Question

At a given temperature, 1.43 atm of H2 and 3.15 atm of I2 are mixed and allowed to come to equilibrium. The equilibrium pressure of HI is found to be 1.844 atm. Calculate Kp for the reaction at this temperature. H2(g) + I2(g) <=> 2 HI(g).

Answer 1

construct the ICE table

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

I         1.43      3.15                0

C         -x          -x                +2x

E        1.43-x       3.15-x        +2x

Kp = [HI]² / [H2][I2]

but he has alrready given equilibrium pressure which is 1.844 atm

[HI] = 2x = 1.844 atm

x = 1.844 / 2 = 0.922 atm

[H2] = 1.43-x = 1.43-0.922 = 0.508 atm

[I2] = 3.15-x = 3.15 - 0.922 = 2.228 atm

now we have all equilibrium concentratiions

Kp = [1.844]² / [0.508][2.288]

Kp = 3.0043