In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.778 L flask at 1,072 K. At equilibrium, the flask contains 0.206 mol of CO gas, 0.187 mol of H2 gas, and 0.187 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1,072 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.

Question

Chemistry

Lance Burch

30 August, 20:57

In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.778 L flask at 1,072 K. At equilibrium, the flask contains 0.206 mol of CO gas, 0.187 mol of H2 gas, and 0.187 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1,072 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.

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Tessa Frey · Answer 1

the water concentration at equilibrium is

⇒ [ H2O (g) ] = 0.0510 mol/L

Explanation:

CH4 (g) + H2O (g) ↔ CO (g) + 3H2 (g)

∴ Kc = ([ CO (g) ] * [ H2 ]³) / ([ CH4 (g) ] * [ H2O (g) ]) = 0,30

equilibrium:

⇒ [ CO (g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L

⇒ [ H2 (g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

⇒ [ CH4 (g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

replacing in Kc:

⇒ ((0.2648) * (0.2404) ³) / ([ H2O (g) ] * 0.2404) = 0.30

⇒ 0.0721 [ H2O (g) ] = 3.679 E-3

⇒ [ H2O (g) ] = 0.0510 mol/L