Chemistry

Chemistry

Problem

Aqueous iodide can be oxidized by hydrogen peroxide according to the balanced equation H2O2(aq)+2 I−(aq)+2 H+(aq)⇌I2(aq)+2 H2O(l)H_{2}O_{2}(aq) + 2\,I^{-}(aq) + 2\,H^{+}(aq) \rightleftharpoons I_{2}(aq) + 2\,H_{2}O(l) At 298 K298\,\text{K} the equilibrium constant for this reaction is K=3.2×10−4K = 3.2 \times 10^{-4}. The universal gas constant is R=8.314 J mol−1 K−1R = 8.314\,\text{J\,mol}^{-1}\,\text{K}^{-1}. a) Determine the standard Gibbs free-energy change ΔG∘\Delta G^\circ (in kJ mol−1\text{kJ\,mol}^{-1}) for the reaction at 298 K298\,\text{K}. b) A chemist prepares a reaction mixture in which the reaction quotient is Q=2.0×10−2Q = 2.0 \times 10^{-2} at 298 K298\,\text{K}. Calculate the Gibbs free-energy change ΔG\Delta G for these nonstandard conditions and state whether the mixture will spontaneously shift toward reactants or products.
Intuitive explanation

This problem connects equilibrium position to the underlying thermodynamics driving a redox reaction. A small KK tells you the reaction strongly favors reactants, which means the products are thermodynamically unfavorable—so ΔG°\Delta G° must be positive. You'll use the relationship ΔG°=−RTln⁡K\Delta G° = -RT\ln K to quantify this unfavorability and understand why this oxidation doesn't proceed spontaneously under standard conditions.

Intuitive explanation

This problem connects equilibrium position to the underlying thermodynamics driving a redox reaction. A small KK tells you the reaction strongly favors reactants, which means the products are thermodynamically unfavorable—so ΔG°\Delta G° must be positive. You'll use the relationship ΔG°=−RTln⁡K\Delta G° = -RT\ln K to quantify this unfavorability and understand why this oxidation doesn't proceed spontaneously under standard conditions.

Browse the problem library