Partial Pressure

Question

An empty steel vessel is charged with 0.200 atm of C and 0.200 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

It is probably easiest to follow this if you set up an ICE table

A (g) ⇌ B (g) + C (g)

0.2 atm.....0 atm.....0.2 atm.......Initial

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

0.2-x.........x.........0.2+x.............Equilibrium

Write the Kp expression

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

11.2 = (x)(0.2+x) / 0.2-x

11.2 = x² + 0.2x / 0.2-x

x² + 11.4x - 2.24 = 0

x = 0.193 atm

Now that we know x, we can solve for the pressure of C, which according to the ICE table is 0.2+x

(C) = 0.200 + 0.193 = 0.393 atm

Answer 2

Let the equilibrium pressure of A, P_A = 0.2 - x, then the equilibrium pressure of B, P_B = x, and that of C, P_C = 0.2 + x

Now we can set up the equation

P_B P_C / P_A = K_P

x (0.2 + x) / (0.2 -x) = 11.2

multiplying both sides by (0.2 - x) and opening the parentheses gives

x² + 0.2 x = 2.24 - 11.2 x

This can be rewritten as a traditional quadratic equation

x² + 11.4 x - 2.24 = 0

x = - 5.7 + sqrt ( 5.7² + 2.24) = 5.893 - 5.7 = 0.193 atm

Hence

P_C = 0.200 + 0.193 = 0.393 atm