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nuclear fission reactions</a></li><li><a href="../19429/solar-energy-1.html">Solar energy</a></li><li><a href="../19437/environmental-impact-global-warming-1.html">Environmental impact - global warming</a></li><li><a href="../19453/electrochemistry-rechargeable-batteries-fuel-cells-1.html">Electrochemistry, rechargeable batteries & fuel cells</a></li><li><a href="../19458/nuclear-fusion-fission-hl--1.html">Nuclear fusion & fission (HL) </a></li><li><a href="../19470/photovoltaic-cells-and-dsscs-1.html">Photovoltaic cells and DSSCs</a></li></ul></li><li><a href="../19078/option-d-1.html" class="father">Option D</a><ul class="level-2"><li><a href="../19079/pharmaceutical-products-drug-action--1.html">Pharmaceutical products & drug action </a></li><li><a href="../19158/aspirin-penicillin--1.html">Aspirin & penicillin </a></li><li><a href="../19174/opiates-1.html">Opiates</a></li><li><a href="../19184/ph-regulation-of-the-stomach-1.html">pH regulation of the stomach</a></li><li><a href="../19194/antiviral-medications--1.html">Antiviral medications </a></li><li><a href="../19211/environmental-impact-of-some-medications-1.html">Environmental impact of some medications</a></li><li><a href="../19254/taxol-a-chiral-auxiliary-case-study-1.html">Taxol - a chiral auxiliary case study</a></li><li><a href="../19258/nuclear-medicine--1.html">Nuclear medicine </a></li><li><a href="../19273/drug-detection-analysis--1.html">Drug detection & analysis </a></li></ul></li></ul></li><li><a href="../16592/practical-scheme-of-work-ia--1.html">Practical scheme of work & IA </a><ul class="level-1"><li><a href="../16682/internal-assessment-1.html" class="father">Internal Assessment</a><ul class="level-2"><li><a href="../18892/scaffolding-the-investigation-1.html">'Scaffolding' the investigation</a></li><li><a href="../18896/timing-organisation--1.html">Timing & organisation </a></li><li><a href="../18905/choosing-the-research-question-1.html">Choosing the research question</a></li><li><a href="../18946/personal-engagement-1.html">Personal engagement</a></li><li><a href="../18950/exploration-1.html">Exploration</a></li><li><a href="../18957/analysis-1.html">Analysis</a></li><li><a href="../18965/evaluation-1.html">Evaluation</a></li><li><a href="../18966/communication-1.html">Communication</a></li><li><a href="../19882/internal-standardization-of-the-ia-1.html">Internal standardization of the IA</a></li><li><a href="../19901/submitting-the-samples-for-moderation-2.html">Submitting the samples for moderation</a></li><li><a href="../33754/ten-suggestions-for-ias-using-secondary-data-1.html">Ten suggestions for IAs using secondary data</a></li><li><a href="../37544/gaining-full-marks-for-a-databased-ia--1.html">Gaining full marks for a databased IA </a></li><li><a href="../19506/ia-example-marking-exercise--1.html">IA example & marking exercise </a></li><li><a href="../23929/genuine-examples-of-moderated-ia-reports-1.html">Genuine examples of moderated IA reports</a></li><li><a href="../25234/examples-of-teacher-marked-ia-reports--1.html">Examples of teacher-marked IA reports </a></li><li><a href="../37542/history-of-internal-assessment-1.html">History of Internal Assessment</a></li></ul></li><li><a href="../16598/mandatory-laboratory-components-1.html" class="father">Mandatory laboratory components</a><ul class="level-2"><li><a href="../16613/formula-of-magnesium-oxide--1.html">Formula of magnesium oxide </a></li><li><a href="../16612/determining-the-mr-of-an-unknown-gas-1.html">Determining the <i>M</i><sub>r</sub> of an unknown gas</a></li><li><a href="../16615/acid-base-titrations--1.html">Acid-base titrations </a></li><li><a href="../16616/a-green-acid-base-practical-1.html">A green acid-base practical</a></li><li><a href="../16617/analysis-of-aspirin-tablets-1.html">Analysis