For a particular reaction, \(\Delta H=-32 \mathrm{kJ}\) and \(\Delta S=-98 \mathrm{J} / \mathrm{K}\) . Assume that \(\Delta H\) and \(\Delta S\) do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0\) ? (b) If \(T\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

We have the equation for Gibbs free energy change: \[ \Delta G = \Delta H - T\Delta S \] We're given \(\Delta H = -32\,\text{kJ}\) and \(\Delta S = -98\,\text{J/K}\). Note that we must convert \(\Delta H\) to J so that the units are consistent in the equation: \(-32\,\text{kJ} = -32,000\,\text{J}\). Now we can write the equation with the given values: \[ \Delta G = -32,000\,\text{J} - T(-98\,\text{J/K}) \]

We want to find the temperature \(T\) at which the free energy change \(\Delta G = 0\). Thus, we can set the equation to zero and solve for \(T\): \[ 0 = -32,000\,\text{J} + 98\,\text{J/K} \cdot T \] Now, isolate the temperature on one side of the equation: \[ T = \frac{32{,}000\,\text{J}}{98\,\text{J/K}} \] Divide to find the temperature: \[ T \approx 326.53\,\text{K} \]

To find if the reaction is spontaneous or nonspontaneous, we must first look at the sign of \(\Delta G\) as the temperature increases. Recall that for this problem, both \(\Delta H\) and \(\Delta S\) are negative. When the temperature increases from \(T \approx 326.53\,\text{K}\), the magnitude of the term \(T\Delta S\) will also increase, because the product of two negative numbers will be positive. Since \(\Delta H\) is already negative, adding more positive values (increasing the magnitude of \(T\Delta S\)) will make \(\Delta G\) more positive. When \(\Delta G\) is positive, the reaction is nonspontaneous. In summary, as the temperature increases from \(T \approx 326.53\,\text{K}\), the reaction will become nonspontaneous.