Question

Discuss the effect of temperature on the spontaneity of reactions with the following values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ} .\) $$ \begin{array}{l} \text { (a) } \Delta H^{\circ}=128 \mathrm{~kJ} ; \Delta S^{\circ}=89.5 \mathrm{~J} / \mathrm{K} \\ \text { (b) } \Delta H^{\circ}=-20.4 \mathrm{~kJ} ; \Delta S^{\circ}=-156.3 \mathrm{~J} / \mathrm{K} \end{array} $$ (c) \(\Delta H^{\circ}=-127 \mathrm{~kJ} ; \Delta S^{\circ}=43.2 \mathrm{~J} / \mathrm{K}\)

Short answer

Question: Discuss the effect of temperature on the spontaneity of the given reactions: (a) \(\Delta H^{\circ} = 128\,\mathrm{kJ}\) and \(\Delta S^{\circ} = 89.5\,\mathrm{J/K}\) (b) \(\Delta H^{\circ} = -20.4\,\mathrm{kJ}\) and \(\Delta S^{\circ} = -156.3\,\mathrm{J/K}\) (c) \(\Delta H^{\circ} = -127\,\mathrm{kJ}\) and \(\Delta S^{\circ} = 43.2\,\mathrm{J/K}\) Answer: (a) The reaction will be spontaneous at high temperatures. (b) The reaction will be spontaneous at low temperatures. (c) The reaction will be spontaneous at all temperatures.

Step-by-step solution

Recall the Gibbs free energy equation

The Gibbs free energy equation is: $$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$$ Where \(\Delta G^{\circ}\) is the change in Gibbs free energy, \(\Delta H^{\circ}\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S^{\circ}\) is the change in entropy.

Analyze each reaction given the values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\)

We are given 3 reactions to analyze: (a) \(\Delta H^{\circ} = 128\,\mathrm{kJ}\) and \(\Delta S^{\circ} = 89.5\,\mathrm{J/K}\) (b) \(\Delta H^{\circ} = -20.4\,\mathrm{kJ}\) and \(\Delta S^{\circ} = -156.3\,\mathrm{J/K}\) (c) \(\Delta H^{\circ} = -127\,\mathrm{kJ}\) and \(\Delta S^{\circ} = 43.2\,\mathrm{J/K}\)

Discuss the effect of temperature on spontaneity for each reaction

(a) For the reaction with \(\Delta H^{\circ} > 0\) and \(\Delta S^{\circ} > 0\), the reaction will be spontaneous at high temperatures. As the temperature increases, the \(T\Delta S^{\circ}\) term will become larger, and since \(\Delta H^{\circ}\) is positive, the reaction will become spontaneous when the temperature is sufficiently high such that the \(T\Delta S^{\circ}\) term overcomes the positive \(\Delta H^{\circ}\), making \(\Delta G^{\circ} < 0\). (b) For the reaction with \(\Delta H^{\circ} < 0\) and \(\Delta S^{\circ} < 0\), the reaction will be spontaneous at low temperatures. In this case, as the temperature decreases, the \(T\Delta S^{\circ}\) term will become less negative since the product of two negative values is positive. When the temperature is low enough, the reaction becomes spontaneous as \(\Delta G^{\circ} < 0\). (c) For the reaction with \(\Delta H^{\circ} < 0\) and \(\Delta S^{\circ} > 0\), the reaction will always be spontaneous regardless of the temperature. As the temperature increases, the \(T\Delta S^{\circ}\) term becomes larger, and \(\Delta G^{\circ}\) becomes more negative. Since both \(\Delta H^{\circ}\) and \(T\Delta S^{\circ}\) terms in the Gibbs free energy equation have a negative contribution, the reaction remains spontaneous at all temperatures.