Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
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
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
