Question

Consider the following reaction: H2(g)+I2(g)?2HI(g) A reaction mixture at equilibrium at 175 K contains PH2=0.958atm, PI2=0.877atm, and PHI=0.020atm. A second reaction mixture, also at 175 K, contains PH2=PI2= 0.630 atm , and PHI= 0.102 atm .

Is the second reaction at equilibrium?

If not, what will be the partial pressure of HI when the reaction reaches equilibrium at 175 K?

Answer 1

The reaction:

H₂(g)+I₂(g)⇌2HI(g)

Reaction mixture at equilibrium at 175 K

Partial pressure of H_2, P_H2=0.958atm

Partial pressure of I₂ , P_I2=0.877atm

Partial pressure of HI, P_HI=0.020atm.

K_P1 = [HI]² / [H₂][I₂]

        = [0.020]²/ [0.958][0.877]

K_P1   = 4.7609 x 10^-4

The second reaction mixture at 175 K,

Partial pressures of H₂ and I₂, P_H2=P_I2= 0.630 atm

Partial pressure of HI, P_HI= 0.102 atm

K_P2 = [0.102]² / [0.630][0.630]

K_P2   = 0.026213 atm

Since K_P1 and K_P2 are not equal in values, it indicates that HI in second reaction is not at equilibrium.

Since

K_P = [HI]² / [H₂][I₂]

The partial pressure of HI at equilibrium will be

[HI]² = K_P x [H₂][I₂]

        = 4.7609 x 10^-4 x [0.630][0.630]

[HI]² = 1.8896 x 10^-4

[HI] = 0.013736 atm

The partial pressure of HI at equilibrium will be 0.013736 atm.