{"id":7686,"date":"2021-06-08T21:57:17","date_gmt":"2021-06-08T21:57:17","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/calculating-equilibrium-constant-values\/"},"modified":"2021-10-07T17:46:32","modified_gmt":"2021-10-07T17:46:32","slug":"calculating-equilibrium-constant-values","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/calculating-equilibrium-constant-values\/","title":{"raw":"Calculating Equilibrium Constant Values","rendered":"Calculating Equilibrium Constant Values"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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    \r\n \t
  1. Calculate equilibrium concentrations from the values of the initial amounts and the K<\/em>eq<\/sub>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nThere are some circumstances in which, given some initial amounts and the K<\/em>eq<\/sub>, you will have to determine the concentrations of all species when equilibrium is achieved. Such calculations are not difficult to do, especially if a consistent approach is applied. We will consider such an approach here.\r\n\r\nSuppose we have this simple equilibrium. Its associated K<\/em>eq<\/sub> is 4.0, and the initial concentration of each reactant is 1.0 M:\r\n

    [latex]\\begin{array}{rl}\r\n\\begin{array}{ccccc}\r\n\\ce{H2(g)}&+&\\ce{Cl2(g)}&\\rightleftharpoons &\\ce{2HCl(g)} \\\\\r\n1.0\\text{ M}&&1.0\\text{ M}&&\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=4.0\r\n\\end{array}[\/latex]<\/p>\r\nBecause we have concentrations for the reactants but not the products, we presume that the reaction will proceed in the forward direction to make products. But by how much will it proceed? We don\u2019t know, so let us assign it a variable. Let us assume that x<\/em> M H2<\/sub> reacts as the reaction goes to equilibrium. This means that at equilibrium, we have (1.0 \u2212 x<\/em>) M H2<\/sub> left over.\r\n\r\nAccording to the balanced chemical equation, H2<\/sub> and Cl2<\/sub> react in a 1:1 ratio. How do we know that? The coefficients of these two species in the balanced chemical equation are 1 (unwritten, of course). This means that if x<\/em> M H2<\/sub> reacts, x<\/em> M Cl2<\/sub> reacts as well. If we start with 1.0 M Cl2<\/sub> at the beginning and we react x<\/em> M, we have (1.0 \u2212 x<\/em>) M Cl2<\/sub> left at equilibrium.\r\n\r\nHow much HCl is made? We start with zero, but we also see that 2 mol of HCl are made for every mole of H2<\/sub> (or Cl2<\/sub>) that reacts (from the coefficients in the balanced chemical equation), so if we lose x<\/em> M H2<\/sub>, we gain 2x<\/em> M HCl. So now we know the equilibrium concentrations of our species:\r\n

    [latex]\\begin{array}{rl}\r\n\\begin{array}{ccccc}\r\n\\ce{H2(g)}&+&\\ce{Cl2(g)}&\\rightleftharpoons &\\ce{2HCl(g)} \\\\\r\n(1.0-x)\\text{M}&&(1.0-x)\\text{M}&&2x\\text{M}\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=4.0\r\n\\end{array}[\/latex]<\/p>\r\nWe can substitute these concentrations into the K<\/em>eq<\/sub> expression for this reaction and combine it with the known value of K<\/em>eq<\/sub>:\r\n

    [latex]K_{\\text{eq}}=\\dfrac{[\\ce{HCl}]^2}{[\\ce{H2}][\\ce{Cl2}]}=\\dfrac{(2x)^2}{(1-x)(1-x)}=4.0[\/latex]<\/p>\r\nThis is an equation in one variable, so we should be able to solve for the unknown value. This expression may look formidable, but first we can simplify the denominator and write it as a perfect square as well:\r\n

    [latex]\\dfrac{(2x)^2}{(1-x)^2}=4.0[\/latex]<\/p>\r\nThe fraction is a perfect square, as is the 4.0 on the right. So we can take the square root of both sides:\r\n

    [latex]\\dfrac{2x}{1-x}=2.0[\/latex]<\/p>\r\nNow we rearrange and solve (be sure you can follow each step):\r\n

    [latex]\\begin{array}{rcccc}\r\n2x&=&2.0&-&2x \\\\\r\n+2x&&&+&2x \\\\\r\n\\hline\r\n4x&=&2.0&& \\\\ \\\\\r\n\\dfrac{4x}{4}&=&\\dfrac{2.0}{4}&& \\\\ \\\\\r\nx&=&0.50&&\r\n\\end{array}[\/latex]<\/p>\r\nNow we have to remind ourselves what x<\/em> is \u2014 the amount of H2<\/sub> and Cl2<\/sub> that reacted \u2014 and 2x<\/em> is the equilibrium [HCl]. To determine the equilibrium concentrations, we need to go back and evaluate the expressions 1 \u2212 x<\/em> and 2x<\/em> to get the equilibrium concentrations of our species:\r\n

