#### Question

At 700 K, equilibrium constant for the reaction

`H_(2(g)) + I_(2(g)) ↔ 2HI_(g)` is 54.8. If 0.5 molL^{–1} of HI_{(}_{g}_{)} is present at equilibrium at 700 K, what are the concentration of H_{2(}_{g}_{)} and I_{2(}_{g}_{)} assuming that we initially started with HI_{(}_{g}_{)} and allowed it to reach equilibrium at 700 K?

#### Solution

It is given that equilibrium constant `K_C` for the reaction

`H_(2(g)) + I_(2(g)) ↔ 2HI_((g))` is 54.8.

Therefore, at equilibrium, the equilibrium constant `K'_C` for the reaction

2HI_(g) ↔ H_(2(g)) + I_(2(g)) will be `1/54.8`

[HI] = 0.5 `molL^(-1)`

Let the concentrations of hydrogen and iodine at equilibrium be *x* molL^{–1}

`[H_2] = [I_2] = xmolL^(-1)`

Therefore, `([H_2][I_2])/[HI]^2 = K'_C`

`=> (x xx x)/(0.5)^2 = 1/54.8`

`x^2 = 0.25/54.8`

`=> x = 0.06754`

x = 0.068 `molL^(-1)` (approximatley)

Hence, at equilibrium, `[H_2] = [I_2] = 0.068 mol L^(-1)`