At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
