To determine the Gibbs Free Energy change [tex]\(\Delta G^\circ\)[/tex] for the combustion of methane at a temperature of 298 K, we need to use the following thermodynamic relationship:
[tex]\[
\Delta G^\circ = \Delta H - T \Delta S
\][/tex]
where:
- [tex]\(\Delta H\)[/tex] is the enthalpy change,
- [tex]\(T\)[/tex] is the temperature in Kelvin,
- [tex]\(\Delta S\)[/tex] is the entropy change.
Given the values:
- [tex]\(\Delta H = -890.3 \ \text{kJ}/\text{mol}\)[/tex]
- [tex]\(\Delta S = -242 \ \text{J}/(\text{mol} \cdot \text{K})\)[/tex]
- [tex]\(T = 298 \ \text{K}\)[/tex]
Step 1: Convert the entropy change to the same units as the enthalpy change.
Since [tex]\(\Delta H\)[/tex] is given in kJ/mol, we need to convert [tex]\(\Delta S\)[/tex] from J/(mol·K) to kJ/(mol·K).
[tex]\[
1 \ \text{kJ} = 1000 \ \text{J}
\][/tex]
Thus,
[tex]\[
\Delta S = -242 \ \text{J}/(\text{mol} \cdot \text{K}) = \frac{-242}{1000} \ \text{kJ}/(\text{mol} \cdot \text{K}) = -0.242 \ \text{kJ}/(\text{mol} \cdot \text{K})
\][/tex]
Step 2: Substitute the given values into the Gibbs Free Energy formula.
Using the formula:
[tex]\[
\Delta G^\circ = \Delta H - T \Delta S
\][/tex]
we have:
[tex]\[
\Delta G^\circ = -890.3 \ \text{kJ}/\text{mol} - (298 \ \text{K} \times -0.242 \ \text{kJ}/(\text{mol} \cdot \text{K}))
\][/tex]
Step 3: Simplify the expression inside the parenthesis.
[tex]\[
298 \ \text{K} \times -0.242 \ \text{kJ}/(\text{mol} \cdot \text{K}) = -72.116 \ \text{kJ}/\text{mol}
\][/tex]
Step 4: Subtract this value from the enthalpy change.
[tex]\[
\Delta G^\circ = -890.3 \ \text{kJ}/\text{mol} - (-72.116 \ \text{kJ}/\text{mol})
\][/tex]
[tex]\[
\Delta G^\circ = -890.3 \ \text{kJ}/\text{mol} + 72.116 \ \text{kJ}/\text{mol}
\][/tex]
[tex]\[
\Delta G^\circ = -818.184 \ \text{kJ}/\text{mol}
\][/tex]
So, the Gibbs Free Energy change [tex]\(\Delta G^\circ\)[/tex] for the combustion of methane at 298 K is:
[tex]\[
\Delta G^\circ = -818.184 \ \text{kJ}/\text{mol}
\][/tex]