of aspirin tablets</a></li><li><a href="../16618/caco3-in-egg-shells-1.html">CaCO<sub>3</sub> in egg shells</a></li><li><a href="../16644/enthalpy-changes-1.html">Enthalpy changes</a></li><li><a href="../16631/reaction-rates-1.html">Reaction rates</a></li><li><a href="../16635/rate-dependent-factors-1.html">Rate-dependent factors</a></li><li><a href="../16636/determining-ea-for-a-reaction-1.html">Determining <i>E</i><sub>a</sub> for a reaction</a></li><li><a href="../16637/iodination-of-propanone-1.html">Iodination of propanone</a></li><li><a href="../18144/titrations-with-a-ph-meter-1.html">Titrations with a pH meter</a></li><li><a href="../18355/redox-titration-with-kmno4--1.html">Redox titration with KMnO<sub>4</sub> </a></li><li><a href="../16640/voltaic-cells-1.html">Voltaic cells</a></li><li><a href="../16659/3-d-molecular-modelling--1.html">3-D molecular modelling </a></li></ul></li><li><a href="../17168/other-good-practicals-1.html" class="father">Other good practicals</a><ul class="level-2"><li><a href="../17196/common-chemical-reactions-1.html">Common chemical reactions</a></li><li><a href="../227/elements-oxides-of-the-third-period--1.html">Elements & oxides of the third period </a></li><li><a href="../17239/the-halogens-1.html">The halogens</a></li><li><a href="../17162/boiling-points-of-mixtures-1.html">Boiling points of mixtures</a></li><li><a href="../17170/polarity-of-molecules-1.html">Polarity of molecules</a></li><li><a href="../18083/le-chateliers-principle--1.html">Le Chatelier's principle </a></li><li><a href="../19582/determining-kc-for-an-esterification-reaction-1.html">Determining <i>K</i><sub>c</sub> for an esterification reaction</a></li><li><a href="../18164/redox-reactions-of-vanadium--1.html">Redox reactions of vanadium </a></li><li><a href="../18351/chlorine-in-swimming-pools--1.html">Chlorine in swimming pools </a></li><li><a href="../18178/analysis-of-cuii-ions-in-solution--1.html">Analysis of Cu(II) ions in solution </a></li><li><a href="../18179/percentage-of-copper-in-brass-1.html">Percentage of copper in brass</a></li><li><a href="../18356/electrolytic-cells--1.html">Electrolytic cells </a></li><li><a href="../18598/reactions-of-organic-compounds-1.html">Reactions of organic compounds</a></li><li><a href="../18752/hydrolysis-of-halogenoalkanes-1.html">Hydrolysis of halogenoalkanes</a></li><li><a href="../18784/preparation-of-13-dinitrobenzene--1.html">Preparation of 1,3-dinitrobenzene </a></li><li><a href="../19022/determination-of-an-organic-structure-1.html">Determination of an organic structure</a></li><li><a href="../19887/preparation-of-nylon-66-1.html">Preparation of nylon 6,6</a></li><li><a href="../19343/determination-of-vitamin-c-content--1.html">Determination of vitamin C content </a></li><li><a href="../19342/hydrolysis-of-starch--1.html">Hydrolysis of starch </a></li><li><a href="../19159/preparation-purification-of-aspirin--1.html">Preparation & purification of aspirin </a></li></ul></li><li><a href="../18851/ict-in-practical-work-1.html" class="father">ICT in practical work</a><ul class="level-2"><li><a href="../18852/data-logging--1.html">Data logging </a></li><li><a href="../18853/software-for-graph-plotting--1.html">Software for graph plotting </a></li><li><a href="../18854/spreadsheets-for-data-processing--1.html">Spreadsheets for data processing </a></li><li><a href="../18855/databases-1.html">Databases</a></li><li><a href="../18856/computer-modelling-simulations--1.html">Computer modelling & simulations </a></li></ul></li><li><a href="../16594/group-4-project-1.html" class="father">Group 4 Project</a><ul class="level-2"><li><a href="../16595/practicalities-1.html">Practicalities</a></li><li><a href="../16596/reflective-statement--2.html">Reflective statement </a></li></ul></li><li><a href="../16666/a-new-laboratory-1.html">A new laboratory?