Question

Consider the chemical equation and equilibrium constant at 25∘C:

H2(g)+I2(g)⇌2HI(g), K=6.2×10^2
Calculate the equilibrium constant for the following reaction at 25∘C:
HI(g) ⇌ 1/2 H2(g)+1/2 I2(g)

Answer 1

The equilibrium constant for the reaction HI(g) ⇌ 1/2 H2(g) + 1/2 I2(g) is 1.61 × 10^(-3).

Step-by-step explanation:

The equilibrium constant (K) for a reaction is determined by the concentrations of the reactants and products at equilibrium. In the given chemical equation H2(g) + I2(g) ⇌ 2HI(g), the equilibrium constant (K) is 6.2 × 10^2 at 25°C. To calculate the equilibrium constant for the reaction HI(g) ⇌ 1/2 H2(g) + 1/2 I2(g), we can use the relationship between the equilibrium constants of reverse reactions.

Since the reverse reaction is obtained by inverting the original equation, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the original reaction. Therefore, the equilibrium constant for HI(g) ⇌ 1/2 H2(g) + 1/2 I2(g) is 1/6.2 × 10^2, which can be simplified as 1.61 × 10^(-3).