At [tex]298 K[/tex], [tex]\Delta H^0=46 \, \text{kJ/mol}[/tex] and [tex]\Delta S^0=0.097 \, \text{kJ/(K \cdot mol)}[/tex]. What is the Gibbs free energy of the reaction?

A. [tex]75 \, \text{kJ}[/tex]
B. [tex]17 \, \text{kJ}[/tex]
C. [tex]1300 \, \text{kJ}[/tex]
D. [tex]0.63 \, \text{kJ}[/tex]



Answer :

Certainly! To determine the Gibbs free energy ([tex]\(\Delta G\)[/tex]) of a reaction, we can use the following thermodynamic relation:

[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]

Where:
- [tex]\(\Delta G\)[/tex] is the Gibbs free energy change
- [tex]\(\Delta H\)[/tex] is the enthalpy change ([tex]\(46 \, \text{kJ/mol}\)[/tex])
- [tex]\(T\)[/tex] is the temperature in Kelvin ([tex]\(298 \, \text{K}\)[/tex])
- [tex]\(\Delta S\)[/tex] is the entropy change ([tex]\(0.097 \, \text{kJ/(K·mol)}\)[/tex])

Substituting the given values into the equation:

[tex]\[ \Delta G = 46 \, \text{kJ/mol} - 298 \, \text{K} \times 0.097 \, \text{kJ/(K·mol)} \][/tex]

Next, we compute the product of the temperature and the entropy change:

[tex]\[ 298 \, \text{K} \times 0.097 \, \text{kJ/(K·mol)} = 28.906 \, \text{kJ/mol} \][/tex]

Now, subtract this result from the enthalpy change:

[tex]\[ 46 \, \text{kJ/mol} - 28.906 \, \text{kJ/mol} = 17.094 \, \text{kJ/mol} \][/tex]

Therefore, the Gibbs free energy of the reaction is approximately [tex]\(17.094 \, \text{kJ/mol}\)[/tex]. Since we generally round to a sensible number of significant figures based on the given data, we round it to [tex]\(17 \, \text{kJ/mol}\)[/tex].

Thus, the answer is:

B. [tex]\(17 \, \text{kJ}\)[/tex]