Question

In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.669 L flask at 1,020 K. At equilibrium, the flask contains 0.276 mol of CO gas, 0.207 mol of H2 gas, and 0.231 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1,020 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:

Concentration of water at equilibrium is 0.1177 M.    

Explanation:

Balanced equation: $CH_{4}(g)+H_{2}O(g)\rightleftharpoons CO(g)+3H_{2}(g)$

Equilibrium concentration of $CH_{4}$, [$CH_{4}$] = $\frac{0.231}{0.669}$ M = 0.345 M

Equilibrium concentration of CO, [CO] = $\frac{0.276}{0.669}$ M = 0.413 M

Equilibrium concentration of $H_{2}$, [$H_{2}$] = $\frac{0.207}{0.669}$ M = 0.309 M

Equilibrium constant for the given reaction in terms of concentration, $K_{c}$ is expressed as:           $K_{c}=\frac{[CO][H_{2}]^{3}}{[CH_{4}][H_{2}O]}$

                          $\Rightarrow [H_{2}O]=\frac{[CO][H_{2}]^{3}}{[CH_{4}].K_{c}}=\frac{(0.413)\times (0.309)^{3}}{(0.345)\times (0.30)}= 0.1177$

Hence, concentration of water at equilibrium is 0.1177 M