At a certain temperature, the equilibrium constant, KC, for this reaction is 53.3. H2(g)+I2(g)↽−−⇀2HI(g)𝐾c=53.3 H 2 ( g ) + I 2 ( g ) ↽ − − ⇀ 2 HI ( g ) K c = 53.3

Question

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

H2(g)+I2(g)↽−−⇀2HI(g)𝐾c=53.3

At this temperature, 0.600 mol H2 and 0.600 mol I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

Answer 1

aA + bB <-> cC

Kc = [C]^c / [A]^a[B]^b at equilibrium

Moles are conserved since 1 mol H2 + 1 mol I2 -> 2 mol HI

The total moles in the system are 0.6 + 0.6 mol = 1.2 mol, so [HI] + [H2] + [I2] = 1.2 mol/L

H2 and I2 are reacted at the same rate, so [H2] = [I2]

53.3 = [HI}^2 / [H2][I2] = [HI]^2 / [H2]^2

[HI] = sqrt(53.3) * [H2]

[HI] + [H2] + [I2] = 1.2 mol/L = [H2] + [H2] + sqrt(53.3) * [H2] = 1.2 mol/L = 9.3[H2]

[H2] = 0.129 mol/L

[I2] = 0.129 mol/L

[HI] = 0.942 mol/L