Question

In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.379 L flask at 1191 K. At equilibrium, the flask contains 0.145 mol of CO gas, 0.218 mol of H2 gas, and 0.25 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1191 K)?
Enter to 4 decimal places.

HINT: Look at sample problem 17.7 in the 8th ed Silberberg book. Write a balanced chemical equation. Write the Kc expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into Kc expression, solve for the unknown.

Answer 1

Answer: The concentration of water at equilibrium is 0.3677 mol/L

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by $K_{c}$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The $K_{c}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The chemical equation for the conversion of methane to carbon monoxide and hydrogen gas follows:

$CH_4+H_2O\rightleftharpoons 3H_2+CO$

The $K_{c}$ is represented as:

$K_{c}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}$      ....(1)

To calculate the concentration, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

$[CO]=\frac{0.145mol}{0.379L}=0.383mol/L$

$[H_2]=\frac{0.218mol}{0.379L}=0.575mol/L$

$[CH_4]=\frac{0.25mol}{0.379L}=0.660mol/L$

$K_c=0.30$

Putting values in equation 1, we get:

$0.30=\frac{(0.575)^3\times 0.383}{0.660\times [H_2O]}$

$[H_2O]=\frac{(0.575)^3\times 0.383}{0.660\times 0.30}=0.3677$

Hence, the concentration of water at equilibrium is 0.3677 mol/L