</a></li></ul></li><li class="selected"><a href="../16284/exams-1.html">Exams</a><ul class="level-1"><li><a href="../16286/essential-facts-1.html" class="father">Essential Facts</a><ul class="level-2"><li><a 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concepts</a></li><li><a href="../3507/why-tok-chemistry-1.html">Why TOK & Chemistry?</a></li><li><a href="../3505/a-modern-paradigm-1.html">A modern paradigm</a></li></ul></li></ul></li><li><a href="../17783/fast-track-to-tests-questions-1.html">Fast track to tests & questions</a><ul class="level-1"><li><a href="../17784/sl-multiple-choice-tests-on-individual-topics-1.html">SL Multiple choice tests on individual topics</a></li><li><a href="../24351/sl-multiple-choice-quizzes-on-sub-topics-1.html">SL Multiple choice 'quizzes' on sub-topics</a></li><li><a href="../31519/sl-practice-paper-1-exams-1.html">SL Practice Paper 1 exams</a></li><li><a href="../17786/sl-questions-on-each-sub-topic-1.html">SL Questions on each sub-topic</a></li><li><a href="../20431/sl-paper-3-section-a-questions-1.html">SL Paper 3 Section A questions</a></li><li><a href="../17792/sl-questions-on-the-options-1.html">SL Questions on the options</a></li><li><a href="../24897/sl-quizzes-on-option-sub-topics--1.html">SL Quizzes on option sub-topics </a></li><li><a href="../17785/hl-multiple-choice-tests-on-individual-topics-1.html">HL Multiple choice tests on individual topics</a></li><li><a href="../24352/hl-multiple-choice-quizzes-on-sub-topics-1.html">HL Multiple choice 'quizzes' on sub-topics</a></li><li><a href="../31520/hl-practice-paper-1-exams--1.html">HL Practice Paper 1 exams </a></li><li><a href="../17787/hl-questions-on-each-sub-topic--1.html">HL Questions on each sub-topic </a></li><li><a href="../20428/hl-paper-3-section-a-questions-1.html">HL Paper 3 Section A questions</a></li><li><a href="../17793/hl-questions-on-the-options-1.html">HL Questions on the options</a></li><li><a href="../24973/hl-quizzes-on-option-sub-topics-1.html">HL Quizzes on option sub-topics</a></li><li><a href="../41573/printable-versions-of-written-tasks--1.html">Printable versions of written tasks </a></li></ul></li><li><a 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</div><section id="main-content"><p class="MsoNormal">The importance of the relationship between the equilibrium constant and the balanced stoichiometric equilibrium reaction it relates to is stressed. What happens to the value of the equilibrium constant when the coefficients of the reactants and products are altered, the reaction is reversed or when reactions are combined is covered with examples. Finally a worked example demonstrate why the ability to manipulate known equilibrium constants is a useful tool for finding the value of an unknown equilibrium constant.</p><div class="blueBg"><h1><img alt="" height="21" src="../../../img/tib-icons/standard-level-32-1.png" title="" width="21"> <img alt="" height="21" src="../../../img/tib-icons/higher-level-32-1.png" title="" width="21"> Introduction</h1><p style="text-align: center;"><img alt="" class="noborder" src="../../../ib/chemistry/images/2014%20Exams/static-equilibrium(1).jpeg" style="width: 550px; height: 413px;" title="Static equilibrium at my local beach"></p><hr class="hidden"><p>Students often have difficulty with manipulating equilibrium constants. One of the reasons may be a lack of understanding as to what equilibrium actually represents.