Kp and Equilibrium Pressures

Question

Consider the following reaction:

2HI(g) H₂(g) + I₂(g)

If 1.90 moles of HI, 0.333 moles of H₂, and 0.271 moles of I₂ are at equilibrium in a 19.1 L container at 854 K, the value of the equilibrium constant, K_p, is _______ .

Answer 1

Kp is the equilibrium constant used when partial pressures are provided. First we must solve for the partial pressures in PV=nRT. (Note: I am waiting until the last step to round to significant digits)

P(pressure) = n(moles)*R(gas constant)*T(temperature)/ V(volume)

V = 19.1 L T = 854 K R = 0.0821 L*atm/K*mol

For the first gas HI:

n = 1.90 mol

P_HI = (1.90mol)*(0.0821 L*atm/K*mol)*(854 K)/(19.1 L) = 6.975 atm

Repeat for the second gas H₂:

n = 0.333 mol

P_H2 = (0.333 mol)*(0.0821 L*atm/K*mol)*(854 K)/(19.1 L) = 1.222 atm

And for the third gas I₂:

n = 0.271 mol

P_I2 = (0.271 mol)*(0.0821 L*atm/K*mol)*(854 K)/(19.1 L) = 0.995 atm

Now we can solve for Kp.

Kp = P_products / P_reactants

Kp = (P_H2 * P_I2) / (P_HI)² = (1.222*0.995)/(6.975²) = 0.02499 = 0.0250

1. P_HI is squared because it has a coefficient of 2 in the chemical reaction

The equilibrium constant is unitless so your answer is simply 0.0250 or 2.5e-2