Question

Consider the equilibrium C(s)+ CO₂(g) ⇌2 CO(g)

A 2.0 L flask contains a mixture of 0.10 mol of CO(g), 0.20 mol of CO₂(g) and 0.40 mol of C(s) in equilibrium at 700°C. *Note that only atm pressure values are allowed in the K expression.* The value of the equilibrium constant K is

Answer 1

C(s)+ CO₂(g) ⇌2 CO(g)

T = (273 + 700) = 973 K

V = 2.0 L

P (CO) = nRT / V = 0.10 * 0.0821 * 973 / 2.0 = 3.994 atm

and

P (CO2) = nRT / V = 0.20 * 0.0821 * 973 / 2.0 = 7.988 atm.

K = P(CO)^2 / p(CO2) [activity of C(s) = 1]

or

K = (3.994)^2 / (7.988) = 1.997

or

K = 1.997