Sure, let's solve each part step-by-step using the Gibbs free energy equation:
[tex]\[
\Delta G = \Delta H - T \Delta S
\][/tex]
### Part (a)
Given:
- [tex]\(\Delta H = -638.4 \, \text{kJ}\)[/tex]
- [tex]\(\Delta S = 156.9 \, \text{J/K}\)[/tex]
First, convert the entropy change [tex]\(\Delta S\)[/tex] to kJ/K (since the enthalpy change [tex]\(\Delta H\)[/tex] is in kJ):
[tex]\[
\Delta S = \frac{156.9 \, \text{J/K}}{1000} = 0.1569 \, \text{kJ/K}
\][/tex]
The standard temperature [tex]\(T\)[/tex] is [tex]\(298 \, \text{K}\)[/tex]. Now, substitute these values into the Gibbs free energy equation:
[tex]\[
\Delta G = -638.4 \, \text{kJ} - 298 \, \text{K} \times 0.1569 \, \text{kJ/K}
\][/tex]
Calculate the term [tex]\(298 \times 0.1569\)[/tex]:
[tex]\[
298 \times 0.1569 = 46.7562 \, \text{kJ}
\][/tex]
Now, calculate [tex]\(\Delta G\)[/tex]:
[tex]\[
\Delta G = -638.4 \, \text{kJ} - 46.7562 \, \text{kJ} = -685.1562 \, \text{kJ}
\][/tex]
Since [tex]\(\Delta G\)[/tex] is negative ([tex]\(-685.1562 \, \text{kJ}\)[/tex]), the reaction is spontaneous.
### Part (b)
Given:
- [tex]\(\Delta H = -57.2 \, \text{kJ}\)[/tex]
- [tex]\(\Delta S = -175.9 \, \text{J/K}\)[/tex]
Again, convert the entropy change [tex]\(\Delta S\)[/tex] to kJ/K:
[tex]\[
\Delta S = \frac{-175.9 \, \text{J/K}}{1000} = -0.1759 \, \text{kJ/K}
\][/tex]
With the standard temperature [tex]\(T = 298 \, \text{K}\)[/tex], substitute these values into the Gibbs free energy equation:
[tex]\[
\Delta G = -57.2 \, \text{kJ} - 298 \, \text{K} \times (-0.1759 \, \text{kJ/K})
\][/tex]
Calculate the term [tex]\(298 \times -0.1759\)[/tex]:
[tex]\[
298 \times -0.1759 = -52.0182 \, \text{kJ}
\][/tex]
Since we have a negative times a negative, this term becomes positive:
[tex]\[
\Delta G = -57.2 \, \text{kJ} + 52.0182 \, \text{kJ} = -5.1818 \, \text{kJ}
\][/tex]
Since [tex]\(\Delta G\)[/tex] is negative ([tex]\(-5.1818 \, \text{kJ}\)[/tex]), the reaction is again spontaneous.
