In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.501 L flask at 1043 K. At equilibrium, the flask contains 0.169 mol of CO gas, 0.257 mol of H2 gas, and 0.255 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1043 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 K expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into K expression, solve for the unknown. (MTS 5/16/2018)

Question

Chemistry

Bronson Alexander

7 September, 22:18

In a study of the conversion of methane to other fuels, a chemical engineer mixes gaseous methane and gaseous water in a 0.501 L flask at 1043 K. At equilibrium, the flask contains 0.169 mol of CO gas, 0.257 mol of H2 gas, and 0.255 mol of methane. What is the water concentration at equilibrium (Kc = 0.30 for this process at 1043 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 K expression. Calculate the equilibrium concentrations of all the species given (moles/liter). Put values into K expression, solve for the unknown. (MTS 5/16/2018)

+1

Answers (1)

Know the Answer?

Kailee Hood · Answer 1

[ H2O ]eq = 0.298 mol/L

Explanation:

CH4 (g) + H2O (g) ↔ CO (g) + 3H2 (g)

∴ V = 0.501 L

∴ T = 1043 K

at equilibrium:

∴ n CH4 = 0.255 mol

⇒ [ CH4 ]eq = 0.255 mol / 0.501 L = 0.509 mol/L

∴ n CO = 0.169 mol

⇒ [ CO ]eq = 0.169 / 0.501 = 0.337 mol/L

∴ n H2 = 0.257

⇒ [ H2 ] eq = 0.257 / 0.501 = 0.513 mol/L

∴ Kc = [ H2 ]³ * [ CO ] / [ CH4 ] * [ H2O ] = 0.30

⇒ [ H2O ] = [ H2 ]³ * [ CO ] / [ CH4 ] * 0.30

replacing the value of the concentration in Kc:

⇒ [ H2O ] = (0.513) ³ * (0.337) / (0.509) * 0.30

⇒ [ H2O ] = 0.298 mol/L