Answer :
To determine the standard free energy change ([tex]\( \Delta G^\circ \)[/tex]) of the given reaction at 298 K, we use the relationship between the equilibrium constant ([tex]\( K_p \)[/tex]) and the standard free energy change. The relationship is given by the formula:
[tex]\[ \Delta G^\circ = -RT \ln(K_p) \][/tex]
where:
- [tex]\( R \)[/tex] is the gas constant, which is 8.314 J/(mol·K)
- [tex]\( T \)[/tex] is the temperature, which is 298 K
- [tex]\( K_p \)[/tex] is the equilibrium constant for the reaction, which is 0.142
Using the values provided:
1. Identify the given quantities:
- [tex]\( R = 8.314 \)[/tex] J/(mol·K)
- [tex]\( T = 298 \)[/tex] K
- [tex]\( K_p = 0.142 \)[/tex]
2. Write down the formula for [tex]\(\Delta G^\circ\)[/tex]:
[tex]\[ \Delta G^\circ = -RT \ln(K_p) \][/tex]
3. Substitute the given values into the formula:
[tex]\[ \Delta G^\circ = -(8.314 \, \text{J/(mol·K)}) \times (298 \, \text{K}) \times \ln(0.142) \][/tex]
4. Calculate [tex]\( \ln(0.142) \)[/tex]:
[tex]\[ \ln(0.142) \approx -1.950355 \][/tex]
5. Complete the substitution:
[tex]\[ \Delta G^\circ = -(8.314 \, \text{J/(mol·K)}) \times (298 \, \text{K}) \times (-1.950355) \][/tex]
[tex]\[ \Delta G^\circ = 4836.0427073030605 \, \text{J/mol} \][/tex]
Therefore, the standard free energy change ([tex]\( \Delta G^\circ \)[/tex]) for the reaction at 298 K is approximately [tex]\( 4836 \)[/tex] J/mol or [tex]\( 4.84 \)[/tex] kJ/mol (upon rounding).
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### Description of the System Approaching Equilibrium
When the reaction [tex]\( N_2O_4(g) \leftrightarrow 2 NO_2(g) \)[/tex] is at standard state conditions, it means that initially, the reactants and products are each present at 1 atm pressure.
As the system approaches equilibrium:
- The reaction will adjust such that the ratio of the concentrations of products to reactants reaches the equilibrium constant, [tex]\( K_p = 0.142 \)[/tex].
Since [tex]\( \Delta G^\circ \)[/tex] is positive (4836 J/mol), the formation of [tex]\( NO_2 \)[/tex] from [tex]\( N_2O_4 \)[/tex] is non-spontaneous under standard conditions. Energy is required to convert [tex]\( N_2O_4 \)[/tex] into [tex]\( NO_2 \)[/tex].
- The initial state with [tex]\( N_2O_4 \)[/tex] and [tex]\( NO_2 \)[/tex] each at 1 atm is not at equilibrium, given that [tex]\( K_p = 0.142 \)[/tex]. This indicates that there is a higher initial ratio of [tex]\( N_2O_4 \)[/tex] compared to [tex]\( NO_2 \)[/tex] at standard conditions.
- As the system progresses towards equilibrium, [tex]\( N_2O_4 \)[/tex] will increasingly dissociate into [tex]\( NO_2 \)[/tex] until the partial pressures adjust to satisfy the equilibrium constant.
In summary, despite the positive [tex]\( \Delta G^\circ \)[/tex], meaning the standard reaction is not spontaneous, the system under standard conditions will spontaneously move to the state where [tex]\( K_p = 0.142 \)[/tex], which involves the production of more [tex]\( NO_2 \)[/tex] from [tex]\( N_2O_4 \)[/tex].
[tex]\[ \Delta G^\circ = -RT \ln(K_p) \][/tex]
where:
- [tex]\( R \)[/tex] is the gas constant, which is 8.314 J/(mol·K)
- [tex]\( T \)[/tex] is the temperature, which is 298 K
- [tex]\( K_p \)[/tex] is the equilibrium constant for the reaction, which is 0.142
Using the values provided:
1. Identify the given quantities:
- [tex]\( R = 8.314 \)[/tex] J/(mol·K)
- [tex]\( T = 298 \)[/tex] K
- [tex]\( K_p = 0.142 \)[/tex]
2. Write down the formula for [tex]\(\Delta G^\circ\)[/tex]:
[tex]\[ \Delta G^\circ = -RT \ln(K_p) \][/tex]
3. Substitute the given values into the formula:
[tex]\[ \Delta G^\circ = -(8.314 \, \text{J/(mol·K)}) \times (298 \, \text{K}) \times \ln(0.142) \][/tex]
4. Calculate [tex]\( \ln(0.142) \)[/tex]:
[tex]\[ \ln(0.142) \approx -1.950355 \][/tex]
5. Complete the substitution:
[tex]\[ \Delta G^\circ = -(8.314 \, \text{J/(mol·K)}) \times (298 \, \text{K}) \times (-1.950355) \][/tex]
[tex]\[ \Delta G^\circ = 4836.0427073030605 \, \text{J/mol} \][/tex]
Therefore, the standard free energy change ([tex]\( \Delta G^\circ \)[/tex]) for the reaction at 298 K is approximately [tex]\( 4836 \)[/tex] J/mol or [tex]\( 4.84 \)[/tex] kJ/mol (upon rounding).
---
### Description of the System Approaching Equilibrium
When the reaction [tex]\( N_2O_4(g) \leftrightarrow 2 NO_2(g) \)[/tex] is at standard state conditions, it means that initially, the reactants and products are each present at 1 atm pressure.
As the system approaches equilibrium:
- The reaction will adjust such that the ratio of the concentrations of products to reactants reaches the equilibrium constant, [tex]\( K_p = 0.142 \)[/tex].
Since [tex]\( \Delta G^\circ \)[/tex] is positive (4836 J/mol), the formation of [tex]\( NO_2 \)[/tex] from [tex]\( N_2O_4 \)[/tex] is non-spontaneous under standard conditions. Energy is required to convert [tex]\( N_2O_4 \)[/tex] into [tex]\( NO_2 \)[/tex].
- The initial state with [tex]\( N_2O_4 \)[/tex] and [tex]\( NO_2 \)[/tex] each at 1 atm is not at equilibrium, given that [tex]\( K_p = 0.142 \)[/tex]. This indicates that there is a higher initial ratio of [tex]\( N_2O_4 \)[/tex] compared to [tex]\( NO_2 \)[/tex] at standard conditions.
- As the system progresses towards equilibrium, [tex]\( N_2O_4 \)[/tex] will increasingly dissociate into [tex]\( NO_2 \)[/tex] until the partial pressures adjust to satisfy the equilibrium constant.
In summary, despite the positive [tex]\( \Delta G^\circ \)[/tex], meaning the standard reaction is not spontaneous, the system under standard conditions will spontaneously move to the state where [tex]\( K_p = 0.142 \)[/tex], which involves the production of more [tex]\( NO_2 \)[/tex] from [tex]\( N_2O_4 \)[/tex].