    [latex]\\begin{array}{rcccrcl}\r\n1.0 - x &=& 1.0 - 0.50&=&0.50\\text{ M}&=&[\\ce{H2}]=[\\ce{Cl2}] \\\\\r\n2x &=&2(0.50)&=&1.0\\text{ M}&=&[\\ce{HCl}]\r\n\\end{array}[\/latex]<\/p>\r\nThe units are assumed to be molarity. To check, we simply substitute these concentrations and verify that we get the numerical value of the K<\/em>eq<\/sub>, in this case 4.0:\r\n

    [latex]\\dfrac{(1.0)^2}{(0.50)(0.50)}=4.0[\/latex]<\/p>\r\nWe formalize this process by introducing the ICE chart, where ICE stands for initial, change, and equilibrium.\r\n\r\nThe initial values go in the first row of the chart. The change values, usually algebraic expressions because we do not yet know their exact numerical values, go in the next row. However, the change values must<\/em> be in the proper stoichiometric ratio as indicated by the balanced chemical equation. Finally, the equilibrium expressions in the last row are a combination of the initial value and the change value for each species. The expressions in the equilibrium row are substituted into the K<\/em>eq<\/sub> expression, which yields an algebraic equation that we try to solve.\r\n\r\nThe ICE chart for the above example would look like this:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nH2<\/sub>(g)<\/th>\r\n+<\/th>\r\nCl2<\/sub>(g)<\/th>\r\n\u21c4<\/th>\r\n2HCl(g)<\/th>\r\nK<\/em>eq<\/sub> = 4.0<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    I<\/td>\r\n1.0<\/td>\r\n<\/td>\r\n1.0<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    C<\/td>\r\n\u2212x<\/em><\/td>\r\n\u2212x<\/em><\/td>\r\n+2x<\/em><\/td>\r\n<\/tr>\r\n
    E<\/td>\r\n1.0 \u2212 x<\/em><\/td>\r\n1.0 \u2212 x<\/em><\/td>\r\n+2x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the last row into the expression for the K<\/em>eq<\/sub> yields\r\n

    [latex]K_{\\text{eq}}=\\dfrac{[\\ce{HCl}]^2}{[\\ce{H2}][\\ce{Cl2}]}=\\dfrac{(2x)^2}{(1-x)(1-x)}=4.0[\/latex]<\/p>\r\nwhich, of course, is the same expression we have already solved and yields the same answers for the equilibrium concentrations. The ICE chart is a more formalized way to do these types of problems. The +\u00a0sign is included explicitly in the change row of the ICE chart to avoid any confusion.\r\n\r\nSometimes when an ICE chart is set up and the K<\/em>eq<\/sub> expression is constructed, a more complex algebraic equation will result. One of the more common equations has an x<\/em>2<\/sup> term in it and is called a quadratic equation<\/em>. There will be two values possible for the unknown x<\/em>, and for a quadratic equation with the general formula ax<\/em>2<\/sup> +\u00a0bx<\/em> +\u00a0c<\/em> = 0 (where a<\/em>, b<\/em>, and c<\/em> are the coefficients<\/em> of the quadratic equation), the two possible values are as follows:\r\n

    [latex]x=\\dfrac{-b\\pm \\sqrt{b^2-4ac}}{2a}[\/latex]<\/p>\r\nOne value of x<\/em> is the +\u00a0sign used in the numerator, and the other value of x<\/em> is the \u2212 sign used in the numerator. In this case, one value of x<\/em> typically makes no sense as an answer and can be discarded as physically impossible, leaving only one possible value and the resulting set of concentrations. Example 13.9 illustrates this.\r\n

    \r\n

    Example 13.9<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nSet up an ICE chart and solve for the equilibrium concentrations in this chemical reaction.\r\n

