Answer :
To determine whether a chemical reaction is spontaneous under given conditions, we can use the Gibbs free energy change formula:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
Where:
- [tex]\(\Delta G\)[/tex] is the change in Gibbs free energy,
- [tex]\(\Delta H\)[/tex] is the enthalpy change,
- [tex]\(T\)[/tex] is the temperature in Kelvin (K), and
- [tex]\(\Delta S\)[/tex] is the entropy change.
Let's analyze the given data:
[tex]\(\Delta H = 620 \, \text{kJ/mol}\)[/tex] and [tex]\(\Delta S = -0.46 \, \text{kJ/(mol·K)}\)[/tex].
1. Calculation of [tex]\(\Delta G\)[/tex] at [tex]\(T = 298 \, \text{K}\)[/tex]:
[tex]\[ \Delta G = 620 \, \text{kJ/mol} - (298 \, \text{K}) \times (-0.46 \, \text{kJ/(mol·K)}) \][/tex]
[tex]\[ \Delta G = 620 \, \text{kJ/mol} + 137.08 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 757.08 \, \text{kJ/mol} \][/tex]
[tex]\(\Delta G\)[/tex] at [tex]\(T = 298 \, \text{K}\)[/tex] is [tex]\(757.08 \, \text{kJ/mol}\)[/tex].
Since [tex]\(\Delta G\)[/tex] is positive, the reaction is not spontaneous at [tex]\(298 \, \text{K}\)[/tex].
2. Determining the temperature at which the reaction becomes spontaneous ([tex]\(\Delta G = 0\)[/tex]):
Set [tex]\(\Delta G = 0\)[/tex]:
[tex]\[ 0 = \Delta H - T \Delta S \][/tex]
Solving for [tex]\(T\)[/tex]:
[tex]\[ T = \frac{\Delta H}{\Delta S} \][/tex]
[tex]\[ T = \frac{620 \, \text{kJ/mol}}{-0.46 \, \text{kJ/(mol·K)}} \][/tex]
[tex]\[ T = -1347.83 \, \text{K} \][/tex]
Since we generally consider temperatures in the positive range for practical purposes, this negative value implies that the reaction will never reach a condition where [tex]\(\Delta G\)[/tex] becomes zero or negative under ordinary conditions.
Therefore, considering all the analysis above, we can conclude:
B. It is never spontaneous.
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
Where:
- [tex]\(\Delta G\)[/tex] is the change in Gibbs free energy,
- [tex]\(\Delta H\)[/tex] is the enthalpy change,
- [tex]\(T\)[/tex] is the temperature in Kelvin (K), and
- [tex]\(\Delta S\)[/tex] is the entropy change.
Let's analyze the given data:
[tex]\(\Delta H = 620 \, \text{kJ/mol}\)[/tex] and [tex]\(\Delta S = -0.46 \, \text{kJ/(mol·K)}\)[/tex].
1. Calculation of [tex]\(\Delta G\)[/tex] at [tex]\(T = 298 \, \text{K}\)[/tex]:
[tex]\[ \Delta G = 620 \, \text{kJ/mol} - (298 \, \text{K}) \times (-0.46 \, \text{kJ/(mol·K)}) \][/tex]
[tex]\[ \Delta G = 620 \, \text{kJ/mol} + 137.08 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 757.08 \, \text{kJ/mol} \][/tex]
[tex]\(\Delta G\)[/tex] at [tex]\(T = 298 \, \text{K}\)[/tex] is [tex]\(757.08 \, \text{kJ/mol}\)[/tex].
Since [tex]\(\Delta G\)[/tex] is positive, the reaction is not spontaneous at [tex]\(298 \, \text{K}\)[/tex].
2. Determining the temperature at which the reaction becomes spontaneous ([tex]\(\Delta G = 0\)[/tex]):
Set [tex]\(\Delta G = 0\)[/tex]:
[tex]\[ 0 = \Delta H - T \Delta S \][/tex]
Solving for [tex]\(T\)[/tex]:
[tex]\[ T = \frac{\Delta H}{\Delta S} \][/tex]
[tex]\[ T = \frac{620 \, \text{kJ/mol}}{-0.46 \, \text{kJ/(mol·K)}} \][/tex]
[tex]\[ T = -1347.83 \, \text{K} \][/tex]
Since we generally consider temperatures in the positive range for practical purposes, this negative value implies that the reaction will never reach a condition where [tex]\(\Delta G\)[/tex] becomes zero or negative under ordinary conditions.
Therefore, considering all the analysis above, we can conclude:
B. It is never spontaneous.
