Question

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?

Short answer

(a) The reaction will have \(\Delta G = 0\) at 326.53 K. (b) If the temperature is increased from 326.53 K, the reaction will be nonspontaneous.

Step-by-step solution

Calculate the Temperature at which ΔG = 0

We are given the values for \(\Delta H\) and \(\Delta S\) and asked to find the temperature at which \(\Delta G = 0\). We can use the relation between these three quantities to solve for the temperature: \(\Delta G = \Delta H - T \Delta S\) 0 = -32 kJ - T (-98 J/K) Now, convert all the units to the same (in this case, J, J/K, and K) and solve for T: 0 = -32000 J + T (98 J/K) #Step 2: Solve for Temperature

Solve for Temperature

Now we have a linear equation with one variable, T: 32000 J = T (98 J/K) To find T, simply divide both sides of the equation by (98 J/K): T = \(\frac{32000~\text{J}}{98~\text{J/K}}\) = 326.53 K So, the reaction will have \(\Delta G = 0\) at 326.53 K. #Step 3: Evaluate the Spontaneity of the Reaction

Evaluate the spontaneity of the reaction

We are asked to determine whether the reaction will be spontaneous or nonspontaneous if the temperature is increased from 326.53 K. To do this, we have to check how the value of \(\Delta G\) changes with the increment in temperature. Remember, when: - \(\Delta G < 0\), the reaction is spontaneous. - \(\Delta G > 0\), the reaction is nonspontaneous. - \(\Delta G = 0\), the reaction is at equilibrium. In our case: \(\Delta G = \Delta H - T \Delta S\) Since \(\Delta H\) is negative and \(\Delta S\) is also negative, when the temperature (T) increases: \(- T \Delta S\) This term will become less negative, causing the value of \(\Delta G\) to get closer to zero or become positive. Thus, increasing the temperature will make the reaction nonspontaneous.