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General Chemistry/Equilibrium Concepts/Calculating Kc and Kp

Equilibrium Concepts

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Calculating Kc and Kp

Equilibrium calculations can be done by writing the balanced equation, identifying the number of moles of reactants and products, designing a table to find concentrations or partial pressures, and applying the formula.

The calculation of Kc from experimental data involves the following steps:

  • Write the correct and balanced chemical equation.
  • Construct a table with initial number of moles of the reactants and products in the first column.
  • Calculate the amounts by which the moles have changed in the second column.
  • Calculate the amounts of moles at the equilibrium position.
  • Apply the values in the formula of Kc to calculate the value.

    The calculation of Kp from experimental data involves similar steps with slight changes:

    • Write the correct and balanced chemical equation.
    • Construct a table with initial number of moles of the reactants and products in the first column.
    • Calculate the amounts by which the moles have changed as the second column.
    • Calculate the amounts of moles at the equilibrium position.
    • Determine the mole fraction of gases by dividing the number of moles at equilibrium by the total number of moles.
    • Calculate the partial pressures by multiplying the mole fraction by the total pressure.
    • Apply the values in the formula of Kp to calculate the value.

      An ICE table can be used to simplify calculations involving changes in concentration during an equilibrium reaction. ICE stands for Initial, Change, and Equilibrium. Given initial values of either number of moles or molar concentrations of the reactants as well as information about changes during the reaction, the final number of moles or concentrations of the reactants and products can be calculated.

      Step-By-Step Example
      Using an ICE Table in an Equilibrium Constant Calculation
      A 2.00-liter vessel is filled with 0.050 moles of methane (CH4), 0.060 moles of steam (H2O), and 0.010 moles of hydrogen (H2). The reaction that takes place in the vessel is:
      CH4(g)+H2O(g)⇌CO(g)+3H2(g){\rm{CH_4}}(g)+{\rm{H_2O}}(g) \rightleftharpoons{\rm{CO}}(g)+3{\rm{H_2}}(g)
      The system is allowed to reach equilibrium at a specific temperature. At equilibrium, 80.0% of the methane is consumed. Calculate the equilibrium constant.
      Step 1

      Set up an ICE table. The reaction should be shown in the first row. Fill in the second row using the given initial number of moles of each reactant and product in the system.

      To fill in the second row, note that 80.0% of the methane is consumed, so the change in number of moles is the product of 0.800 and the initial number of moles. The changes of H2O, CO, and H2 are determined from the stoichiometric coefficients of the balanced equation. The amount of CH4 that is consumed must equal the amount of H2O that is consumed, the amount of CO that is produced, and three times the amount of H2 that is produced.

      Use the values in the second and third rows to calculate the number of moles of each reactant and product at equilibrium.
      Step 2
      The volume of the system is given as 2.00 L. Use this volume and the number of moles at equilibrium calculated in the ICE table to calculate the equilibrium concentrations of each reactant and product.
      [CH4]eq=0.010mol2.00L=0.0050mol/L{\left[{{\rm{CH_4}}}\right]_{{\rm{eq}}}}=\frac{{0.010{\;\rm{mol}}}}{{2.00{\;\rm{L}}}}=0.0050{\;\rm{mol/L}}
      [H2O]eq=0.020mol2.00L=0.010mol/L{\left[{{\rm{H_2O}}}\right]_{{\rm{eq}}}}=\frac{{0.020{\;\rm{mol}}}}{{2.00{\;\rm{L}}}}=0.010{\;\rm{mol/L}}
      [CO]eq=0.040mol2.00L=0.020mol/L{\left[{{\rm{CO}}}\right]_{{\rm{eq}}}}=\frac{{0.040{\;\rm{mol}}}}{{2.00{\;\rm{L}}}}=0.020{\;\rm{mol/L}}
      [H2]eq=0.130mol2.00L=0.065mol/L{\left[{{{\rm{H_2}}}}\right]_{{\rm{eq}}}}=\frac{{0.130{\;\rm{mol}}}}{{2.00{\;\rm{L}}}}=0.065{\;\rm{mol/L}}
      Solution
      Use the equilibrium concentrations to calculate the equilibrium constant.
      Kc=[CO]eq[H2]eq3[CH4]eq[H2O]eq=(0.020)(0.065)3(0.0050)(0.010)=0.11\begin{aligned}K_{\rm{c}}&=\frac{{\left[\rm{CO}\right]}_{\rm{eq}}{\left[{\rm{H}}_2\right]_{\rm{eq}}}^3}{{\left[{\rm{CH_4}}\right]}_{\rm{eq}}{\left[{\rm{H_2O}}\right]}_{\rm{eq}}}\\&=\frac{(0.020)(0.065)^3}{(0.0050)(0.010)}\\&=0.11\end{aligned}
      Step-By-Step Example
      Calculation of Equilibrium Constant from Equilibrium Concentrations of Reactants and Products
      Fragrances are made up of organic molecules that frequently contain esters. The ester ethyl acetate (CH3COOC2H5) is responsible for the sweet smell of some nail polish remover and glues. It is formed by the reaction of acetic acid (CH3COOH) and ethyl alcohol (C2H5OH), with water as the solvent. Calculate the equilibrium constant Kc for this reaction given these equilibrium concentrations:
      [CH3COOH]=0.0700mol/L[C2H5OH]=0.0800mol/L[CH3COOC2H5]=0.200mol/L[H2O]=0.100mol/L\begin{aligned}\left[\rm{CH_3COOH}\right]&=0.0700\;\rm{ mol/L}\\\left[\rm{C_2H_5OH}\right]&=0.0800\;\rm{ mol/L}\\\left[\rm{CH_3COOC_2H_5}\right]&=0.200\;\rm{mol/L}\\\left[\rm{H_2O}\right]&=0.100\;\rm{mol/L}\end{aligned}
      Step 1
      Write the balanced equation for the reaction.
      CH3COOH(aq)+C2H5OH(aq)⇌CH3COOC2H5(aq)+H2O(l){\rm{CH_3COOH}}(aq)+{\rm{C_2H_5OH}}(aq)\rightleftharpoons{\rm{CH_3 COOC_2H_5}}(aq)+{\rm{H_2O}}(l)
      Step 2
      Use the formula for the equilibrium constant.
      Kc=[CH3COOC2H5]eq[CH3COOH]eq[C2H5OH]eq=(0.200)(0.0700)(0.0800)= 35.7\begin{aligned}K_{\rm{c}}=&\frac{{\left[{\rm{CH_3COOC_2H_5}}\right]}_{\rm{eq}}}{{\left[{\rm{CH_3COOH}}\right]}_{\rm{eq}}{\left[{\rm{C_2H_5OH}}\right]}_{\rm{eq}}}\\=&\frac{(0.200)}{(0.0700)(0.0800)} \\ =&\ 35.7\end{aligned}
      Solution
      The equilibrium constant for the reaction of acetic acid (CH3COOH) and ethyl alcohol (C2H5OH), with water as the solvent, is 35.7. Recall that when water is a byproduct of a reaction, such as in this esterification process, it should not be ignored. However, because the water produced in this reaction is negligible compared with the water solvent, the concentration of water does not change and therefore can be omitted in the equation.

      As an example of calculating Kp, consider the contact process for the manufacture of sulfuric acid, which happens in a series of steps.

      • Sulfur dioxide formation
      S(s)+O2(g)→SO2(g){\rm{S}}(s)+{\rm{O_2}}(g)\rightarrow{\rm{SO_2}}(g)
      • Conversion of sulfur dioxide to sulfur trioxide
      2SO2(g)+O2(g)⇌2SO3(g)ΔH=−196kJ/mol2{\rm{SO_2}}(g)+{\rm{O_2}}(g)\rightleftharpoons2{\rm{SO_3}}(g)\;\;\;\Delta H=-196\;\rm{kJ/mol}
      • Conversion of sulfur trioxide to oleum
      H2SO4(l)+SO3(g)→H2S2O7(l){\rm{H_2SO_4}}(l)+{\rm{SO_3}}(g)\rightarrow{\rm{H_2S_2O_7}}(l)
      • Dissolving oleum in water
      H2S2O7(l)+H2O(l)→2H2SO4(l){\rm{H_2S_2O_7}}(l)+{\rm{H_2O}}(l)\rightarrow2{\rm{H_2SO_4}}(l)
      Three steps can be taken to optimize the amount of sulfur trioxide that is produced. First, according to Le Chatelier's principle, increasing the concentration of oxygen will shift the reaction to the right, producing a greater amount of sulfur trioxide. Also, since the reaction is exothermic, Le Chatelier's principle also indicates that lowering the temperature will produce more product.

      In addition, notice that the conversion of sulfur dioxide to sulfur trioxide, there are two molecules on the right side and three molecules on the left side. Therefore, if the pressure is lowered, the reaction will shift to the side with fewer molecules. In this case, it will shift to the right. These are theoretical assumptions. In practice, optimum conditions are chosen by considering all criteria and constraints.