    [latex]\\begin{array}{rl}\r\n\\begin{array}{ccccc}\r\n\\ce{COI2(g)}&\\leftrightharpoons&\\ce{CO(g)}&+&\\ce{I2(g)} \\\\\r\n0.55\\text{ M}&&0&&0\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=0.00088\r\n\\end{array}[\/latex]<\/p>\r\nSolution<\/em>\r\nThe ICE chart is set up like this. First, the initial values:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nCOI2<\/sub>(g)<\/th>\r\n\u21c4<\/th>\r\nCO(g)<\/th>\r\n+<\/th>\r\nI2<\/sub>(g)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    I<\/td>\r\n0.55<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    C<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    E<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSome of the COI2<\/sub> will be lost, but how much? We don\u2019t know, so we represent it by the variable x<\/em>. So x<\/em> M COI2<\/sub> will be lost, and for each COI2<\/sub> that is lost, x<\/em> M CO and x<\/em> M I2<\/sub> will be produced. These expressions go into the change row:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nCOI2<\/sub>(g)<\/th>\r\n\u21c4<\/th>\r\nCO(g)<\/th>\r\n+<\/th>\r\nI2<\/sub>(g)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    I<\/td>\r\n0.55<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    C<\/td>\r\n\u2212x<\/em><\/td>\r\n+x<\/em><\/td>\r\n+x<\/em><\/td>\r\n<\/tr>\r\n
    E<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAt equilibrium, the resulting concentrations will be a combination of the initial amount and the changes:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nCOI2<\/sub>(g)<\/th>\r\n\u21c4<\/th>\r\nCO(g)<\/th>\r\n+<\/th>\r\nI2<\/sub>(g)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    I<\/td>\r\n0.55<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    C<\/td>\r\n\u2212x<\/em><\/td>\r\n+x<\/em><\/td>\r\n+x<\/em><\/td>\r\n<\/tr>\r\n
    E<\/td>\r\n0.55 \u2212 x<\/em><\/td>\r\n+x<\/em><\/td>\r\n+x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe expressions in the equilibrium row go into the K<\/em>eq<\/sub> expression:\r\n

    [latex]K_{\\text{eq}}=\\dfrac{[\\ce{CO}][\\ce{I2}]}{\\ce{COI2}}=0.00088=\\dfrac{(x)(x)}{(0.55-x)}[\/latex]<\/p>\r\nWe rearrange this into a quadratic equation that equals 0:\r\n

    [latex]\\begin{array}{rcl}\r\n(0.55-x)0.00088&=&\\dfrac{(x)(x)}{(\\cancel{0.55-x})}\\times (\\cancel{0.55-x}) \\\\ \\\\\r\n0.000484-0.00088x&=&x^2 \\\\\r\nx^2+0.00088x-0.000484&=&0\r\n\\end{array}[\/latex]<\/p>\r\nNow we use the quadratic equation to solve for the two possible values of x<\/em>:\r\n

    [latex]x=\\dfrac{-0.00088\\pm \\sqrt{(0.00088)^2-4(1)(-0.000484)}}{2(1)}[\/latex]<\/p>\r\nEvaluate for both signs in the numerator \u2014 first the + sign and then the \u2212 sign:\r\n

    [latex]x=0.0216\\text{ or }x=-0.0224[\/latex]<\/p>\r\nBecause x<\/em> is the final concentration of both CO and I2<\/sub>, it cannot be negative, so we discount the second numerical answer as impossible. Thus x<\/em> = 0.0216.\r\n\r\nGoing back to determine the final concentrations using the expressions in the E row of our ICE chart, we have:\r\n

    [latex]\\begin{align*}\r\n[\\ce{COI2}]&=0.55-x=0.55-0.0216=0.53\\text{ M} \\\\\r\n{[\\ce{CO}]}&=x=0.0216\\text{ M} \\\\\r\n{[\\ce{I2}]}&=x=0.0216\\text{ M}\r\n\\end{align*}[\/latex]<\/p>\r\nYou can verify that these numbers are correct by substituting them into the K<\/em>eq<\/sub> expression and evaluating and comparing to the known K<\/em>eq<\/sub> value.\r\n\r\nTest Yourself<\/em>\r\nSet up an ICE chart and solve for the equilibrium concentrations in this chemical reaction.\r\n

    [latex]\\begin{array}{rl}\r\n\\begin{array}{ccccc}\r\n\\ce{N2H2(g)}&\\leftrightharpoons&\\ce{N2(g)}&+&\\ce{H2(g)} \\\\\r\n0.075\\text{ M}&&0&&0\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=0.052\r\n\\end{array}[\/latex]<\/p>\r\nAnswer<\/em>\r\nThe completed ICE chart is as follows:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nN2<\/sub>H2<\/sub>(g)<\/th>\r\n\u21c4<\/th>\r\nN2<\/sub>(g)<\/th>\r\n+<\/th>\r\nH2<\/sub>(g)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    I<\/td>\r\n0.075<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
    C<\/td>\r\n\u2212x<\/em><\/td>\r\n+x<\/em><\/td>\r\n+x<\/em><\/td>\r\n<\/tr>\r\n
    E<\/td>\r\n0.075 \u2212 x<\/em><\/td>\r\n+x<\/em><\/td>\r\n+x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSolving for x<\/em> gives the equilibrium concentrations as [N2<\/sub>H2<\/sub>] = 0.033 M; [N2<\/sub>] = 0.042 M; and [H2<\/sub>] = 0.042 M\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

    \r\n
      \r\n \t
    • An ICE chart is a convenient way to determine equilibrium concentrations from starting amounts.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n
      \r\n

      Exercises<\/p>\r\n\r\n<\/header>\r\n

      \r\n

      Questions<\/h1>\r\n
        \r\n \t
      1. Describe the three parts of an ICE chart.<\/li>\r\n \t
      2. What is the relationship between the equilibrium row in an ICE chart and the other two rows?<\/li>\r\n \t
      3. Set up (but do not solve) an ICE chart for this reaction, given the initial conditions.\r\n

        [latex]\\begin{array}{ccc}\r\n\\ce{3O2(g)}&\\leftrightharpoons&\\ce{2O3(g)} \\\\\r\n0.075\\text{ M}&&\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

      4. Set up (but do not solve) an ICE chart for this reaction, given the initial conditions.\r\n

        [latex]\\begin{array}{ccccccc}\r\n\\ce{CH4(g)}&+&\\ce{2O2(g)}&\\leftrightharpoons&\\ce{CO2(g)}&+&\\ce{2H2O(g)} \\\\\r\n0.750\\text{ M}&&0.450\\text{ M}&&&&\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

      5. Given that pure solids and liquids do not appear in K<\/em>eq<\/sub> expressions, set up the ICE chart for this reaction, given the initial conditions.\r\n

        [latex]\\begin{array}{ccccccc}\r\n\\ce{CH4(g)}&+&\\ce{2O2(g)}&\\leftrightharpoons&\\ce{CO2(g)}&+&\\ce{2H2O(\\ell)} \\\\\r\n0.0060\\text{ M}&&0.055\\text{ M}&&&&\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

      6. Given that pure solids and liquids do not appear in K<\/em>eq<\/sub> expressions, set up the ICE chart for this reaction, given the initial conditions.\r\n

        [latex]\\begin{array}{ccccccc}\r\n\\ce{N2H4(\\ell)}&+&\\ce{O2(g)}&\\leftrightharpoons&\\ce{N2(g)}&+&\\ce{2H2O(\\ell)} \\\\\r\n2.33\\text{ M}&&1.09\\text{ M}&&&&\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

      7. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\r\n

        [latex]\\begin{array}{rl}\r\n\\begin{array}{ccc}\r\n\\ce{HCN(g)}&\\leftrightharpoons&\\ce{HNC(g)} \\\\\r\n2.00\\text{ M}&&\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=4.50\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

      8. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\r\n

        [latex]\\begin{array}{rl}\r\n\\begin{array}{ccccc}\r\n\\ce{IF3(g)}&+&\\ce{F2(g)}&\\leftrightharpoons&\\ce{IF5(g)} \\\\\r\n1.0\\text{ M}&&0.50\\text{ M}&&\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=7.59\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

      9. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\r\n

        [latex]\\begin{array}{rl}\r\n\\begin{array}{ccccc}\r\n\\ce{N2O3(g)}&\\leftrightharpoons&\\ce{NO(g)}&+&\\ce{NO2(g)} \\\\\r\n0.0663\\text{ M}&&&&\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=2.50\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

      10. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\r\n

        [latex]\\begin{array}{rl}\r\n\\begin{array}{ccccccc}\r\n\\ce{CO(g)}&+&\\ce{H2O(g)}&\\leftrightharpoons&\\ce{CO2(g)}&+&\\ce{H2(g)} \\\\\r\n0.750\\text{ M}&&0.750\\text{ M}&&&&\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=16.0\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

      11. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\r\n

        [latex]\\begin{array}{rl}\r\n\\begin{array}{ccccc}\r\n\\ce{H2S(g)}&\\leftrightharpoons&\\ce{H2(g)}&+&\\ce{S(s)} \\\\\r\n0.882\\text{ M}&&&&\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=0.055\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

      12. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\r\n

        [latex]\\begin{array}{rl}\r\n\\begin{array}{ccccccc}\r\n\\ce{AgCl2(s)}&+&\\ce{F2(g)}&\\leftrightharpoons&\\ce{2AgF(s)}&+&\\ce{Cl2(g)} \\\\\r\n&&1.99\\text{ M}&&&&\r\n\\end{array}\r\n&\\hspace{2em}\r\nK_{\\text{eq}}=1.2\\times 10^2\r\n\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n<\/ol>\r\n

