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]
