\(K_{a}\) for acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is \(1.754 \times 10^{-5} .\) At \(50^{\circ} \mathrm{C}, K_{\mathrm{a}}\) is \(1.633 \times 10^{-5}\). Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are not affected by a change in temperature, calculate \(\Delta S^{\circ}\) for the ionization of acetic acid.

To get the \(\Delta G^{\circ}\) values for the given two temperatures we will use \(\Delta G^{\circ}=-RT \ln \left(K_{a}\right)\) for each one, where K is the equilibrium constant, R is the gas constant, and T is the temperature in Kelvin. For \(25^{\circ} \mathrm{C}\) (or \(298\,\mathrm{K}\)): \(\Delta G_1^{\circ}=-R(298\,\mathrm{K})\ln\left(1.754 \times 10^{-5}\right)\) For \(50^{\circ} \mathrm{C}\) (or \(323\,\mathrm{K}\)): \(\Delta G_2^{\circ}=-R(323\,\mathrm{K})\ln\left(1.633 \times 10^{-5}\right)\)

Now we are going to set up equations for each of them using \(\Delta G^{\circ}= \Delta H^{\circ} - T\Delta S^{\circ}\). For \(25^{\circ} \mathrm{C}\) (\(298\,\mathrm{K}\)): \(\Delta G_1^{\circ}=\Delta H^{\circ}-298\,\mathrm{K}\Delta S^{\circ}\) For \(50^{\circ} \mathrm{C}\) (\(323\,\mathrm{K}\)): \(\Delta G_2^{\circ}=\Delta H^{\circ}-323\,\mathrm{K}\Delta S^{\circ}\)

Now we have a system of two linear equations with two variables, \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\). \(\Delta G_1^{\circ}-\Delta G_2^{\circ}=-25\,\mathrm{K}\Delta S^{\circ}\) Next, we can substitute the \(\Delta G^{\circ}\) values calculated in Step 1 and solve the equation for \(\Delta S^{\circ}\). \((-R(298\,\mathrm{K})\ln\left(1.754 \times 10^{-5}\right))-(-R(323\,\mathrm{K})\ln\left(1.633 \times 10^{-5}\right))=-25\,\mathrm{K}\Delta S^{\circ}\) Now, we can isolate \(\Delta S^{\circ}\) and get its value: \(\Delta S^{\circ}=\frac{-R(298\,\mathrm{K})\ln\left(1.754 \times 10^{-5}\right)+R(323\,\mathrm{K})\ln\left(1.633 \times 10^{-5}\right)}{-25\,\mathrm{K}}\) Finally, plugging the values of R(\(8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}\)) to get the numerical value for \(\Delta S^{\circ}\): \(\Delta S^{\circ}=\frac{-8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}(298\,\mathrm{K})\ln\left(1.754 \times 10^{-5}\right)+8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}(323\,\mathrm{K})\ln\left(1.633 \times 10^{-5}\right)}{-25\,\mathrm{K}}\) After calculating, we get: \(\Delta S^{\circ}\approx -109.94\, \mathrm{J\,K^{-1}\,mol^{-1}}\)