What is the equilibrium constant, K eq , for the following reaction at 25 C, given that D G rxn = - 16.8 kJ/mol?
H₂ (g) + I₂ (g) 2 HI (g)

Question

What is the equilibrium constant, K eq , for the following reaction at 25 C, given that D G rxn = - 16.8 kJ/mol? H₂ (g) + I₂ (g) 2 HI (g)

Quizplus · Accepted Answer

The equilibrium constant, K(eq), can be calculated from the Gibbs free energy, ΔG(rxn), using the equation: ΔG(rxn) = -RTlnK(eq) where R is the gas constant (8.314 J/mol.K), T is the temperature in Kelvin, and ln is the natural logarithm. Substituting the given values: -16,800 J/mol = -(8.314 J/mol.K)(298 K)lnK(eq) K(eq) = e^(-16,800 J/mol / (8.314 J/mol.K × 298 K)) K(eq) = e^(-6.687) K(eq) = 0.00127 M^-2 or 1.27 × 10^-3 M^-2 However, the balanced equation shows that there are two moles of reactant (H₂ and I₂) and two moles of product (HI), which means that the equilibrium constant should be squared: K(eq)^2 = (1.27 × 10^-3)^2 K(eq)^2 = 1.61 × 10^-9 M^-4 Taking the square root gives: K(eq) = 1.27 × 10^-4 M^-2, which is closest to answer choice C (884).

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