Answer :
To understand when a high temperature could make a reaction that was nonspontaneous at low temperature become spontaneous, we need to analyze the Gibbs free energy equation:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
Here:
- [tex]\(\Delta G\)[/tex] is the change in Gibbs free energy.
- [tex]\(\Delta H\)[/tex] is the change in enthalpy.
- [tex]\(T\)[/tex] is the temperature in Kelvin.
- [tex]\(\Delta S\)[/tex] is the change in entropy.
A reaction is spontaneous if [tex]\(\Delta G < 0\)[/tex] (negative).
Let's analyze the given options:
Option A: [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive.
- [tex]\(\Delta H < 0\)[/tex]
- [tex]\(\Delta S > 0\)[/tex]
In this case, both terms [tex]\(\Delta H\)[/tex] and [tex]\( -T \Delta S\)[/tex] are negative (because [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive). Thus, [tex]\(\Delta G\)[/tex] will be negative at any temperature, meaning the reaction is already spontaneous at low temperatures and will remain spontaneous at high temperatures.
Option B: [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both negative.
- [tex]\(\Delta H < 0\)[/tex]
- [tex]\(\Delta S < 0\)[/tex]
Here, [tex]\(\Delta H\)[/tex] is negative, but [tex]\( -T \Delta S\)[/tex] is positive since [tex]\(\Delta S\)[/tex] is negative. At low temperatures, [tex]\(\Delta G\)[/tex] might be negative (spontaneous) because the [tex]\(\Delta H\)[/tex] term dominates. However, at high temperatures, the positive [tex]\( -T \Delta S\)[/tex] term grows larger, making [tex]\(\Delta G\)[/tex] less negative or even positive, thus making the reaction less spontaneous or nonspontaneous.
Option C: [tex]\(\Delta H\)[/tex] is positive and [tex]\(\Delta S\)[/tex] is negative.
- [tex]\(\Delta H > 0\)[/tex]
- [tex]\(\Delta S < 0\)[/tex]
In this case, both [tex]\(\Delta H\)[/tex] and [tex]\( -T \Delta S\)[/tex] are positive. Thus, [tex]\(\Delta G\)[/tex] is positive at all temperatures, meaning the reaction is nonspontaneous regardless of the temperature.
Option D: [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both positive.
- [tex]\(\Delta H > 0\)[/tex]
- [tex]\(\Delta S > 0\)[/tex]
Here, [tex]\(\Delta H\)[/tex] is positive, but [tex]\( -T \Delta S\)[/tex] is negative. At low temperatures, [tex]\(\Delta G\)[/tex] might be positive (nonspontaneous) because the [tex]\(\Delta H\)[/tex] term dominates. At higher temperatures, the negative [tex]\( -T \Delta S\)[/tex] term becomes larger, potentially making [tex]\(\Delta G\)[/tex] negative and making the reaction spontaneous.
Thus, the correct answer is:
D. When [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both positive.
At high temperatures, the [tex]\(-T\Delta S\)[/tex] term can become large enough to make [tex]\(\Delta G\)[/tex] negative, turning a nonspontaneous reaction at low temperature into a spontaneous one.
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
Here:
- [tex]\(\Delta G\)[/tex] is the change in Gibbs free energy.
- [tex]\(\Delta H\)[/tex] is the change in enthalpy.
- [tex]\(T\)[/tex] is the temperature in Kelvin.
- [tex]\(\Delta S\)[/tex] is the change in entropy.
A reaction is spontaneous if [tex]\(\Delta G < 0\)[/tex] (negative).
Let's analyze the given options:
Option A: [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive.
- [tex]\(\Delta H < 0\)[/tex]
- [tex]\(\Delta S > 0\)[/tex]
In this case, both terms [tex]\(\Delta H\)[/tex] and [tex]\( -T \Delta S\)[/tex] are negative (because [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive). Thus, [tex]\(\Delta G\)[/tex] will be negative at any temperature, meaning the reaction is already spontaneous at low temperatures and will remain spontaneous at high temperatures.
Option B: [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both negative.
- [tex]\(\Delta H < 0\)[/tex]
- [tex]\(\Delta S < 0\)[/tex]
Here, [tex]\(\Delta H\)[/tex] is negative, but [tex]\( -T \Delta S\)[/tex] is positive since [tex]\(\Delta S\)[/tex] is negative. At low temperatures, [tex]\(\Delta G\)[/tex] might be negative (spontaneous) because the [tex]\(\Delta H\)[/tex] term dominates. However, at high temperatures, the positive [tex]\( -T \Delta S\)[/tex] term grows larger, making [tex]\(\Delta G\)[/tex] less negative or even positive, thus making the reaction less spontaneous or nonspontaneous.
Option C: [tex]\(\Delta H\)[/tex] is positive and [tex]\(\Delta S\)[/tex] is negative.
- [tex]\(\Delta H > 0\)[/tex]
- [tex]\(\Delta S < 0\)[/tex]
In this case, both [tex]\(\Delta H\)[/tex] and [tex]\( -T \Delta S\)[/tex] are positive. Thus, [tex]\(\Delta G\)[/tex] is positive at all temperatures, meaning the reaction is nonspontaneous regardless of the temperature.
Option D: [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both positive.
- [tex]\(\Delta H > 0\)[/tex]
- [tex]\(\Delta S > 0\)[/tex]
Here, [tex]\(\Delta H\)[/tex] is positive, but [tex]\( -T \Delta S\)[/tex] is negative. At low temperatures, [tex]\(\Delta G\)[/tex] might be positive (nonspontaneous) because the [tex]\(\Delta H\)[/tex] term dominates. At higher temperatures, the negative [tex]\( -T \Delta S\)[/tex] term becomes larger, potentially making [tex]\(\Delta G\)[/tex] negative and making the reaction spontaneous.
Thus, the correct answer is:
D. When [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both positive.
At high temperatures, the [tex]\(-T\Delta S\)[/tex] term can become large enough to make [tex]\(\Delta G\)[/tex] negative, turning a nonspontaneous reaction at low temperature into a spontaneous one.