      The equilibrium constant Kp for the reaction is the square of the partial pressure of the product divided by the square of the partial pressures of the reactants multiplied.
      Kp=PSO32PSO22PO2{K_{\rm{p}}} =\frac{{P_{{\rm{SO_3}}}}^2}{{{P_{{\rm{SO_2}}}}^2}{P_{\rm{O_2}}}}
      Next, consider the dissociation of dichlorine pentoxide (Cl2O5) to form chlorine gas (Cl2) and oxygen gas (O2).
      2Cl2O5(g)⇌2Cl2(g)+5O2(g)2{\rm{Cl_2O_5}}(g)\rightleftharpoons2{\rm{Cl_2}}(g)+5{\rm{O_2}}(g)
      Assume a closed vessel contains the gases with mole fractions of 0.220 Cl2, 0.254 O2, and 0.526 Cl2O5, and equilibrium is reached at a total pressure of 3.00 atm. To find Kp, use the partial pressure of each gas, where PA=(mole fraction of A)(total pressure)P_{\rm{A}}=(\text{mole fraction of }{\rm{A})\text{(total pressure)}}.
      Kp=(PCl2)2(PO2)5(PCl2O5)2=(χCl2Ptotal)2(χO2Ptotal)5(χCl2O5Ptotal)2=[(0.220)(3.00)]2[(0.254)(3.00)]5[(0.526)(3.00)]2=0.0449\begin{aligned}{K_{\rm{p}}}&= \frac{{{{\left( {{P_{{\rm{Cl_2}}}}}\right)}^2}{{\left({{P_{{{\rm{O_2}}}}}} \right)}^5}}}{{{{\left( {{P_{{\rm{Cl_2O_5}}}}} \right)}^2}}}\\ &= \frac{{{{\left( {{\chi_{{\rm{Cl_2}}}}{P_{{\rm{total}}}}}\right)}^2}{{\left({{\chi_{{\rm{O_2}}}}{P_{{\rm{total}}}}} \right)}^5}}}{{{{\left( {{\chi _{{\rm{Cl_2O_5}}}}{P_{{\rm{total}}}}} \right)}^2}}}\\&= \frac{{{{\left[ {\left({0.220}\right)\left(3.00 \right)} \right]}^2}{{\left[ {\left({0.254}\right)\left(3.00 \right)} \right]}^5}}}{{{{\left[ {\left( {0.526} \right)\left( 3.00 \right)} \right]}^2}}}\\&= 0.0449\end{aligned}
      Step-By-Step Example
      Calculation of Kp Using Mole Fractions
      Consider the decomposition of phosphorus pentachloride (PCl5) to produce phosphorus trichloride (PCl3) and chlorine (Cl2) when heated. Assume the reaction occurs in a closed flask at a temperature of 250°C and a pressure of 3.00 atm, and 59.0% of phosphorus pentachloride is decomposed. Calculate Kp.
      Step 1
      Write the balanced equation for the reaction.
      PCl5(g)⇌PCl3(g)+Cl2(g){\rm{PC}}{{\rm{l}}_5}(g) \rightleftharpoons {\rm{PC}}{{\rm{l}}_3}(g) + {\rm{C}}{{\rm{l}}_2}(g)
      Step 2
      Use an ICE table to calculate the number of moles of each species at equilibrium.
      An ICE table is used to calculate the number of moles of each reactant and product at equilibrium. The calculations are performed according to numbers of moles. Since the initial number of moles is unknown, the calculations are written per mole of each species. The coefficient for each species in the balanced equation is 1. Since 59.0% of PCl5 decomposes, an equal number of moles are produced for PCl3 and Cl2.
      Step 3
      Calculate the number of moles at equilibrium.
      Total number of moles at equilibrium=0.410mol+0.590mol+0.590mol=1.59mol\begin{aligned}\text{Total number of moles at equilibrium}&=0.410\;\rm{mol} + 0.590\;\rm{mol} + 0.590\;\rm{mol}\\&=1.59\;\rm{mol}\end{aligned}
      Step 4
      In order to calculate the equilibrium constant, the partial pressure of each species must be determined.
      χPCI5=0.410mol1.59mol=0.258χPCI3=0.590mol1.59mol=0.371χCl2=0.590mol1.59mol=0.371\begin{aligned}\chi_{{\rm{PCI_5}}} &=\frac{0.410\;\rm{mol}}{1.59\;\rm{mol}}=0.258\\\chi_{{\rm{PCI_3}}} &=\frac{0.590\;\rm{mol}}{1.59\;\rm{mol}}=0.371\\\chi_{{\rm {Cl_2}}} &=\frac{0.590\;\rm{mol}}{1.59\;\rm{mol}}=0.371\end{aligned}
      Solution
      The mole fractions can be used to calculate the equilibrium constant.
      Kp=(χPCI3Ptotal)(χCI2Ptotal)(χPCI5Ptotal)=[(0.371)(3.00)][(0.371)(3.00)](0.258)(3.00)=1.60\begin{aligned}K_{\rm{p}}&= \frac{\left(\chi_{{\rm{PCI_3}}}P_{\rm{total}}\right)\left(\chi_{{\rm{CI_2}}}P_{\rm{total}}\right)}{\left(\chi_{{\rm{PCI_5}}} P_{\rm{total}}\right)}\\&=\frac{\left[(0.371)(3.00)\right]\left[(0.371)(3.00)\right]}{(0.258)(3.00)}\\&=1.60\end{aligned}
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