</p><p> A state of chemical equilibrium is reached in a closed system when the rate of the forward reaction is equal to the rate of the reverse reaction. That is for the reaction:</p><p class="MsoNormal"><span style="font-family:Arial"><em>w</em>A<span style="mso-spacerun:yes"> </span>+<span style="mso-spacerun:yes"> </span><em>x</em>B<span style="mso-spacerun:yes"> </span></span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> <span style="mso-spacerun:yes"> </span><em>y</em>C + <em>z</em>D</span></p><p class="MsoNormal"><span style="font-family:Arial">the rate at which A and B are reacting to form C and D is equal to the rate that C and D are reacting to form A and B. Unlike the image of static equilibrium above that I made from stones and driftwood on my local beach, chemical equilibrium is dynamic, i.e. when the position of equilibrium has been reached, the forward and reverse reactions are still occurring but there is no change in the concentrations of reactants and products. The IB is only concerned with homogenous equlibria, i.e. the reactants and products are all in the same phase and also the IB only requires equilibrium constants to be expressed in concentrations (rather than in terms of partial pressures).</span></p><p class="MsoNormal"><span style="font-family:Arial">For the above reaction the equilibrium law states</span></p><p class="MsoNormal"><em>K</em><sub>c</sub> = <img align="middle" alt="begin mathsize 12px style fraction numerator left square bracket C right square bracket to the power of y space x space left square bracket D right square bracket to the power of z over denominator left square bracket A right square bracket to the power of w space x space left square bracket B right square bracket to the power of x end fraction end style" class="Wirisformula" data-mathml="«math style=¨font-family:Arial¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨12px¨»«mfrac»«mrow»«mo»[«/mo»«mi»C«/mi»«msup»«mo»]«/mo»«mi»y«/mi»«/msup»«mo»§#xA0;«/mo»«mi»x«/mi»«mo»§#xA0;«/mo»«mo»[«/mo»«mi»D«/mi»«msup»«mo»]«/mo»«mi»z«/mi»«/msup»«/mrow»«mrow»«mo»[«/mo»«mi»A«/mi»«msup»«mo»]«/mo»«mi»w«/mi»«/msup»«mo»§#xA0;«/mo»«mi»x«/mi»«mo»§#xA0;«/mo»«mo»[«/mo»«mi»B«/mi»«msup»«mo»]«/mo»«mi»x«/mi»«/msup»«/mrow»«/mfrac»«/mstyle»«/math»" src="../../../ckeditor/plugins/wiris/integration/showimage-11.php?formula=fa12e9e8700ae0f7b6744dc99b24861d.png"> where <em>K</em><sub>c</sub> is known as the equilibrium constant at a stated temperature.</p><p class="MsoNormal"><span style="font-family:Arial">Although expressed in terms of concentrations, equilibrium constants have no units as technically the square brackets represent activities rather than concentrations and activities have no units. If the system is not in equilibrium then the reaction quotient, <i>Q</i> can be used. The reaction quotient measures the relative amount of products and reactants present during a reaction at a particular point in time by using the non-equilibrium concentrations in the equilibrium expression. </span></p><p class="MsoNormal"><span style="font-family:Arial">The definition of the equilibrium expression and the way in which it <strong>relates to a stated stoichiometric equation</strong> is important and underlies how equilibrium constants can be manipulated. </span></p><hr class="hidden"></div><hr class="hidden"><div class="yellowBg"><h2 class="MsoNormal"><span lang="EN-US" style="font-family:Arial;mso-ansi-language:
EN-US"><img alt="" height="21" src="../../../img/tib-icons/standard-level-32-1.png" title="" width="21"> <img alt="" height="21" src="../../../img/tib-icons/higher-level-32-1.png" title="" width="21"> The relationship between different equilibrium constants for the same reaction (at the same temperature) when represented by equations written in different ways.</span><span style="font-family:Arial"></span><br><br><span lang="EN-US" style="font-family:Arial;mso-ansi-language:
EN-US"></span></h2><h4 class="MsoNormal">1. Changing the coefficients</h4><p class="MsoNormal"><span style="font-family:Arial">Consider the equilibrium reaction between gaseous hydrogen and gaseous iodine to from gaseous hydrogen iodide.</span></p><p class="MsoNormal"><span style="font-family:Arial"><span style="mso-tab-count:
