Question

If the \(\Delta G^{0}\) for a reaction is \(-4.5 \mathrm{kcal} / \mathrm{mol}\) at \(298 \mathrm{~K}\), what is the \(K_{\mathrm{eq}}\) for this reaction? What is the change in entropy of this reaction if \(\Delta H^{0}=-3.2 \mathrm{kcal} / \mathrm{mol}\) ?

Short answer

Question: Calculate the equilibrium constant, K_eq, and the change in entropy, ΔS, for a reaction given the standard Gibbs free energy change, ΔG° = -4.5 kcal/mol, the standard enthalpy change, ΔH° = -3.2 kcal/mol, and the temperature T = 298 K. Answer: The equilibrium constant, K_eq, for this reaction is approximately 1941.6 and the change in entropy, ΔS, is 18.26 J/mol.K.

Step-by-step solution

Conversion to SI units

First, we need to convert the given values of \(\Delta G^0\) and \(\Delta H^0\) to their SI units, which are J/mol. We know that 1 kcal = 4184 J, so: \(\Delta G^0 = -4.5 \,\text{kcal/mol} \times 4184 \frac{\text{J}}{\text{kcal}} = -18828 \,\text{J/mol}\) and \(\Delta H^0 = -3.2 \,\text{kcal/mol} \times 4184 \frac{\text{J}}{\text{kcal}} = -13388.8 \,\text{J/mol}\)

Calculate the equilibrium constant \(K_{eq}\)

Now, we will use the formula \(\Delta G^0 = -RT\ln{K_{eq}}\) to calculate the equilibrium constant \(K_{eq}\): Solving for \(K_{eq}\), we have: \(K_{eq} = e^{\frac{-\Delta G^0}{RT}}\) Plug in the known values: \(K_{eq} = e^{\frac{18828}{(8.314)(298)}} = e^{7.569} \approx 1941.6\) The equilibrium constant, \(K_{eq}\), for this reaction is approximately 1941.6.

Calculate the change in entropy, \(\Delta S\)

To find \(\Delta S\), we'll use the formula \(\Delta G^0 = \Delta H^0 - T\Delta S\): Solving for \(\Delta S\): \(\Delta S = \frac{\Delta H^0 - \Delta G^0}{T}\) Plug in the known values: \(\Delta S = \frac{-13388.8 - (-18828)}{298} = \frac{5441.2}{298} = 18.26 \,\text{J/mol.K}\) The change in entropy for this reaction is 18.26 J/mol.K.