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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]
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