Equilibrium Partial Pressure of C

Question

An empty steel vessel is charged with 0.400 atm of C and 0.400 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

A (g) ⇌ B (g) + C (g)

0.400.....0...........0.400.....I

-x...........+x...........+x.......C

0.4-x.......x..........0.4+x.....E

Kp = 11.2 = (C)(B) / (A)

11.2 = (0.4+x)(x) / 0.4-x

11.2 = 0.4x + x2 / 0.4-x

4.48 - 11.2x = 0.4x + x²

x² + 11.6x - 4.48 = 0

x = 0.374 atm

Partial pressure of C = 0.400 + 0.374 = 0.774 atm