Answer :
To solve for the Gibbs free energy change ([tex]\(\Delta G\)[/tex]), we can use the Gibbs free energy equation:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
where:
- [tex]\(\Delta H\)[/tex] is the change in enthalpy,
- [tex]\(T\)[/tex] is the temperature in Kelvin,
- [tex]\(\Delta S\)[/tex] is the change in entropy.
Given:
- [tex]\(\Delta H = 27 \, \text{kJ/mol}\)[/tex],
- [tex]\(\Delta S = 0.09 \, \text{kJ/(mol \cdot K)}\)[/tex],
- [tex]\(T = 300 \, \text{K}\)[/tex].
First, we substitute these values into the equation:
[tex]\[ \Delta G = 27 \, \text{kJ/mol} - 300 \, \text{K} \times 0.09 \, \text{kJ/(mol \cdot K)} \][/tex]
Next, perform the multiplication:
[tex]\[ 300 \, \text{K} \times 0.09 \, \text{kJ/(mol \cdot K)} = 27 \, \text{kJ/mol} \][/tex]
Then, subtract this result from the enthalpy change:
[tex]\[ \Delta G = 27 \, \text{kJ/mol} - 27 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 0 \, \text{kJ/mol} \][/tex]
Thus, the value for [tex]\(\Delta G\)[/tex] at [tex]\(300 \, \text{K}\)[/tex] is:
C. [tex]\(\Delta G = 0 \, \text{kJ/mol}\)[/tex]
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
where:
- [tex]\(\Delta H\)[/tex] is the change in enthalpy,
- [tex]\(T\)[/tex] is the temperature in Kelvin,
- [tex]\(\Delta S\)[/tex] is the change in entropy.
Given:
- [tex]\(\Delta H = 27 \, \text{kJ/mol}\)[/tex],
- [tex]\(\Delta S = 0.09 \, \text{kJ/(mol \cdot K)}\)[/tex],
- [tex]\(T = 300 \, \text{K}\)[/tex].
First, we substitute these values into the equation:
[tex]\[ \Delta G = 27 \, \text{kJ/mol} - 300 \, \text{K} \times 0.09 \, \text{kJ/(mol \cdot K)} \][/tex]
Next, perform the multiplication:
[tex]\[ 300 \, \text{K} \times 0.09 \, \text{kJ/(mol \cdot K)} = 27 \, \text{kJ/mol} \][/tex]
Then, subtract this result from the enthalpy change:
[tex]\[ \Delta G = 27 \, \text{kJ/mol} - 27 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 0 \, \text{kJ/mol} \][/tex]
Thus, the value for [tex]\(\Delta G\)[/tex] at [tex]\(300 \, \text{K}\)[/tex] is:
C. [tex]\(\Delta G = 0 \, \text{kJ/mol}\)[/tex]