        Answers<\/h1>\r\n
          \r\n \t
        1. I = initial concentrations; C = change in concentrations; E = equilibrium concentrations<\/li>\r\n<\/ol>\r\n
            \r\n \t
          1. \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
            <\/th>\r\n3O2<\/sub><\/th>\r\n\u21c4<\/span><\/th>\r\n2O3<\/sub><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
            I<\/td>\r\n0.075<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
            C<\/td>\r\n\u22123x<\/em><\/td>\r\n+2x<\/em><\/td>\r\n<\/tr>\r\n
            E<\/td>\r\n0.075 \u2212 3x<\/em><\/td>\r\n+2x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n<\/ol>\r\n
              \r\n \t
            1. \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
              <\/th>\r\nCH4<\/sub><\/th>\r\n+<\/th>\r\n2O2<\/sub><\/th>\r\n\u21c4<\/span><\/th>\r\nCO2<\/sub><\/th>\r\n+<\/th>\r\n2H2<\/sub>O<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
              I<\/td>\r\n0.0060<\/td>\r\n<\/td>\r\n0.055<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
              C<\/td>\r\n\u2212x<\/em><\/td>\r\n\u22122x<\/em><\/td>\r\n+x<\/em><\/td>\r\n\u2014<\/td>\r\n<\/tr>\r\n
              E<\/td>\r\n0.0060 \u2212 x<\/em><\/td>\r\n0.055 \u2212 2x<\/em><\/td>\r\n+x<\/em><\/td>\r\n\u2014<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n<\/ol>\r\n
                \r\n \t
              1. [HCN] = 0.364 M; [HNC] = 1.64 M<\/li>\r\n<\/ol>\r\n
                  \r\n \t
                1. [N2<\/sub>O3<\/sub>] = 0.0017 M; [NO] = [NO2<\/sub>] = 0.0646 M<\/li>\r\n<\/ol>\r\n
                    \r\n \t
                  1. [H2<\/sub>S] = 0.836 M; [H2<\/sub>] = 0.046 M<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>","rendered":"
                    \n
                    \n

                    Learning Objectives<\/p>\n<\/header>\n

                    \n
                      \n
                    1. Calculate equilibrium concentrations from the values of the initial amounts and the K<\/em>eq<\/sub>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                      There are some circumstances in which, given some initial amounts and the K<\/em>eq<\/sub>, you will have to determine the concentrations of all species when equilibrium is achieved. Such calculations are not difficult to do, especially if a consistent approach is applied. We will consider such an approach here.<\/p>\n

                      Suppose we have this simple equilibrium. Its associated K<\/em>eq<\/sub> is 4.0, and the initial concentration of each reactant is 1.0 M:<\/p>\n

                      \"\begin{array}{rl}<\/p>\n

                      Because we have concentrations for the reactants but not the products, we presume that the reaction will proceed in the forward direction to make products. But by how much will it proceed? We don\u2019t know, so let us assign it a variable. Let us assume that x<\/em> M H2<\/sub> reacts as the reaction goes to equilibrium. This means that at equilibrium, we have (1.0 \u2212 x<\/em>) M H2<\/sub> left over.<\/p>\n

                      According to the balanced chemical equation, H2<\/sub> and Cl2<\/sub> react in a 1:1 ratio. How do we know that? The coefficients of these two species in the balanced chemical equation are 1 (unwritten, of course). This means that if x<\/em> M H2<\/sub> reacts, x<\/em> M Cl2<\/sub> reacts as well. If we start with 1.0 M Cl2<\/sub> at the beginning and we react x<\/em> M, we have (1.0 \u2212 x<\/em>) M Cl2<\/sub> left at equilibrium.<\/p>\n

                      How much HCl is made? We start with zero, but we also see that 2 mol of HCl are made for every mole of H2<\/sub> (or Cl2<\/sub>) that reacts (from the coefficients in the balanced chemical equation), so if we lose x<\/em> M H2<\/sub>, we gain 2x<\/em> M HCl. So now we know the equilibrium concentrations of our species:<\/p>\n

                      \"\begin{array}{rl}<\/p>\n

                      We can substitute these concentrations into the K<\/em>eq<\/sub> expression for this reaction and combine it with the known value of K<\/em>eq<\/sub>:<\/p>\n

                      \"K_{\text{eq}}=\dfrac{[\ce{HCl}]^2}{[\ce{H2}][\ce{Cl2}]}=\dfrac{(2x)^2}{(1-x)(1-x)}=4.0\"<\/p>\n

                      This is an equation in one variable, so we should be able to solve for the unknown value. This expression may look formidable, but first we can simplify the denominator and write it as a perfect square as well:<\/p>\n

                      \"\dfrac{(2x)^2}{(1-x)^2}=4.0\"<\/p>\n

                      The fraction is a perfect square, as is the 4.0 on the right. So we can take the square root of both sides:<\/p>\n

