Hydrogen iodide decomposes according to the equation 2HI(g) $\rightleftharpoons$ H₂(g) + I₂(g), for which K_c = 0.0156 at 400ºC. 0.550 mol HI was injected into a 2.00 L reaction vessel at 400ºC. Calculate the concentration of H₂ at equilibrium.

Question

Quizplus · Accepted Answer

The initial concentration of HI is 0.275 M (0.550 mol / 2.00 L). Let x be the change in concentration at equilibrium for H2 and I2. At equilibrium, [HI] = 0.275 - 2x, [H2] = x, and [I2] = x. The equilibrium expression is Kc = ([H2][I2]) / ([HI]^2) = 0.0156. Substituting the equilibrium concentrations gives 0.0156 = (x^2) / (0.275 - 2x)^2. Solving for x gives x ≈ 0.0275 M, which is the concentration of H2 at equilibrium.

Hydrogen Iodide Decomposes According to the Equation 2HI(g) ⇌\rightleftharpoons⇌ H2(g) + I2(g), for Which Kc = 0

Hydrogen Iodide Decomposes According to the Equation 2HI(g) $\rightleftharpoons$ H₂(g) + I₂(g), for Which K_c = 0