p><p><label class="radio"> <input type="radio"> [H<sub>2</sub>] = 0.7 and [NH<sub>3</sub>] = 0.4<span></span></label> </p></div><div class="q-explanation"><p>Using the initial-change-equilibrum (ICE) moles table (or RICE if you include the mole <strong>ratio</strong>):</p><table border="0" cellpadding="0" cellspacing="0" style="width:100%;"><tbody><tr><td style="text-align: center;"> </td><td style="text-align: center;">N<sub>2 </sub>(g)</td><td style="text-align: center;">+</td><td style="text-align: center;">3H<sub>2</sub></td><td style="text-align: center;">⇌</td><td style="text-align: center;">2NH<sub>3 </sub>(g)</td></tr><tr><td style="text-align: center;">R</td><td style="text-align: center;">1</td><td style="text-align: center;">:</td><td style="text-align: center;">3</td><td style="text-align: center;">:</td><td style="text-align: center;">2</td></tr><tr><td style="text-align: center;">I</td><td style="text-align: center;">1.0</td><td style="text-align: center;"> </td><td style="text-align: center;">1.0</td><td style="text-align: center;"> </td><td style="text-align: center;">1.0</td></tr><tr><td style="text-align: center;">C</td><td style="text-align: center;"><strong>−0.1 </strong></td><td style="text-align: center;"> </td><td style="text-align: center;"><strong>−0.3</strong></td><td style="text-align: center;"> </td><td style="text-align: center;"><strong>+0.2</strong></td></tr><tr><td style="text-align: center;">E</td><td style="text-align: center;">0.9</td><td style="text-align: center;"> </td><td style="text-align: center;"><strong>0.7</strong></td><td style="text-align: center;"> </td><td style="text-align: center;"><strong>1.2</strong></td></tr></tbody></table><p>The non-bold information is given in the question.</p><p>The total volume is 1.0 dm<sup>3</sup>, so the moles values will also be the concentration values.</p><p>We have initial and equilibrium moles of N<sub>2</sub> so the<strong> 1st </strong>step is to calculate the change in N<sub>2</sub>: <strong>1.0 to 0.9 is a change of −0.1</strong></p><p>If we have the change in N<sub>2</sub> we can then calculate the change in H<sub>2</sub> (<strong>2nd</strong>): The mole ratio is 1:3, so the change in H<sub>2 </sub>will be <strong>−0.3</strong> and [H<sub>2</sub>] = <strong>0.7</strong>.</p><p>Lastly (<strong>3rd</strong>) we can calculate the change in moles of NH<sub>3</sub>. The mole ratio of N<sub>2</sub> : NH<sub>3</sub> is 1:2, so the change in NH<sub>3 </sub>will be <strong>+0.2</strong> and [NH<sub>3</sub>] = <strong>1.2</strong>.</p><p>Therefore the correct answer is<strong> [H<sub>2</sub>] = 0.7 and [NH<sub>3</sub>] = 1.2</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>Components X and Y were mixed and allowed to reach equilibrium. The relative concentrations of X, Y, W and Z at equilibrium are 2, 2, 1 and 4 respectively.</p><p>What is the value of the equilibrium constant?</p><p style="text-align: center;">2X + Y<sub> </sub>⇌<sub> </sub>W + 2Z</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> 2 </label></p><p><label class="radio"> <input type="radio"> <span>½</span></label> </p><p><label class="radio"> <input type="radio"> 1 </label></p><p><label class="radio"> <input type="radio"> 4 </label></p></div><div class="q-explanation"><p>Using the equilibrium expression:</p><p><span class="math-tex">\(K_c = {[W][Z]^2 \over [X]^2[Y]}\)</span></p><p><span class="math-tex">\(K_c = {1 \times4^2 \over 2 \times2^2}\)</span></p><p><span class="math-tex">\(K_c = {16 \over 8}\)</span>= 2</p><p>The correct answer is<strong> 2</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">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>1 mol of A and 1 mol of B were mixed and allowed to reach equilibrium.</p><p>What is the value of the equilibrium constant?</p><p style="text-align: center;">A + B<sub> </sub>⇌<sub> </sub>C + D</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> K<sub>c</sub> = <span class="math-tex">\( {x^2 \over (1-x)^2}\)</span></label></p><p><label class="radio"> <input type="radio"> <span>K<sub>c</sub> = <span class="math-tex">\( {(1-x)^2 \over x^2}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> K<sub>c</sub> = <span class="math-tex">\( {2x \over (1-x)^2}\)</span></label> </p><p><label class="radio"> <input type="radio"> K<sub>c</sub> = <span class="math-tex">\( {(1-x)^2 \over 2x}\)</span></label> </p></div><div class="q-explanation"><p>As the mole ratio for the equilibrium is 1 : 1 : 1 : 1, the moles of A and the moles of B that are used up, will be equal to the moles of C and moles of D that are produced. Hence, the relative concentrations of A, B, C and D at equilibrium are 1−<span class="math-tex">\(x\)</span>, 1−<span class="math-tex">\(x\)</span>, <span class="math-tex">\(x\)</span> and <span class="math-tex">\(x\)</span> respectively.</p><p>Using the equilibrium expression:</p><p><span class="math-tex">\(K_c = {[C][D] \over [A][B]}\)</span></p><p><span class="math-tex">\(K_c = {x.x \over (1-x)(1-x)}\)</span></p><p><span class="math-tex">\(K_c = {x^2 \over (1-x)^2}\)</span></p><p>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">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>Components J and K were mixed and allowed to reach equilibrium. The forward reaction is exothermic.</p><p>At 300K, the relative concentrations of J, K, L and M at equilibrium are 4, 1, 4 and 2 respectively.</p><p>What can be deduced about the value of the equilibrium constant at <strong>600K</strong>?</p><p style="text-align: center;">J + K<sub> </sub>⇌<sub> </sub>2L + M</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> K<sub>c</sub> < 8 </label></p><p><label class="radio"> <input type="radio"> <span>K<sub>c</sub> = 8</span></label> </p><p><label class="radio"> <input type="radio"> K<sub>c</sub> > 16 </label></p><p><label class="radio"> <input type="radio"> K<sub>c</sub> = 16</label></p></div><div class="q-explanation"><p>Using the equilibrium expression:</p><p><span class="math-tex">\(K_c = {[L]^2[M] \over [J][K]}\)</span></p><p><span class="math-tex">\(K_c = {4^2 \times2 \over 4 \times1}\)</span></p><p><span class="math-tex">\(K_c = {32 \over 4}\)</span>= 8 at 300K</p><p>But 600K is higher than 300K, and at higher temperature the equilibrium will favour the reverse (endothermic) direction which will decrease the value of the equilibrium constant.</p><p>The correct answer is therefore<strong> K<sub>c</sub> < 8</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="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="2750"> <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>Equilibrium AHL (HL only) paper 1 questions</strong> have you understood?</p><div class="slider-container text-center"><div 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