Question

A mixture of 0.2000 mol of CO2, 0.1000 mol of H2, and 0.1600 mol of H2O is placed in a 2.000-L vessel. The following equilibrium is established at 500 K:
CO2(g)+H2(g)⇌CO(g)+H2O(g)

Calculate Kc for the reaction.

Answer 1

Calculate Kc for the reaction.

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

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

First we calculate the P of all species as follows:

PV= nRT

R= 0.08206 L atm / K mol, T= 500K V= 2.000L

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

Here n for CO not given means initially CO not formed thus
P(CO) = 0 (initially)

Here equilibrium pressure not given so we cannot calculate the Kp.