At a certain temperature, the equilibrium constant, K, for the given reaction is 53.3

Question

H2 (g) + I2 (g) ---> (and the reverse arrow) 2 HI (g) K = 53.3

At this temperature, the reactants were placed in a container to react. If the partial pressures of H2 and I2 were each 0.700 bar initially, what is the partial pressure of HI at equilibrium?

Answer 1

H₂(g) + I₂(g) <==> 2HI(g)

0.700....0.700..........0.........Initial

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

0.7-x.....0.7-x.........2x.........Equilibrium

Kp = (HI)² / (H₂)(I₂)

53.3 = (2x)² / (0.7-x)(0.7-x)

53.3 = 4x² / x² -1.4x + 0.49

Solving for x using the quadratic formula...

x = 0.631

Equilibrium pressure of HI = 2x = 2 x 0.631 bar

Equilibrium pressure of HI = 1.26 bar

(be sure to check the math, especially the quadratic solving for x)