Answer :
To determine when a reaction is always spontaneous according to the Gibbs free energy equation, [tex]\(\Delta G = \Delta H - T\Delta S\)[/tex], we need to analyze how changes in enthalpy ([tex]\(\Delta H\)[/tex]) and entropy ([tex]\(\Delta S\)[/tex]) affect the Gibbs free energy ([tex]\(\Delta G\)[/tex]).
A reaction is considered spontaneous if [tex]\(\Delta G < 0\)[/tex].
Let's examine each scenario:
### A. When [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both positive
- [tex]\(\Delta H > 0\)[/tex]
- [tex]\(\Delta S > 0\)[/tex]
- [tex]\(\Delta G = \Delta H - T\Delta S\)[/tex]
Since both [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are positive, the term [tex]\(\Delta H\)[/tex] increases [tex]\(\Delta G\)[/tex], while the term [tex]\(-T\Delta S\)[/tex] decreases [tex]\(\Delta G\)[/tex]. Whether [tex]\(\Delta G\)[/tex] is negative depends on the temperature [tex]\(T\)[/tex]. At high temperatures, [tex]\(T\Delta S\)[/tex] might be large enough to make [tex]\(\Delta G\)[/tex] negative, but at low temperatures, [tex]\(\Delta H\)[/tex] might dominate and make [tex]\(\Delta G\)[/tex] positive. Therefore, a reaction is not always spontaneous in this case.
### B. When [tex]\(\Delta H\)[/tex] is positive and [tex]\(\Delta S\)[/tex] is negative
- [tex]\(\Delta H > 0\)[/tex]
- [tex]\(\Delta S < 0\)[/tex]
- [tex]\(\Delta G = \Delta H - T(\Delta S)\)[/tex]
With a positive [tex]\(\Delta H\)[/tex] and a negative [tex]\(\Delta S\)[/tex], the expression [tex]\(-T(\Delta S)\)[/tex] turns into [tex]\(+T|\Delta S|\)[/tex]. This means both terms [tex]\(\Delta H\)[/tex] and [tex]\(T|\Delta S|\)[/tex] will increase [tex]\(\Delta G\)[/tex], making [tex]\(\Delta G\)[/tex] always positive. Therefore, a reaction is never spontaneous in this case.
### C. When [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both negative
- [tex]\(\Delta H < 0\)[/tex]
- [tex]\(\Delta S < 0\)[/tex]
- [tex]\(\Delta G = \Delta H - T\Delta S\)[/tex]
When both [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are negative, the term [tex]\(-T\Delta S\)[/tex] turns into [tex]\(+T|\Delta S|\)[/tex]. The [tex]\(\Delta H\)[/tex] term decreases [tex]\(\Delta G\)[/tex], but the [tex]\(T|\Delta S|\)[/tex] term increases [tex]\(\Delta G\)[/tex]. Depending on the temperature, [tex]\(\Delta G\)[/tex] can either be negative (low temperatures) or positive (high temperatures). Therefore, a reaction is not always spontaneous in this case.
### D. When [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive
- [tex]\(\Delta H < 0\)[/tex]
- [tex]\(\Delta S > 0\)[/tex]
- [tex]\(\Delta G = \Delta H - T\Delta S\)[/tex]
With a negative [tex]\(\Delta H\)[/tex] and a positive [tex]\(\Delta S\)[/tex], both terms work to decrease [tex]\(\Delta G\)[/tex]. The [tex]\(\Delta H\)[/tex] term decreases [tex]\(\Delta G\)[/tex], and the [tex]\( -T\Delta S\)[/tex] term further decreases [tex]\(\Delta G\)[/tex]. Therefore, [tex]\(\Delta G\)[/tex] will always be negative, regardless of the temperature.
Thus, the reaction is always spontaneous in scenario D. So, the correct answer is:
[tex]\(\boxed{D}\)[/tex] When [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive
A reaction is considered spontaneous if [tex]\(\Delta G < 0\)[/tex].
Let's examine each scenario:
### A. When [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both positive
- [tex]\(\Delta H > 0\)[/tex]
- [tex]\(\Delta S > 0\)[/tex]
- [tex]\(\Delta G = \Delta H - T\Delta S\)[/tex]
Since both [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are positive, the term [tex]\(\Delta H\)[/tex] increases [tex]\(\Delta G\)[/tex], while the term [tex]\(-T\Delta S\)[/tex] decreases [tex]\(\Delta G\)[/tex]. Whether [tex]\(\Delta G\)[/tex] is negative depends on the temperature [tex]\(T\)[/tex]. At high temperatures, [tex]\(T\Delta S\)[/tex] might be large enough to make [tex]\(\Delta G\)[/tex] negative, but at low temperatures, [tex]\(\Delta H\)[/tex] might dominate and make [tex]\(\Delta G\)[/tex] positive. Therefore, a reaction is not always spontaneous in this case.
### B. When [tex]\(\Delta H\)[/tex] is positive and [tex]\(\Delta S\)[/tex] is negative
- [tex]\(\Delta H > 0\)[/tex]
- [tex]\(\Delta S < 0\)[/tex]
- [tex]\(\Delta G = \Delta H - T(\Delta S)\)[/tex]
With a positive [tex]\(\Delta H\)[/tex] and a negative [tex]\(\Delta S\)[/tex], the expression [tex]\(-T(\Delta S)\)[/tex] turns into [tex]\(+T|\Delta S|\)[/tex]. This means both terms [tex]\(\Delta H\)[/tex] and [tex]\(T|\Delta S|\)[/tex] will increase [tex]\(\Delta G\)[/tex], making [tex]\(\Delta G\)[/tex] always positive. Therefore, a reaction is never spontaneous in this case.
### C. When [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are both negative
- [tex]\(\Delta H < 0\)[/tex]
- [tex]\(\Delta S < 0\)[/tex]
- [tex]\(\Delta G = \Delta H - T\Delta S\)[/tex]
When both [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] are negative, the term [tex]\(-T\Delta S\)[/tex] turns into [tex]\(+T|\Delta S|\)[/tex]. The [tex]\(\Delta H\)[/tex] term decreases [tex]\(\Delta G\)[/tex], but the [tex]\(T|\Delta S|\)[/tex] term increases [tex]\(\Delta G\)[/tex]. Depending on the temperature, [tex]\(\Delta G\)[/tex] can either be negative (low temperatures) or positive (high temperatures). Therefore, a reaction is not always spontaneous in this case.
### D. When [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive
- [tex]\(\Delta H < 0\)[/tex]
- [tex]\(\Delta S > 0\)[/tex]
- [tex]\(\Delta G = \Delta H - T\Delta S\)[/tex]
With a negative [tex]\(\Delta H\)[/tex] and a positive [tex]\(\Delta S\)[/tex], both terms work to decrease [tex]\(\Delta G\)[/tex]. The [tex]\(\Delta H\)[/tex] term decreases [tex]\(\Delta G\)[/tex], and the [tex]\( -T\Delta S\)[/tex] term further decreases [tex]\(\Delta G\)[/tex]. Therefore, [tex]\(\Delta G\)[/tex] will always be negative, regardless of the temperature.
Thus, the reaction is always spontaneous in scenario D. So, the correct answer is:
[tex]\(\boxed{D}\)[/tex] When [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive
