Question

A mixture of 0.200 mol of CO2, 0.100 mol of H2, and 0.1600 mol of H2O is placed in a 2.00 L vessel. The following equilibrium is established at 500 K: CO2 (g) + H2 (g) -> CO (g) + H2O (g) (A) calculate the initial partial pressures of CO2, H2, and H2O. (B) At equilibrium PH2O= 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO. (C) Calculate Kp for the reaction

Answer 1

CO2 (g) + H2 (g) <===> CO (g) + H2O (g)

Kp = (P(CO) x P(H2O)) / (P(CO2) x P(H2))

a.)

P(CO2) = nRT / V = 0.2000 mol x 0.08206 L atm / K mol x 500 K / 2.000 L = 4.103 atm

P(H2) = nRT / V = 0.1000 mol x 0.08206 L atm / K mol x 500 K / 2.000 L = 2.052 atm

P(H2O) = nRT / V = 0.1600 mol x 0.08206 L atm / K mol x 500 K / 2.000 L = 3.282 atm

P(CO) = 0 (initially)

b.)
The partial pressure of H2O increases from 3.282 to 3.51 atm, an increase of 0.228 atm.

If H2O increases by 0.228 atm, CO also increases by 0.228 atm, CO2 decreases by 0.228 atm, and H2 decreases by 0.228 atm.

P(CO2) = 4.103 - 0.228 = 3.875 atm
P(H2) = 2.052 - 0.228 = 1.824 atm
P(CO) = 0 + 0.228 = 0.228 atm
P(H2O) = 3.51 atm (given)

c.)

Kp = (P(CO) x P(H2O)) / (P(CO2) x P(H2))
Kp = (0.228 atm x 3.51 atm) / (3.875 atm x 1.824 atm) = 0.113 (no units)
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Answer 2

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