Answer :
To determine at which temperature the reaction is spontaneous, we need to calculate the Gibbs free energy change ([tex]\(\Delta G\)[/tex]) for each temperature given in the choices. The Gibbs free energy change can be calculated using the formula:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
where:
- [tex]\(\Delta H\)[/tex] is the enthalpy change of the reaction.
- [tex]\(T\)[/tex] is the temperature in Kelvin.
- [tex]\(\Delta S\)[/tex] is the entropy change of the reaction.
Given:
- [tex]\(\Delta H = -92 \)[/tex] kJ/mol
- [tex]\(\Delta S = -0.199 \)[/tex] kJ/(mol·K)
We need to evaluate [tex]\(\Delta G\)[/tex] for each of the following temperatures: [tex]\(500 K\)[/tex], [tex]\(400 K\)[/tex], [tex]\(700 K\)[/tex], and [tex]\(600 K\)[/tex].
1. At [tex]\(500 K\)[/tex]:
[tex]\[ \Delta G = -92 - (500 \times -0.199) \][/tex]
[tex]\[ \Delta G = -92 + 99.5 \][/tex]
[tex]\[ \Delta G = 7.5 \text{ kJ/mol} \][/tex]
2. At [tex]\(400 K\)[/tex]:
[tex]\[ \Delta G = -92 - (400 \times -0.199) \][/tex]
[tex]\[ \Delta G = -92 + 79.6 \][/tex]
[tex]\[ \Delta G = -12.4 \text{ kJ/mol} \][/tex]
3. At [tex]\(700 K\)[/tex]:
[tex]\[ \Delta G = -92 - (700 \times -0.199) \][/tex]
[tex]\[ \Delta G = -92 + 139.3 \][/tex]
[tex]\[ \Delta G = 47.3 \text{ kJ/mol} \][/tex]
4. At [tex]\(600 K\)[/tex]:
[tex]\[ \Delta G = -92 - (600 \times -0.199) \][/tex]
[tex]\[ \Delta G = -92 + 119.4 \][/tex]
[tex]\[ \Delta G = 27.4 \text{ kJ/mol} \][/tex]
A reaction is spontaneous when [tex]\(\Delta G\)[/tex] is less than zero (i.e., [tex]\(\Delta G < 0\)[/tex]).
From our calculations:
- At [tex]\(500 K\)[/tex], [tex]\(\Delta G = 7.5\)[/tex] kJ/mol (not spontaneous).
- At [tex]\(400 K\)[/tex], [tex]\(\Delta G = -12.4\)[/tex] kJ/mol (spontaneous).
- At [tex]\(700 K\)[/tex], [tex]\(\Delta G = 47.3\)[/tex] kJ/mol (not spontaneous).
- At [tex]\(600 K\)[/tex], [tex]\(\Delta G = 27.4\)[/tex] kJ/mol (not spontaneous).
Therefore, the reaction is only spontaneous at [tex]\(400 K\)[/tex].
Thus, the correct answer is:
B. [tex]\[400 K\][/tex]
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
where:
- [tex]\(\Delta H\)[/tex] is the enthalpy change of the reaction.
- [tex]\(T\)[/tex] is the temperature in Kelvin.
- [tex]\(\Delta S\)[/tex] is the entropy change of the reaction.
Given:
- [tex]\(\Delta H = -92 \)[/tex] kJ/mol
- [tex]\(\Delta S = -0.199 \)[/tex] kJ/(mol·K)
We need to evaluate [tex]\(\Delta G\)[/tex] for each of the following temperatures: [tex]\(500 K\)[/tex], [tex]\(400 K\)[/tex], [tex]\(700 K\)[/tex], and [tex]\(600 K\)[/tex].
1. At [tex]\(500 K\)[/tex]:
[tex]\[ \Delta G = -92 - (500 \times -0.199) \][/tex]
[tex]\[ \Delta G = -92 + 99.5 \][/tex]
[tex]\[ \Delta G = 7.5 \text{ kJ/mol} \][/tex]
2. At [tex]\(400 K\)[/tex]:
[tex]\[ \Delta G = -92 - (400 \times -0.199) \][/tex]
[tex]\[ \Delta G = -92 + 79.6 \][/tex]
[tex]\[ \Delta G = -12.4 \text{ kJ/mol} \][/tex]
3. At [tex]\(700 K\)[/tex]:
[tex]\[ \Delta G = -92 - (700 \times -0.199) \][/tex]
[tex]\[ \Delta G = -92 + 139.3 \][/tex]
[tex]\[ \Delta G = 47.3 \text{ kJ/mol} \][/tex]
4. At [tex]\(600 K\)[/tex]:
[tex]\[ \Delta G = -92 - (600 \times -0.199) \][/tex]
[tex]\[ \Delta G = -92 + 119.4 \][/tex]
[tex]\[ \Delta G = 27.4 \text{ kJ/mol} \][/tex]
A reaction is spontaneous when [tex]\(\Delta G\)[/tex] is less than zero (i.e., [tex]\(\Delta G < 0\)[/tex]).
From our calculations:
- At [tex]\(500 K\)[/tex], [tex]\(\Delta G = 7.5\)[/tex] kJ/mol (not spontaneous).
- At [tex]\(400 K\)[/tex], [tex]\(\Delta G = -12.4\)[/tex] kJ/mol (spontaneous).
- At [tex]\(700 K\)[/tex], [tex]\(\Delta G = 47.3\)[/tex] kJ/mol (not spontaneous).
- At [tex]\(600 K\)[/tex], [tex]\(\Delta G = 27.4\)[/tex] kJ/mol (not spontaneous).
Therefore, the reaction is only spontaneous at [tex]\(400 K\)[/tex].
Thus, the correct answer is:
B. [tex]\[400 K\][/tex]
