At a certain temperature, the equilibrium constant, 𝐾c, for this reaction is 53.3.

Question

H2(g)+I2(g)↽−−⇀2HI(g) 𝐾c=53.3

At this temperature, 0.800 mol H2 and 0.800 mol I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

Answer 1

H₂(g) + I₂(g) <==> 2HI(g) .. Kc = 53.3

0.8........0.8..............0........Initial

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

0.8-x....0.8-x...........2x.......Equilibrium

Kc = 53.3 = [HI]² / [H₂][I₂]

53.3 = (0.8-x)(0.8-x) / (2x)²

Solve for x

[H₂] and [I₂] @ equilibrium = 0.8 - x molar