A (g) + 2B (g) → C (g) + D (g)

If you initially start with 1.00 atm of both A and B and find that at equilibrium 0.211 atm of C is present, what is the value of Kp for the reaction at the temperature the reaction was run?

0.169

Explanation:

Let's consider the following reaction.

A (g) + 2B (g) ⇄ C (g) + D (g)

We can find the pressures at equilibrium using an ICE chart.

A (g) + 2 B (g) ⇄ C (g) + D (g)

I 1.00 1.00 0 0

C - x - 2x + x + x

E 1.00-x 1.00-2x x x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2 (0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC * pD / pA * pB²

Kp = 0.211 * 0.211 / 0.789 * 0.578²

Kp = 0.169