Kp with Initial and Equilibrium Pressures

Question

A student ran the following reaction in the laboratory at 692 K:

H₂(g) + I₂(g) 2HI(g)

When she introduced H₂(g) and I₂(g) into a 1.00 L evacuated container, so that the initial partial pressure of H₂ was 4.39 atm and the initial partial pressure of I₂ was 3.19 atm, she found that the equilibrium partial pressure of HI was 5.65 atm.

Calculate the equilibrium constant, K_p, she obtained for this reaction.

K_p = 

Answer 1

H₂(g) + I₂(g) <==> 2HI(g)

Kp = (HI)² / (H₂)(I₂)

H₂(g) + I₂(g) <==> 2HI(g)

4.39.....3.19.............0........Initial

-x.........-x...............+2x......Change

4.39-x....3.19-x......5.65......Equilibrium

Since 2x = 5.65, then x must be 2.83 and we can now find equilibrium pressures of H₂ and I₂:

Equilibrium pressures are as follows:

H2 = 4.39 - 2.83 = 1.56 atm

I2 = 3.19 - 2.83 = 0.36 atm

HI = 5.65 atm

Solving for Kp we have...

Kp = (5.65)² / (1.56)(0.36)

Kp = 56.8