equilibrium calculations with partial pressures and kp

Question

A student ran the following reaction in the laboratory at 651 K:

2NH₃(g) N₂(g) + 3H₂(g)

When she introduced NH₃(g) at a pressure of 1.07 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of H₂(g) to be 1.57 atm.

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

K_p = 

Answer 1

It may be helpful to set up an ICE table

2NH3(g) <===> N2(g) + 3H2(g)

1.07......................0.............0.............Initial

-2x.......................+x..........+3x...........Change

1.07-2x.................x.............3x...........Equilibrium

Since @ equilibrium we have 1.57 atm H2, we can set 3x = 1.52 atm.

3x = 1.52 atm

x = 0.507 atm

At equilibrium we will have the following partial pressures:

P_NH3 = 1.07 - 2x = 1.07 - 1.01 = 0.060 atm

P_N2 = x = 0.507 atm

P_H2 = 1.57 atm

Kp = (H₂)³ (N₂) / (NH₃)²

Kp = (1.57)³ (0.507) / (0.06)²

Kp = 2120