Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
For a reaction to be always spontaneous, we need to consider the Gibbs free energy change (ΔG). The formula for ΔG is given by:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
where:
- ΔG is the change in Gibbs free energy,
- ΔH is the change in enthalpy,
- T is the temperature in Kelvin, and
- ΔS is the change in entropy.
For a reaction to be spontaneous, ΔG must be negative ([tex]\( \Delta G < 0 \)[/tex]). Let's analyze the conditions under which this can be true at all temperatures:
1. Case 1: [tex]\(+\Delta H\)[/tex], [tex]\(-\Delta S\)[/tex]
If [tex]\(\Delta H\)[/tex] is positive and [tex]\(\Delta S\)[/tex] is negative:
[tex]\[ \Delta G = (+\Delta H) - T(-\Delta S) = \Delta H + T\Delta S \][/tex]
Since both [tex]\( \Delta H \)[/tex] and [tex]\(\Delta S\)[/tex] are positive, [tex]\( \Delta G \)[/tex] will be positive at all temperatures. Therefore, the reaction will never be spontaneous under these conditions. This option is not correct.
2. Case 2: [tex]\(+\Delta H\)[/tex], [tex]\(+\Delta S\)[/tex]
If [tex]\(\Delta H\)[/tex] is positive and [tex]\(\Delta S\)[/tex] is positive:
[tex]\[ \Delta G = (+\Delta H) - T(+\Delta S) = \Delta H - T\Delta S \][/tex]
This indicates that the reaction can be spontaneous at high temperatures, but not at low temperatures. Therefore, the reaction is not always spontaneous, and this option is not correct.
3. Case 3: [tex]\(\Delta H\)[/tex], [tex]\(-\Delta S\)[/tex]
If [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is negative:
[tex]\[ \Delta G = (-\Delta H) - T(-\Delta S) = -\Delta H + T\Delta S \][/tex]
If [tex]\( -\Delta H \)[/tex] is negative and [tex]\( T\Delta S \)[/tex] is positive, the reaction can be spontaneous at low temperatures but not at high temperatures. Therefore, the reaction is not always spontaneous, and this option is not correct.
4. Case 4: [tex]\(-\Delta H\)[/tex], [tex]\(+\Delta S\)[/tex]
If [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive:
[tex]\[ \Delta G = (-\Delta H) - T(+\Delta S) = -\Delta H - T\Delta S \][/tex]
Both terms are negative, making [tex]\(\Delta G\)[/tex] negative regardless of the temperature. Therefore, this condition guarantees that the reaction will always be spontaneous.
Hence, the correct relationship for a reaction that is always spontaneous is:
[tex]\[ \Delta H < 0 \quad \text{and} \quad \Delta S > 0 \][/tex]
Therefore, the correct choice is:
[tex]\[ \boxed{D. -\Delta H, +\Delta S} \][/tex]
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
where:
- ΔG is the change in Gibbs free energy,
- ΔH is the change in enthalpy,
- T is the temperature in Kelvin, and
- ΔS is the change in entropy.
For a reaction to be spontaneous, ΔG must be negative ([tex]\( \Delta G < 0 \)[/tex]). Let's analyze the conditions under which this can be true at all temperatures:
1. Case 1: [tex]\(+\Delta H\)[/tex], [tex]\(-\Delta S\)[/tex]
If [tex]\(\Delta H\)[/tex] is positive and [tex]\(\Delta S\)[/tex] is negative:
[tex]\[ \Delta G = (+\Delta H) - T(-\Delta S) = \Delta H + T\Delta S \][/tex]
Since both [tex]\( \Delta H \)[/tex] and [tex]\(\Delta S\)[/tex] are positive, [tex]\( \Delta G \)[/tex] will be positive at all temperatures. Therefore, the reaction will never be spontaneous under these conditions. This option is not correct.
2. Case 2: [tex]\(+\Delta H\)[/tex], [tex]\(+\Delta S\)[/tex]
If [tex]\(\Delta H\)[/tex] is positive and [tex]\(\Delta S\)[/tex] is positive:
[tex]\[ \Delta G = (+\Delta H) - T(+\Delta S) = \Delta H - T\Delta S \][/tex]
This indicates that the reaction can be spontaneous at high temperatures, but not at low temperatures. Therefore, the reaction is not always spontaneous, and this option is not correct.
3. Case 3: [tex]\(\Delta H\)[/tex], [tex]\(-\Delta S\)[/tex]
If [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is negative:
[tex]\[ \Delta G = (-\Delta H) - T(-\Delta S) = -\Delta H + T\Delta S \][/tex]
If [tex]\( -\Delta H \)[/tex] is negative and [tex]\( T\Delta S \)[/tex] is positive, the reaction can be spontaneous at low temperatures but not at high temperatures. Therefore, the reaction is not always spontaneous, and this option is not correct.
4. Case 4: [tex]\(-\Delta H\)[/tex], [tex]\(+\Delta S\)[/tex]
If [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive:
[tex]\[ \Delta G = (-\Delta H) - T(+\Delta S) = -\Delta H - T\Delta S \][/tex]
Both terms are negative, making [tex]\(\Delta G\)[/tex] negative regardless of the temperature. Therefore, this condition guarantees that the reaction will always be spontaneous.
Hence, the correct relationship for a reaction that is always spontaneous is:
[tex]\[ \Delta H < 0 \quad \text{and} \quad \Delta S > 0 \][/tex]
Therefore, the correct choice is:
[tex]\[ \boxed{D. -\Delta H, +\Delta S} \][/tex]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
