Question

Consider the following reaction:
H2(g)+I2(g)⇌2HI(g)
A reaction mixture at equilibrium at 175 K containsPH2=0.958atm, PI2=0.877atm, and PHI=0.020atm. A second reaction mixture, also at 175 K, contains PH2=PI2= 0.628 atm , and PHI= 0.107 atm .

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

Answer 1

i) Calculation of K from first reaction mixture

H2(g) + I2(g) <------> 2HI(g)

K = (P_HI)²/(P_H2 × P_I2)

K = (0.020atm)²/(0.958atm × 0.877atm ) = 4.76 ×10^-4

ii) Calculation of partial pressure of HI when second reaction mixture reaches equillibrium

Reaction quotient , Q = (0.107atm)²/(0.628atm)² = 0.02903atm

Q is not equal to K , therefore , reaction mixture is not at equillbrium

Initial partial pressure

P_H2 = 0.628

P_I2 = 0.628

P_HI = 0.107

Change in concentration

P_H2 = - x

P_I2 = - x

P_HI = + 2x

Equillibrium concentration

P_H2 = 0.628 - x

P_I2 = 0.628 - x

P_HI = 0.107 + 2x

so,

(0.107 + 2x)²/(0.628 -x)² = 4.76 ×10^-4

0.107 + 2x / 0.628 - x = 0.02182

0.107 + 2x =_I 0.013703 - 0.02182x

2.02182x = - 0.093297

x = - 0.04615

Therefore , when equillibrium is established

P_H2 = 0.628 - (-0.04615) = 0.6742atm

P_I2 = 0.628 - (-0.04615) = 0.6742atm

P_HI = 0.107 + (2 × - 0.04615) = 0.0147atm