Question

The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K: H2(g) + I2(g) 2HI(g) Calculate the equilibrium concentrations of reactants and product when 0.276 moles of H2 and 0.276 moles of I2 are introduced into a 1.00 L vessel at 698 K.

[H2] = _____M

[I2] =______M

[HI] =______M

Answer 1

Answer 2

The balanced equation for the reaction is:

H2(g) + I2(g) ⇌ 2HI(g)

The equilibrium constant expression for the reaction is:

Kc = [HI]^2 / [H2][I2]

At equilibrium, the concentrations of the reactants and products will be:

[H2] = initial concentration - x [I2] = initial concentration - x [HI] = 2x

where x is the change in concentration at equilibrium.

Substituting these expressions into the equilibrium constant expression and solving for x, we get:

55.6 = (2x)^2 / (0.276 - x)(0.276 - x) x = 0.143 M

Therefore, the equilibrium concentrations are:

[H2] = 0.276 - x = 0.133 M [I2] = 0.276 - x = 0.133 M [HI] = 2x = 0.286 M

So the final answers are:

[H2] = 0.133 M [I2] = 0.133 M [HI] = 0.286 M