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 the conditions under which [tex]\(\Delta G\)[/tex] is always negative, since a reaction is spontaneous when [tex]\(\Delta G < 0\)[/tex].
### Step-by-Step Analysis:
1. Understanding the Equation:
- [tex]\(\Delta G\)[/tex]: Change in Gibbs free energy
- [tex]\(\Delta H\)[/tex]: Change in enthalpy
- [tex]\(T\)[/tex]: Absolute temperature (in Kelvin)
- [tex]\(\Delta S\)[/tex]: Change in entropy
2. Conditions for Spontaneity:
- For a reaction to be spontaneous at any temperature, [tex]\(\Delta G\)[/tex] must always be less than zero ([tex]\(\Delta G < 0\)[/tex]).
- This requires careful consideration of the signs of [tex]\(\Delta H\)[/tex] (enthalpy change) and [tex]\(\Delta S\)[/tex] (entropy change).
3. Evaluating Sign Combinations:
- Case 1: [tex]\(\Delta H > 0\)[/tex] and [tex]\(\Delta S > 0\)[/tex]:
- Here, whether [tex]\(\Delta G\)[/tex] is negative depends on the temperature [tex]\(T\)[/tex].
- At low temperatures, [tex]\(\Delta G\)[/tex] might be positive, making the reaction non-spontaneous.
- Case 2: [tex]\(\Delta H < 0\)[/tex] and [tex]\(\Delta S < 0\)[/tex]:
- Here, at higher temperatures, [tex]\(-T\Delta S\)[/tex] can be a large positive number, which might make [tex]\(\Delta G\)[/tex] positive.
- Case 3: [tex]\(\Delta H > 0\)[/tex] and [tex]\(\Delta S < 0\)[/tex]:
- In this scenario, [tex]\(\Delta G\)[/tex] is positive in most conditions, leading to non-spontaneity.
- Case 4: [tex]\(\Delta H < 0\)[/tex] and [tex]\(\Delta S > 0\)[/tex]:
- Here, [tex]\(\Delta H\)[/tex] is negative and [tex]\( -T\Delta S\)[/tex] is also negative for all [tex]\(T\)[/tex].
- This ensures [tex]\(\Delta G\)[/tex] is negative no matter the value of [tex]\(T\)[/tex], making the reaction always spontaneous.
4. Conclusion:
- The reaction is always spontaneous when [tex]\(\Delta H\)[/tex] is negative ([tex]\(\Delta H < 0\)[/tex]) and [tex]\(\Delta S\)[/tex] is positive ([tex]\(\Delta S > 0\)[/tex]) because these conditions ensure [tex]\(\Delta G\)[/tex] is negative at all temperatures.
Therefore, the correct answer is:
B. When [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive.
### Step-by-Step Analysis:
1. Understanding the Equation:
- [tex]\(\Delta G\)[/tex]: Change in Gibbs free energy
- [tex]\(\Delta H\)[/tex]: Change in enthalpy
- [tex]\(T\)[/tex]: Absolute temperature (in Kelvin)
- [tex]\(\Delta S\)[/tex]: Change in entropy
2. Conditions for Spontaneity:
- For a reaction to be spontaneous at any temperature, [tex]\(\Delta G\)[/tex] must always be less than zero ([tex]\(\Delta G < 0\)[/tex]).
- This requires careful consideration of the signs of [tex]\(\Delta H\)[/tex] (enthalpy change) and [tex]\(\Delta S\)[/tex] (entropy change).
3. Evaluating Sign Combinations:
- Case 1: [tex]\(\Delta H > 0\)[/tex] and [tex]\(\Delta S > 0\)[/tex]:
- Here, whether [tex]\(\Delta G\)[/tex] is negative depends on the temperature [tex]\(T\)[/tex].
- At low temperatures, [tex]\(\Delta G\)[/tex] might be positive, making the reaction non-spontaneous.
- Case 2: [tex]\(\Delta H < 0\)[/tex] and [tex]\(\Delta S < 0\)[/tex]:
- Here, at higher temperatures, [tex]\(-T\Delta S\)[/tex] can be a large positive number, which might make [tex]\(\Delta G\)[/tex] positive.
- Case 3: [tex]\(\Delta H > 0\)[/tex] and [tex]\(\Delta S < 0\)[/tex]:
- In this scenario, [tex]\(\Delta G\)[/tex] is positive in most conditions, leading to non-spontaneity.
- Case 4: [tex]\(\Delta H < 0\)[/tex] and [tex]\(\Delta S > 0\)[/tex]:
- Here, [tex]\(\Delta H\)[/tex] is negative and [tex]\( -T\Delta S\)[/tex] is also negative for all [tex]\(T\)[/tex].
- This ensures [tex]\(\Delta G\)[/tex] is negative no matter the value of [tex]\(T\)[/tex], making the reaction always spontaneous.
4. Conclusion:
- The reaction is always spontaneous when [tex]\(\Delta H\)[/tex] is negative ([tex]\(\Delta H < 0\)[/tex]) and [tex]\(\Delta S\)[/tex] is positive ([tex]\(\Delta S > 0\)[/tex]) because these conditions ensure [tex]\(\Delta G\)[/tex] is negative at all temperatures.
Therefore, the correct answer is:
B. When [tex]\(\Delta H\)[/tex] is negative and [tex]\(\Delta S\)[/tex] is positive.