To determine how temperature affects the spontaneity of a reaction where both the enthalpy change ([tex]\(\Delta H\)[/tex]) and the entropy change ([tex]\(\Delta S\)[/tex]) are positive, we turn to the Gibbs free energy equation:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
For a reaction to be spontaneous, the change in Gibbs free energy ([tex]\(\Delta G\)[/tex]) must be negative ([tex]\(\Delta G < 0\)[/tex]).
Given:
- [tex]\(\Delta H > 0\)[/tex] (positive)
- [tex]\(\Delta S > 0\)[/tex] (positive)
We need to find when [tex]\(\Delta G\)[/tex] is negative:
[tex]\[ \Delta G = \Delta H - T \Delta S < 0 \][/tex]
Rearranging the inequality:
[tex]\[ \Delta H < T \Delta S \][/tex]
[tex]\[ T \Delta S > \Delta H \][/tex]
Dividing both sides by [tex]\(\Delta S\)[/tex] (which is positive, so the direction of the inequality doesn't change):
[tex]\[ T > \frac{\Delta H}{\Delta S} \][/tex]
This indicates that the temperature [tex]\(T\)[/tex] must be greater than [tex]\(\frac{\Delta H}{\Delta S}\)[/tex] for the reaction to be spontaneous. Therefore, as the temperature increases, the term [tex]\(T \Delta S\)[/tex] becomes larger, making it more likely for the expression [tex]\(\Delta H - T \Delta S\)[/tex] to be negative, which in turn makes [tex]\(\Delta G\)[/tex] negative and the reaction spontaneous.
Therefore, the correct answer is:
B. A high temperature will make it spontaneous.
