Equilbrium Constant Calculations (Edexcel A Level Chemistry): Revision Note

Equilibrium Constant Calculations

Calculations involving K_c

• In the equilibrium expression each figure within a square bracket represents the concentration in mol dm^-3

• The units of K_c therefore depend on the form of the equilibrium expression

• Some questions give the number of moles of each of the reactants and products at equilibrium together with the volume of the reaction mixture

• The concentrations of the reactants and products can then be calculated from the number of moles and total volume

Equation to calculate concentration from number of moles and volume

Answer

• Step 1: Calculate the concentrations of the reactants and products

  • [CH₃COOH] $=\frac{0.235}{0.500}=$0.470 mol dm^-3 

  • [C₂H₅OH] $=\frac{0.035}{0.500}=$0.070 mol dm^-3 

  • [CH₃COOC₂H₅] $=\frac{0.182}{0.500}=$0.364 mol dm^-3 

  • [H₂O] $=\frac{0.182}{0.500}=$0.364 mol dm^-3 

• Step 2: Write out the balanced chemical equation with the concentrations of beneath each substance

• Step 3: Write the equilibrium constant for this reaction in terms of concentration

K_c = $\frac{\left[{\mathrm{H}}_2\mathrm{O}\right]\left[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5\right]}{\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

• Step 4: Substitute the equilibrium concentrations into the expression

K_c = $=\frac{[0.364]\times [0.364]}{[0.070]\times [0.0470]}=$4.03

• Step 5: Deduce the correct units for K_c 

K_c = $\frac{\left[\mathrm{mol}{\mathrm{dm}}^{-3}\right]\times \left[\mathrm{mol}{\mathrm{dm}}^{-3}\right]}{\left[\mathrm{mol}{\mathrm{dm}}^{-3}\right]\times \left[\mathrm{mol}{\mathrm{dm}}^{-3}\right]}$

All units cancel out

Therefore, K_c = 4.03

• Note that the smallest number of significant figures used in the question is 3, so the final answer should also be given to 3 significant figures

• Some questions give the initial and equilibrium concentrations of the reactants but products

• An initial, change and equilibrium table should be used to determine the equilibrium concentration of the products using the molar ratio of reactants and products in the stoichiometric equation

Answer

• Step 1: Write out the balanced chemical equation with the concentrations of beneath each substance using an initial, change and equilibrium table

• Step 2: Calculate the concentrations of the reactants and products

• Step 3: Write the equilibrium constant for this reaction in terms of concentration

• Step 4: Substitute the equilibrium concentrations into the expression

$K_{\mathit{c}}=\frac{(0.0346)\times (0.0346)}{(0.0654)\times (0.0654)}$

K_c = 0.28

• Step 5: Deduce the correct units for K_c

$K_{\mathit{c}}=\frac{\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)\times \left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)}{\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)\times \left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)}$

All units cancel out

Therefore, K_c = 0.288

Calculations involving K_p

• In the equilibrium expression the p represent the partial pressure of the reactants and products in Pa

• The units of K_p therefore depend on the form of the equilibrium expression

Answer

• Step 1: Write the equilibrium constant for the reaction in terms of partial pressures

$K_p=\frac{p^2{\mathrm{SO}}_3}{p^2{\mathrm{SO}}_2\times p{\mathrm{O}}_2}$

• Step 2: Substitute the equilibrium concentrations into the expression

$K_{\mathit{p}}=\frac{{(8.0\times {10}^6)}^2}{{(1.0\times {10}^6)}^2\times (7.0\times {10}^6)}$

K_p = 9.1 x 10^-6

• Step 3: Deduce the correct units of K_p

$K_p=\frac{P{a}^2}{P{a}^2\times Pa}$

The units of K_p are Pa^-1

Therefore, K_p = 9.1 x 10^-6 Pa^-1

• Some questions only give the number of moles of gases present and the total pressure

• The number of moles of each gas should be used to first calculate the mole fractions

• The mole fractions are then used to calculate the partial pressures

• The values of the partial pressures are then substituted in the equilibrium expression

Answer

• Step 1: Calculate the total number of moles

Total number of moles = 1.71 x 10^-3 + 2.91 x 10^-3 + 1.65 x 10^-2

= 2.112 x 10^-2

• Step 2: Calculate the mole fraction of each gas

$\begin{array}{l}{\mathrm{H}}_2=\frac{1.71\times {10}^{-3}}{2.112\times {10}^{-2}}=0.0810\\ {\mathrm{I}}_2=\frac{2.91\times {10}^{-3}}{2.112\times {10}^{-2}}=0.1378\\ \mathrm{HI}=\frac{1.65\times {10}^{-2}}{2.112\times {10}^{-2}}=0.7813\end{array}$

• Step 3: Calculate the partial pressure of each gas

H₂ = 0.0810 x 100 = 8.10 kPa

I₂ = 0.1378 x 100 = 13.78 kPa

HI = 0.7813 x 100 = 78.13 kPa

• Step 4: Write the equilibrium constant in terms of partial pressure

$K_p=\frac{p^2\mathrm{HI}}{p{\mathrm{H}}_2\times p{\mathrm{I}}_2}$

• Step 5: Substitute the values into the equilibrium expression

$K_{\mathit{p}}=\frac{78.{13}^2}{8.10\times 13.78}$

K_p = 54.7

• Step 6: Deduce the correct units for K_p

$K_p=\frac{P{a}^2}{Pa\times Pa}$

All units cancel out

Therefore, K_p = 54.7

• Other questions related to equilibrium expressions may involve calculating quantities present at equilibrium given appropriate data