3"></span>H<sub>2</sub>(g) + I<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2HI(g)</span></p><p class="MsoNormal"><span style="font-family:Arial">By definition the equilibrium constant <i>K</i><sub>c</sub> for this reaction is defined by the equation</span></p><p class="MsoNormal"><em>K</em><sub>c</sub> = <img align="middle" alt="fraction numerator text [HI(g)] end text to the power of text 2 end text end exponent over denominator text [H end text subscript text 2 end text end subscript text (g)] x [I end text subscript text 2 end text end subscript text (g)] end text end fraction" class="Wirisformula" data-mathml="«math style=¨font-family:Arial¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«msup»«mtext»[HI(g)]«/mtext»«mtext»2«/mtext»«/msup»«mrow»«msub»«mtext»[H«/mtext»«mtext»2«/mtext»«/msub»«msub»«mtext»(g)]§#xA0;x§#xA0;[I«/mtext»«mtext»2«/mtext»«/msub»«mtext»(g)]«/mtext»«/mrow»«/mfrac»«/math»" src="../../../ckeditor/plugins/wiris/integration/showimage-12.php?formula=7f20335ec13f5dcc71699f9f9f278657.png"></p><p class="MsoNormal"><span style="font-family:Arial">If we had written the equation as ½H<sub>2</sub>(g) + ½I<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> HI(g) it is still the same reaction but now the new equilibrium constant, <i>K</i><sub>c</sub>’ is given by the expression</span></p><p class="MsoNormal"><em>K</em><sub>c</sub>' = <img align="middle" alt="fraction numerator text [HI(g)] end text over denominator text [H end text subscript text 2 end text end subscript text (g) end text right square bracket to the power of text 1/2 end text end exponent space x space text [I end text subscript text 2 end text end subscript text (g)] end text to the power of text 1/2 end text end exponent end fraction" class="Wirisformula" data-mathml="«math style=¨font-family:Arial¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mtext»[HI(g)]«/mtext»«mrow»«msub»«mtext»[H«/mtext»«mtext»2«/mtext»«/msub»«mtext»(g)«/mtext»«msup»«mo»]«/mo»«mtext»1/2«/mtext»«/msup»«mo»§#xA0;«/mo»«mi»x«/mi»«mo»§#xA0;«/mo»«msub»«mtext»[I«/mtext»«mtext»2«/mtext»«/msub»«msup»«mtext»(g)]«/mtext»«mtext»1/2«/mtext»«/msup»«/mrow»«/mfrac»«/math»" src="../../../ckeditor/plugins/wiris/integration/showimage-13.php?formula=43b1b71641ac5536b255556156a84596.png"></p><p class="MsoNormal">Hence <em>K</em><sub>c</sub>' = √<em>K</em><sub>c</sub></p><hr class="hidden"><h4 class="MsoNormal">2. Reversing the direction of the reaction</h4><p class="MsoNormal"><span style="font-family:Arial">If we look at the reverse reaction</span></p><p class="MsoNormal"><span style="font-family:Arial">2HI(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> H<sub>2</sub>(g) + I<sub>2</sub>(g) </span></p><p class="MsoNormal"><span style="font-family:Arial">The equilibrium constant <i>K</i><sub>c</sub>” for this reverse reaction<span style="mso-spacerun:yes"> </span>is given by the expression</span></p><p><span style="font-family:Arial"></span><i><span style="font-family:Arial">K</span></i><sub><span style="font-family:Arial">c</span></sub><span style="font-family:Arial">” = <img align="middle" alt="fraction numerator text [H end text subscript text 2 end text end subscript text (g)] x [I end text subscript text 2 end text end subscript text (g)] end text over denominator text [HI(g)] end text to the power of text 2 end text end exponent end fraction" class="Wirisformula" data-mathml="«math style=¨font-family:Arial¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«msub»«mtext»[H«/mtext»«mtext»2«/mtext»«/msub»«msub»«mtext»(g)]§#xA0;x§#xA0;[I«/mtext»«mtext»2«/mtext»«/msub»«mtext»(g)]§#xA0;«/mtext»«/mrow»«msup»«mtext»[HI(g)]«/mtext»«mtext»2«/mtext»«/msup»«/mfrac»«/math»" src="../../../ckeditor/plugins/wiris/integration/showimage-14.php?formula=a3873021d0c0f5cf0631e12caca9eb70.png"> </span></p><p class="MsoNormal"><span