                      \"\dfrac{2x}{1-x}=2.0\"<\/p>\n

                      Now we rearrange and solve (be sure you can follow each step):<\/p>\n

                      \"\begin{array}{rcccc}<\/p>\n

                      Now we have to remind ourselves what x<\/em> is \u2014 the amount of H2<\/sub> and Cl2<\/sub> that reacted \u2014 and 2x<\/em> is the equilibrium [HCl]. To determine the equilibrium concentrations, we need to go back and evaluate the expressions 1 \u2212 x<\/em> and 2x<\/em> to get the equilibrium concentrations of our species:<\/p>\n

                      \"\begin{array}{rcccrcl}<\/p>\n

                      The units are assumed to be molarity. To check, we simply substitute these concentrations and verify that we get the numerical value of the K<\/em>eq<\/sub>, in this case 4.0:<\/p>\n

                      \"\dfrac{(1.0)^2}{(0.50)(0.50)}=4.0\"<\/p>\n

                      We formalize this process by introducing the ICE chart, where ICE stands for initial, change, and equilibrium.<\/p>\n

                      The initial values go in the first row of the chart. The change values, usually algebraic expressions because we do not yet know their exact numerical values, go in the next row. However, the change values must<\/em> be in the proper stoichiometric ratio as indicated by the balanced chemical equation. Finally, the equilibrium expressions in the last row are a combination of the initial value and the change value for each species. The expressions in the equilibrium row are substituted into the K<\/em>eq<\/sub> expression, which yields an algebraic equation that we try to solve.<\/p>\n

                      The ICE chart for the above example would look like this:<\/p>\n\n\n\n\n\n\n\n
                      <\/th>\nH2<\/sub>(g)<\/th>\n+<\/th>\nCl2<\/sub>(g)<\/th>\n\u21c4<\/th>\n2HCl(g)<\/th>\nK<\/em>eq<\/sub> = 4.0<\/th>\n<\/tr>\n<\/thead>\n
                      I<\/td>\n1.0<\/td>\n<\/td>\n1.0<\/td>\n<\/td>\n0<\/td>\n<\/td>\n<\/tr>\n
                      C<\/td>\n\u2212x<\/em><\/td>\n\u2212x<\/em><\/td>\n+2x<\/em><\/td>\n<\/tr>\n
                      E<\/td>\n1.0 \u2212 x<\/em><\/td>\n1.0 \u2212 x<\/em><\/td>\n+2x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      Substituting the last row into the expression for the K<\/em>eq<\/sub> yields<\/p>\n

                      \"K_{\text{eq}}=\dfrac{[\ce{HCl}]^2}{[\ce{H2}][\ce{Cl2}]}=\dfrac{(2x)^2}{(1-x)(1-x)}=4.0\"<\/p>\n

                      which, of course, is the same expression we have already solved and yields the same answers for the equilibrium concentrations. The ICE chart is a more formalized way to do these types of problems. The +\u00a0sign is included explicitly in the change row of the ICE chart to avoid any confusion.<\/p>\n

                      Sometimes when an ICE chart is set up and the K<\/em>eq<\/sub> expression is constructed, a more complex algebraic equation will result. One of the more common equations has an x<\/em>2<\/sup> term in it and is called a quadratic equation<\/em>. There will be two values possible for the unknown x<\/em>, and for a quadratic equation with the general formula ax<\/em>2<\/sup> +\u00a0bx<\/em> +\u00a0c<\/em> = 0 (where a<\/em>, b<\/em>, and c<\/em> are the coefficients<\/em> of the quadratic equation), the two possible values are as follows:<\/p>\n

                      \"x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\"<\/p>\n

                      One value of x<\/em> is the +\u00a0sign used in the numerator, and the other value of x<\/em> is the \u2212 sign used in the numerator. In this case, one value of x<\/em> typically makes no sense as an answer and can be discarded as physically impossible, leaving only one possible value and the resulting set of concentrations. Example 13.9 illustrates this.<\/p>\n

                      \n
                      \n

                      Example 13.9<\/p>\n<\/header>\n

                      \n

                      Set up an ICE chart and solve for the equilibrium concentrations in this chemical reaction.<\/p>\n

