For a particular reaction, the equilibrium constant is 1.56 × 10^-2 at 324°C, and ΔH° is +12.9 kJ at 25°C. Assuming ΔH° and ΔS° are temperature independent, calculate ΔS° for the reaction.

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Quizplus · Accepted Answer

-13.0 J/KThe equilibrium constant (K) is related to the Gibbs free energy change (ΔG°) by the equation ΔG° = -RTlnK, where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. Given that K = 1.56 × 10^-2 at 324°C (which is 597K), we can calculate ΔG°. First, convert 324°C to Kelvin: 324 + 273 = 597 K.ΔG° = -8.314 × 597 × ln(1.56 × 10^-2) = 8.314 × 597 × 4.153 = 20571.7 J/mol ≈ 20.6 kJ/mol.The relationship between ΔG°, ΔH°, and ΔS° is given by ΔG° = ΔH° - TΔS°. Rearranging this equation to solve for ΔS° gives ΔS° = (ΔH° - ΔG°) / T.Given ΔH° = +12.9 kJ/mol = 12900 J/mol and ΔG° ≈ 20600 J/mol at 597 K, we can calculate ΔS° as follows:ΔS° = (12900 - 20600) / 597 = -7700 / 597 ≈ -12.9 J/K·mol.Therefore, the closest answer is -13.0 J/K, which is option D.

For a Particular Reaction, the Equilibrium Constant Is 1