For a reaction: , carried out at , the ratio of equilibrium concentrations of product to reactant ch — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$\Delta G^\circ = -RT \ln K_c$. If $K_c$ increases by a factor of $e^4$, the new standard free energy is $\Delta G^{\circ\prime} = -RT \ln(K_c e^4) = \Delta G^\circ - 4RT$. The change is $-4RT = -4(2 \text{ cal/K-mol})(300 \text{ K}) = -2400 \text{ cal} = -2.4 \text{ kcal}$, denoting a 2.4 kcal fall.