Answer :
To determine the spontaneity of the reaction, we need to consider the Gibbs free energy change ([tex]\( \Delta G \)[/tex]), which can be determined using the equation:
[tex]\[ \Delta G = \Delta H - T\Delta S \][/tex]
where:
- [tex]\( \Delta H \)[/tex] is the change in enthalpy (620 kJ/mol),
- [tex]\( \Delta S \)[/tex] is the change in entropy (-0.46 kJ/(mol·K)),
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).
First, let's understand the given thermodynamic parameters:
- [tex]\( \Delta H = 620 \, \text{kJ/mol} \)[/tex] indicates that the reaction is endothermic.
- [tex]\( \Delta S = -0.46 \, \text{kJ/(mol·K)} \)[/tex] suggests that the disorder of the system decreases during the reaction.
For a reaction to be spontaneous, [tex]\( \Delta G \)[/tex] must be negative ([tex]\( \Delta G < 0 \)[/tex]).
To investigate if the reaction is ever spontaneous, we can check at particular temperatures. Specifically, we consider the temperature at which the reaction might transition from non-spontaneous to spontaneous.
Given the reaction is endothermic ([tex]\( \Delta H > 0 \)[/tex]) and the entropy change is negative ([tex]\( \Delta S < 0 \)[/tex]), we need to assess if [tex]\( \Delta G \)[/tex] can ever be negative. Consider a temperature [tex]\( T = 1347 \)[/tex] K (mentioned in option B):
[tex]\[ \Delta G = 620 \, \text{kJ/mol} - (1347 \, \text{K} \times -0.46 \, \text{kJ/(mol·K)}) \][/tex]
Calculating:
[tex]\[ \Delta G = 620 \, \text{kJ/mol} + 619.62 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 1239.62 \, \text{kJ/mol} \][/tex]
This results in a positive [tex]\( \Delta G \)[/tex]. Since [tex]\( \Delta G \)[/tex] remains positive at 1347 K, the reaction is not spontaneous at this temperature.
Therefore, we check whether the reaction is spontaneous at any temperature, say 298 K often used for standard conditions:
[tex]\[ \Delta G = 620 \, \text{kJ/mol} - (298 \, \text{K} \times -0.46 \, \text{kJ/(mol·K)}) \][/tex]
Calculating:
[tex]\[ \Delta G = 620 \, \text{kJ/mol} + 137.08 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 757.08 \, \text{kJ/mol} \][/tex]
Again, [tex]\( \Delta G \)[/tex] is positive, so the reaction is not spontaneous at this standard temperature either.
Since [tex]\( \Delta G \)[/tex] is positive at these temperatures, and considering that both [tex]\( \Delta H \)[/tex] is positive and [tex]\( \Delta S \)[/tex] is negative, thereby making [tex]\( \Delta G \)[/tex] always positive:
The reaction is never spontaneous.
Thus, the correct answer is:
A. It is never spontaneous.
[tex]\[ \Delta G = \Delta H - T\Delta S \][/tex]
where:
- [tex]\( \Delta H \)[/tex] is the change in enthalpy (620 kJ/mol),
- [tex]\( \Delta S \)[/tex] is the change in entropy (-0.46 kJ/(mol·K)),
- [tex]\( T \)[/tex] is the temperature in Kelvin (K).
First, let's understand the given thermodynamic parameters:
- [tex]\( \Delta H = 620 \, \text{kJ/mol} \)[/tex] indicates that the reaction is endothermic.
- [tex]\( \Delta S = -0.46 \, \text{kJ/(mol·K)} \)[/tex] suggests that the disorder of the system decreases during the reaction.
For a reaction to be spontaneous, [tex]\( \Delta G \)[/tex] must be negative ([tex]\( \Delta G < 0 \)[/tex]).
To investigate if the reaction is ever spontaneous, we can check at particular temperatures. Specifically, we consider the temperature at which the reaction might transition from non-spontaneous to spontaneous.
Given the reaction is endothermic ([tex]\( \Delta H > 0 \)[/tex]) and the entropy change is negative ([tex]\( \Delta S < 0 \)[/tex]), we need to assess if [tex]\( \Delta G \)[/tex] can ever be negative. Consider a temperature [tex]\( T = 1347 \)[/tex] K (mentioned in option B):
[tex]\[ \Delta G = 620 \, \text{kJ/mol} - (1347 \, \text{K} \times -0.46 \, \text{kJ/(mol·K)}) \][/tex]
Calculating:
[tex]\[ \Delta G = 620 \, \text{kJ/mol} + 619.62 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 1239.62 \, \text{kJ/mol} \][/tex]
This results in a positive [tex]\( \Delta G \)[/tex]. Since [tex]\( \Delta G \)[/tex] remains positive at 1347 K, the reaction is not spontaneous at this temperature.
Therefore, we check whether the reaction is spontaneous at any temperature, say 298 K often used for standard conditions:
[tex]\[ \Delta G = 620 \, \text{kJ/mol} - (298 \, \text{K} \times -0.46 \, \text{kJ/(mol·K)}) \][/tex]
Calculating:
[tex]\[ \Delta G = 620 \, \text{kJ/mol} + 137.08 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 757.08 \, \text{kJ/mol} \][/tex]
Again, [tex]\( \Delta G \)[/tex] is positive, so the reaction is not spontaneous at this standard temperature either.
Since [tex]\( \Delta G \)[/tex] is positive at these temperatures, and considering that both [tex]\( \Delta H \)[/tex] is positive and [tex]\( \Delta S \)[/tex] is negative, thereby making [tex]\( \Delta G \)[/tex] always positive:
The reaction is never spontaneous.
Thus, the correct answer is:
A. It is never spontaneous.
