An empty steel vessel is charged with 0.140 atm of C and 0.140 atm of A. Once the system reaches equilibrium according to the reaction below, what is the equilibrium partial pressure of C?

Question

An empty steel vessel is charged with 0.140 atm of C and 0.140 atm of A. Once the system reaches equilibrium according to the reaction below, what is the equilibrium partial pressure of C?

Kp for this reaction is 11.2.

A (g) ⇌ B (g) + C (g)

Answer 1

Probably easiest to set up an ICE table:

A (g) ⇌ B (g ) + C (g)

0.14........0..........0.14.......Initial

-x...........+x.........+x..........Change

0.14-x.....x.........0.14+x.....Equilibrium

Based on the positive value of Kp (+11.2), we know that the reaction favors the products and not the reactants, so we can use 0.14-x and 0.14+x for the reactants and products, respectively.

Kp = (B)(C)/(A)

11.2 = (x)(0.14+x) / 0.14-x

1.57 - 11.2x = 0.14x + x2

x2 + 11.34x -1.57 = 0

x = 0.137

At equilibrium, partial pressure of C = 0.14 + 0.137 = 0.277 atm