style="font-family:Arial">Hence<span style="mso-spacerun:yes"> </span><i>K</i><sub>c</sub>” = 1/<i>K</i><sub>c</sub>. </span></p><hr class="hidden"><p class="MsoNormal"><span style="font-family:Arial"></span></p><h4 class="MsoNormal">3. Combining chemical reactions</h4><p class="MsoNormal"><span style="font-family:Arial">A third relationship is when two (or more) equations can be combined to give a new equation. The equilibrium constant for the new reaction is equal to the product of the equilibrium constants for the two (or more) reactions that are being combined, i.e. for the combination of three equations <em>K</em><sub>c</sub> = <em>K</em><sub>c1</sub> x <em>K</em><sub>c2</sub> x <em>K</em><sub>c3</sub>. </span></p><p class="MsoNormal"><span style="font-family:Arial">For example if we use the equations given above for the reaction between hydrogen and iodine then if we take ½H<sub>2</sub>(g) + ½I<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> HI(g) for which <i>K<sub>c</sub>’</i> = √<i>K<sub>c</sub></i> and add it to another ½H<sub>2</sub>(g) + ½I<sub>2</sub>(g) </span><span style="font-family:
"Menlo Regular"">⇌</span><span style="font-family:Arial"> HI(g) we arrive back at the original equation of H<sub>2</sub>(g) + I<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2HI(g). The equilibrium constant for this reaction = <i>K<sub>c</sub>’</i> x <i>K<sub>c</sub>’</i> = √<i>K</i><sub>c</sub> x √<i>K</i><sub>c</sub> = <i>K</i><sub>c</sub>.</span></p><hr class="hidden"><p class="MsoNormal"><span style="font-family:Arial"></span></p><p class="MsoNormal">All three of these ways of manipulating equilibrium constants according to the stated reaction are covered in <span style="font-family:Arial"><a href="../18040/equilibrium--1.html" title="Core & AHL » Teaching each topic & sub-topic » Topics 7 & 17 » Equilibrium ">Topic 7 Equilibrium</a>. </span></p><hr class="hidden"><p class="MsoNormal"><span style="font-family:Arial"></span></p><h4 class="MsoNormal">Worked example</h4><p class="MsoNormal"><span style="font-family:Arial">The ability to utilise and manipulate equilibrium constants in this way can be useful to determine the value of an unknown equilibrium constant for a reaction from the values of the equilibrium constants for known reactions. For example: </span></p><p class="MsoNormal"><span style="font-family:Arial"><strong><span style="mso-spacerun:yes"></span>(a) </strong>The equilibrium constants for the following three reactions are given as <i>K</i><sub>c1</sub>, <i>K</i><sub>c2 </sub>and <i>K</i><sub>c3 </sub>respectively.</span></p><p class="MsoNormal"><span style="font-family:Arial">(1)<span style="mso-tab-count:
1"> </span><span style="mso-spacerun:yes"></span>2N<sub>2</sub>O(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2N<sub>2</sub> (g) + O<sub>2</sub>(g)<span style="mso-spacerun:yes"> </span><i>K</i><sub>c1</sub></span></p><p class="MsoNormal"><span style="font-family:Arial"><br>(2)<span style="mso-spacerun:yes"> </span>2NO<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> N<sub>2</sub>O<sub>4</sub>(g) <span style="mso-tab-count:2"> </span><span style="mso-spacerun:yes"> </span><span style="mso-spacerun:yes"> </span><i>K</i><sub>c2</sub></span></p><p class="MsoNormal"><span style="font-family:Arial"><br>(3)<span style="mso-spacerun:yes"> </span>N<sub>2</sub>(g) + 2O<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2NO<sub>2</sub>(g)<span style="mso-spacerun:yes"> </span><span style="mso-tab-count:1"> </span><span style="mso-spacerun:yes"></span><i>K</i><sub>c3</sub></span></p><p class="MsoNormal"><span style="font-family:Arial"></span><span style="font-family:Arial">Derive the expression for the equilibrium constant, <i>K</i><sub>c</sub> for the reaction</span></p><p class="MsoNormal"><span style="font-family:Arial"><span style="mso-spacerun:yes"></span>2N<sub>2</sub>O(g) + 3O<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2N<sub>2</sub>O<sub>4</sub>(g) </span></p><p class="MsoNormal"><span style="font-family:Arial">in terms of <i>K</i><sub>c1</sub>, <i>K</i><sub>c2 </sub>and <i>K</i><sub>c3</sub>.