Question 14 of 69intermediate🔧 ApplyNumerical3 marks

One mole of H2OH_2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,

H2O(g)+CO(g)⇌H2(g)+CO2(g)H_2O (g) + CO (g) \rightleftharpoons H_2 (g) + CO_2 (g)

Calculate the equilibrium constant for the reaction.

Correct Answer

The equilibrium constant KcK_c for the reaction is 0.4440.444.

Exercise: EXERCISES | Q: 6.14 | (Page 42)
For More Understanding

Explanation

To find the equilibrium constant, we first determine the initial concentrations of the reactants. Since 40% of the water reacts, we can find the equilibrium moles of all species involved. Dividing by the vessel volume gives equilibrium concentrations, which are then substituted into the KcK_c expression.

Solution Steps

  1. Initial Concentrations: Initial moles of H2OH_2O = 1 mol, CO = 1 mol. Volume of vessel = 10 L. Initial concentration, [H2O]=[CO]=1 mol10 L=0.1 M[H_2O] = [CO] = \frac{1 \text{ mol}}{10 \text{ L}} = 0.1 \text{ M}.

  2. Equilibrium Moles: At equilibrium, 40% of water reacts. Moles of H2OH_2O reacted = 40% of 1=0.4 mol40\% \text{ of } 1 = 0.4 \text{ mol}. Based on the stoichiometry of the reaction H2O(g)+CO(g)⇌H2(g)+CO2(g)H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g): Equilibrium moles of H2O=1−0.4=0.6 molH_2O = 1 - 0.4 = 0.6 \text{ mol}. Equilibrium moles of CO=1−0.4=0.6 molCO = 1 - 0.4 = 0.6 \text{ mol}. Equilibrium moles of H2=0.4 molH_2 = 0.4 \text{ mol}. Equilibrium moles of CO2=0.4 molCO_2 = 0.4 \text{ mol}.

  3. Equilibrium Concentrations:

    [H2O]=0.610=0.06 M[H_2O] = \frac{0.6}{10} = 0.06 \text{ M} [CO]=0.610=0.06 M[CO] = \frac{0.6}{10} = 0.06 \text{ M} [H2]=0.410=0.04 M[H_2] = \frac{0.4}{10} = 0.04 \text{ M} [CO2]=0.410=0.04 M[CO_2] = \frac{0.4}{10} = 0.04 \text{ M}

  4. Calculate Equilibrium Constant (KcK_c):

    Kc=[H2][CO2][H2O][CO]K_c = \frac{[H_2][CO_2]}{[H_2O][CO]} Kc=(0.04)×(0.04)(0.06)×(0.06)K_c = \frac{(0.04) \times (0.04)}{(0.06) \times (0.06)} Kc=0.00160.0036=1636=49=0.444K_c = \frac{0.0016}{0.0036} = \frac{16}{36} = \frac{4}{9} = 0.444
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