For a certain reaction,$\Delta$H° = -74.4 kJ and $\Delta$S° = -227 J/K.If n = 3,calculate $\varepsilon$ $\circ$ for the reaction at 25°C.

Question

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The correct answer is A because the Nernst equation is:E° = -RT/nF * ln(K)where E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is the Faraday constant, and K is the equilibrium constant.In this case, n = 3, R = 8.314 J/mol*K, T = 298 K, and K = exp(-ΔG°/RT) = exp(-(-74.4 kJ/mol)/(8.314 J/mol*K * 298 K)) = 1.23 x 10^11.Therefore, E° = -8.314 J/mol*K * 298 K / (3 mol * 96485 C/mol) * ln(1.23 x 10^11) = 0.0233 V.

For a Certain Reaction Δ\DeltaΔ H° = -744 KJ And Reaction Δ\DeltaΔ

For a Certain Reaction $\Delta$ H° = -744 KJ And Reaction $\Delta$