</span></p><p class="MsoNormal"><span style="font-family:Arial"></span></p><h4 class="MsoNormal"><span style="color:#FF0000;"><span style="font-family:Arial">Worked answer</span></span></h4><p class="MsoNormal"><span style="font-family:Arial">Double equations (2) and (3) and add then to equation (1) </span></p><p class="MsoNormal"><span style="font-family:Arial"><span style="mso-spacerun:yes"></span>2N<sub>2</sub>O(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2N<sub>2</sub> (g) + O<sub>2</sub>(g)<span style="mso-spacerun:yes"> </span><i>K</i><sub>c1</sub></span></p><p class="MsoNormal"><span style="font-family:Arial"><span style="mso-spacerun:yes"></span>4NO<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2N<sub>2</sub>O<sub>4</sub>(g) <span style="mso-tab-count:1"> </span><span style="mso-spacerun:yes"> </span><i>K</i><sub>c</sub>’ = (<i>K</i><sub>c2</sub>)<sup>2</sup></span></p><p class="MsoNormal"><span style="font-family:Arial"><span style="mso-spacerun:yes"></span>2N<sub>2</sub>(g) + 4O<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 4NO<sub>2</sub>(g)<span style="mso-spacerun:yes"> </span><i>K</i><sub>c</sub>” = <span style="mso-spacerun:yes"> </span>(<i>K</i><sub>c3</sub>)<sup>2</sup></span><span style="font-family:Arial"><span style="mso-spacerun:yes"></span></span></p><p class="MsoNormal"><span style="font-family:Arial"><span style="mso-spacerun:yes"></span>-------------------------------------</span><span style="font-family:Arial"><span style="mso-spacerun:yes"></span></span></p><p class="MsoNormal"><span style="font-family:Arial"><span style="mso-spacerun:yes"></span>2N<sub>2</sub>O(g) + 3O<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2N<sub>2</sub>O<sub>4</sub>(g)<span style="mso-spacerun:yes"> </span><i>K</i><sub>c</sub> = <i>K</i><sub>c1</sub> x (<i>K</i><sub>c2</sub>)<sup>2</sup> x (<i>K</i><sub>c3</sub>)<sup>2</sup></span></p><hr class="hidden"><p class="MsoNormal"><span style="font-family:Arial"></span></p><p><span style="font-family:Arial"><strong>(b) </strong>Using the following data calculate the value for the equilibrium constant at 298 K for the reaction</span></p><p><span style="font-family:Arial"> 2N<sub>2</sub>O(g) + 3O<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2N<sub>2</sub>O<sub>4</sub>(g)</span></p><p><img alt="" class="noborder" src="../../../ib/chemistry/images/2014%20Exams/values-of-kc.png" style="width: 550px; height: 88px;"></p><hr class="hidden"><h4><span style="color:#FF0000;"><span style="font-family:Arial">Worked answer</span></span><span style="font-family:Arial"></span></h4><p><span style="font-family:Arial">The 1<sup>st</sup> reaction is equation (1) reversed so <i>K</i><sub>c1</sub> = 1/1.2 x 10<sup>−35</sup> = 8.33 x 10<sup>34</sup></span></p><p class="MsoNormal"><span style="font-family:Arial">The 2<sup>nd</sup> reaction is equation (2) reversed so <span style="font-family:Arial"><i>K</i><sub>c2</sub> = 1/(</span><span style="font-family:Arial">4.6 x 10<sup>−3</sup></span>) and (<i>K</i><sub>c2</sub>)<sup>2</sup> = (1/4.6 x 10<sup>−3</sup>)<sup>2</sup> = 4.73 x 10<sup>4</sup></span></p><p class="MsoNormal"><span style="font-family:Arial">The 3<sup>rd</sup> reaction needs to be multiplied by 2 to give equation (3) so <i>K</i><sub>c3 = </sub>(4.1 x 10<sup>−9</sup>)<sup>2</sup> and (<i>K</i><sub>c3</sub>)<sup>2</sup> = (4.1 x 10<sup>−9</sup>)<sup>4 </sup>= 2.83 x 10<sup>−34</sup> <span style="mso-spacerun:yes"> </span></span></p><p class="MsoNormal"><span style="font-family:Arial">Hence <i>K</i><sub>c</sub> for 2N<sub>2</sub>O(g) + 3O<sub>2</sub>(g) </span><span style="font-family:"Menlo Regular"">⇌</span><span style="font-family:Arial"> 2N<sub>2</sub>O<sub>4</sub>(g)</span></p><p class="MsoNormal"><span style="font-family:Arial"> = <i>K</i><sub>c1</sub> x (<i>K</i><sub>c2</sub>)<sup>2</sup> x (<i>K</i><sub>c3</sub>)<sup>2</sup></span><span style="font-family:Arial"><span style="mso-tab-count:
5"> </span>= 8.33 x 10<sup>34 <span style="mso-spacerun:yes"> </span></sup>x<sup> </sup><span style="mso-spacerun:yes"> </span>4.73 x 10<sup>4 </sup><span style="mso-spacerun:yes"> </span>x 2.83 x 10<sup>−34</sup></span><span style="font-family:Arial"><span style="mso-tab-count:
5"> </span>= 1.1 x 10<sup>6</sup><span style="mso-tab-count:4"> </span></span></p><hr class="hidden"></div></section></article><section id="media-extras"><div class="page-actions no-print navbar inline hidden-desktop"><div class="navbar-inner"><ul class="nav"><li><a class="presentation" href="#" onclick="return false;"><i class="fa fa-desktop"></i></a></li><li><a class="print-section-blog" href="#" onclick="return false;"><i class="fa fa-print"></i></a></li><li><a class="page-bookmarker" data-ticket="IB Docs (2) Team" data-pid="32566" href="#" onclick="return false;"><i class="fa fa-star"></i></a></li><li><a class="personal-notes" href="#" onclick="return false;"><i class="fa fa-file-text"></i></a></li><li><a class="" data-toggle="modal" href="#modal-feedback" onclick="return false;"><i class="fa fa-envelope-o"></i></a></li><li class="dropdown"><a class="dropdown-toggle" data-toggle="dropdown" data-target="#" href="#"><i class="fa fa-share-alt"></i></a><ul class="dropdown-menu" role="menu"><li><a class="" target="_blank" title="Share on Twitter" href="https://twitter.com/share?text=%22Manipulating+equilibrium+constants+%22+-+via+%40InThinker+%23chemistry%0A&url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32566%2Fmanipulating-equilibrium-constants-"><i class="fa fa-twitter-square"></i><span>Twitter</span></a></li><li><a class="" target="_blank" title="Share on Facebook" href="https://www.facebook.com/sharer.php?u=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32566%2Fmanipulating-equilibrium-constants-&t=Manipulating+equilibrium+constants+"><i class="fa fa-facebook-square"></i><span>Facebook</span></a></li><li><a class="" href="http://www.linkedin.com/shareArticle?mini=true&url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32566%2Fmanipulating-equilibrium-constants-"><i class="fa fa-linkedin-square"></i><span>LinkedIn</span></a></li><li><a class="" href="https://plus.google.com/share?url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32566%2Fmanipulating-equilibrium-constants-"><i class="fa fa-google-plus-square"></i><span>Google +</span></a></li><li><a class="" href="mailto:?subject=Manipulating equilibrium constants &body=The importance of the relationship between the equilibrium constant and the balanced stoichiometric equilibrium reaction it relates to is stressed. What happens to the value of the equilibrium constant when the coefficients...%0Ahttps%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32566%2Fmanipulating-equilibrium-constants-"><i class="fa fa-envelope"></i><span>Email</span></a></li></ul></li><li><a class="" href="chemistry/teaching-materials"><i class="fa fa-puzzle-piece colored"></i></a></li></ul></div></div><div style="border-top: solid 1px #eee;border-bottom: solid 1px #eee;padding: 4px 0 4px 0;line-height: 1em;color: #666;font-size: .8em"><small><em>All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. Any unauthorised copying or posting of materials on other websites is an infringement of our copyright and could result in your account being blocked and legal action being taken against you.</em></small></div></section><section id="comments"></section></div><!-- /#main-column -->
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