                      \"\begin{array}{rl}<\/p>\n

                      Solution<\/em>
                      \nThe ICE chart is set up like this. First, the initial values:<\/p>\n\n\n\n\n\n\n\n
                      <\/th>\nCOI2<\/sub>(g)<\/th>\n\u21c4<\/th>\nCO(g)<\/th>\n+<\/th>\nI2<\/sub>(g)<\/th>\n<\/tr>\n<\/thead>\n
                      I<\/td>\n0.55<\/td>\n<\/td>\n0<\/td>\n<\/td>\n0<\/td>\n<\/tr>\n
                      C<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
                      E<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      Some of the COI2<\/sub> will be lost, but how much? We don\u2019t know, so we represent it by the variable x<\/em>. So x<\/em> M COI2<\/sub> will be lost, and for each COI2<\/sub> that is lost, x<\/em> M CO and x<\/em> M I2<\/sub> will be produced. These expressions go into the change row:<\/p>\n\n\n\n\n\n\n\n
                      <\/th>\nCOI2<\/sub>(g)<\/th>\n\u21c4<\/th>\nCO(g)<\/th>\n+<\/th>\nI2<\/sub>(g)<\/th>\n<\/tr>\n<\/thead>\n
                      I<\/td>\n0.55<\/td>\n<\/td>\n0<\/td>\n<\/td>\n0<\/td>\n<\/tr>\n
                      C<\/td>\n\u2212x<\/em><\/td>\n+x<\/em><\/td>\n+x<\/em><\/td>\n<\/tr>\n
                      E<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      At equilibrium, the resulting concentrations will be a combination of the initial amount and the changes:<\/p>\n\n\n\n\n\n\n\n
                      <\/th>\nCOI2<\/sub>(g)<\/th>\n\u21c4<\/th>\nCO(g)<\/th>\n+<\/th>\nI2<\/sub>(g)<\/th>\n<\/tr>\n<\/thead>\n
                      I<\/td>\n0.55<\/td>\n<\/td>\n0<\/td>\n<\/td>\n0<\/td>\n<\/tr>\n
                      C<\/td>\n\u2212x<\/em><\/td>\n+x<\/em><\/td>\n+x<\/em><\/td>\n<\/tr>\n
                      E<\/td>\n0.55 \u2212 x<\/em><\/td>\n+x<\/em><\/td>\n+x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      The expressions in the equilibrium row go into the K<\/em>eq<\/sub> expression:<\/p>\n

                      \"K_{\text{eq}}=\dfrac{[\ce{CO}][\ce{I2}]}{\ce{COI2}}=0.00088=\dfrac{(x)(x)}{(0.55-x)}\"<\/p>\n

                      We rearrange this into a quadratic equation that equals 0:<\/p>\n

                      \"\begin{array}{rcl}<\/p>\n

                      Now we use the quadratic equation to solve for the two possible values of x<\/em>:<\/p>\n

                      \"x=\dfrac{-0.00088\pm \sqrt{(0.00088)^2-4(1)(-0.000484)}}{2(1)}\"<\/p>\n

                      Evaluate for both signs in the numerator \u2014 first the + sign and then the \u2212 sign:<\/p>\n

                      \"x=0.0216\text{ or }x=-0.0224\"<\/p>\n

                      Because x<\/em> is the final concentration of both CO and I2<\/sub>, it cannot be negative, so we discount the second numerical answer as impossible. Thus x<\/em> = 0.0216.<\/p>\n

                      Going back to determine the final concentrations using the expressions in the E row of our ICE chart, we have:<\/p>\n

                      \n

                        <\/span>   <\/span>\"\begin{align*}<\/p>\n

                      You can verify that these numbers are correct by substituting them into the K<\/em>eq<\/sub> expression and evaluating and comparing to the known K<\/em>eq<\/sub> value.<\/p>\n

                      Test Yourself<\/em>
                      \nSet up an ICE chart and solve for the equilibrium concentrations in this chemical reaction.<\/p>\n

                      \"\begin{array}{rl}<\/p>\n

                      Answer<\/em>
                      \nThe completed ICE chart is as follows:<\/p>\n\n\n\n\n\n\n\n
                      <\/th>\nN2<\/sub>H2<\/sub>(g)<\/th>\n\u21c4<\/th>\nN2<\/sub>(g)<\/th>\n+<\/th>\nH2<\/sub>(g)<\/th>\n<\/tr>\n<\/thead>\n
                      I<\/td>\n0.075<\/td>\n<\/td>\n0<\/td>\n<\/td>\n0<\/td>\n<\/tr>\n
                      C<\/td>\n\u2212x<\/em><\/td>\n+x<\/em><\/td>\n+x<\/em><\/td>\n<\/tr>\n
                      E<\/td>\n0.075 \u2212 x<\/em><\/td>\n+x<\/em><\/td>\n+x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                      Solving for x<\/em> gives the equilibrium concentrations as [N2<\/sub>H2<\/sub>] = 0.033 M; [N2<\/sub>] = 0.042 M; and [H2<\/sub>] = 0.042 M<\/p>\n<\/div>\n<\/div>\n

                      \n
                      \n

                      Key Takeaways<\/p>\n<\/header>\n

                      \n
                        \n
                      • An ICE chart is a convenient way to determine equilibrium concentrations from starting amounts.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n
                        \n
                        \n

                        Exercises<\/p>\n<\/header>\n

                        \n

                        Questions<\/h1>\n
                          \n
                        1. Describe the three parts of an ICE chart.<\/li>\n
                        2. What is the relationship between the equilibrium row in an ICE chart and the other two rows?<\/li>\n
                        3. Set up (but do not solve) an ICE chart for this reaction, given the initial conditions.\n

                          \"\begin{array}{ccc}<\/p>\n<\/li>\n

                        4. Set up (but do not solve) an ICE chart for this reaction, given the initial conditions.\n

                          \"\begin{array}{ccccccc}<\/p>\n<\/li>\n

                        5. Given that pure solids and liquids do not appear in K<\/em>eq<\/sub> expressions, set up the ICE chart for this reaction, given the initial conditions.\n

                          \"\begin{array}{ccccccc}<\/p>\n<\/li>\n

                        6. Given that pure solids and liquids do not appear in K<\/em>eq<\/sub> expressions, set up the ICE chart for this reaction, given the initial conditions.\n

                          \"\begin{array}{ccccccc}<\/p>\n<\/li>\n

                        7. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\n

                          \"\begin{array}{rl}<\/p>\n<\/li>\n

                        8. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\n

                          \"\begin{array}{rl}<\/p>\n<\/li>\n

                        9. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\n

                          \"\begin{array}{rl}<\/p>\n<\/li>\n

                        10. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\n

                          \"\begin{array}{rl}<\/p>\n<\/li>\n

                        11. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\n

                          \"\begin{array}{rl}<\/p>\n<\/li>\n

                        12. Determine the equilibrium concentrations for this chemical reaction with the given K<\/em>eq<\/sub>.\n

                          \"\begin{array}{rl}<\/p>\n<\/li>\n<\/ol>\n

                          Answers<\/h1>\n
                            \n
                          1. I = initial concentrations; C = change in concentrations; E = equilibrium concentrations<\/li>\n<\/ol>\n
                              \n
                            1. \n\n\n\n\n\n\n\n
                              <\/th>\n3O2<\/sub><\/th>\n\u21c4<\/span><\/th>\n2O3<\/sub><\/th>\n<\/tr>\n<\/thead>\n
                              I<\/td>\n0.075<\/td>\n<\/td>\n0<\/td>\n<\/tr>\n
                              C<\/td>\n\u22123x<\/em><\/td>\n+2x<\/em><\/td>\n<\/tr>\n
                              E<\/td>\n0.075 \u2212 3x<\/em><\/td>\n+2x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n
                                \n
                              1. \n\n\n\n\n\n\n\n
                                <\/th>\nCH4<\/sub><\/th>\n+<\/th>\n2O2<\/sub><\/th>\n\u21c4<\/span><\/th>\nCO2<\/sub><\/th>\n+<\/th>\n2H2<\/sub>O<\/th>\n<\/tr>\n<\/thead>\n
                                I<\/td>\n0.0060<\/td>\n<\/td>\n0.055<\/td>\n<\/td>\n0<\/td>\n<\/td>\n0<\/td>\n<\/tr>\n
                                C<\/td>\n\u2212x<\/em><\/td>\n\u22122x<\/em><\/td>\n+x<\/em><\/td>\n\u2014<\/td>\n<\/tr>\n
                                E<\/td>\n0.0060 \u2212 x<\/em><\/td>\n0.055 \u2212 2x<\/em><\/td>\n+x<\/em><\/td>\n\u2014<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n
                                  \n
                                1. [HCN] = 0.364 M; [HNC] = 1.64 M<\/li>\n<\/ol>\n
                                    \n
                                  1. [N2<\/sub>O3<\/sub>] = 0.0017 M; [NO] = [NO2<\/sub>] = 0.0646 M<\/li>\n<\/ol>\n
                                      \n
                                    1. [H2<\/sub>S] = 0.836 M; [H2<\/sub>] = 0.046 M<\/li>\n<\/ol>\n<\/div>\n<\/div>\n","protected":false},"author":90,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-7686","chapter","type-chapter","status-publish","hentry"],"part":7638,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7686","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":5,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7686\/revisions"}],"predecessor-version":[{"id":8947,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7686\/revisions\/8947"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/7638"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7686\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/media?parent=7686"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=7686"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/contributor?post=7686"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/license?post=7686"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}