Sure! Let's solve the problem step by step.
Given:
- Enthalpy change: [tex]\(\Delta H = -890.3 \, \text{kJ/mol}\)[/tex]
- Entropy change: [tex]\(\Delta S = -242 \, \text{J/(mol⋅K)}\)[/tex]
- Temperature: [tex]\(T = 298 \, \text{K}\)[/tex]
We need to find the Gibbs free energy change, which is given by the equation:
[tex]\[
\Delta G = \Delta H - T \Delta S
\][/tex]
However, notice that the enthalpy change ([tex]\(\Delta H\)[/tex]) is given in [tex]\(\text{kJ/mol}\)[/tex] while the entropy change ([tex]\(\Delta S\)[/tex]) is given in [tex]\(\text{J/(mol⋅K)}\)[/tex]. We need to convert the entropy change from [tex]\(\text{J/(mol⋅K)}\)[/tex] to [tex]\(\text{kJ/(mol⋅K)}\)[/tex] to match the units of enthalpy change.
1. Convert entropy change to [tex]\(\text{kJ/(mol⋅K)}\)[/tex]:
[tex]\[
\Delta S = -242 \, \text{J/(mol⋅K)} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = -0.242 \, \text{kJ/(mol⋅K)}
\][/tex]
2. Now, we can calculate [tex]\(\Delta G\)[/tex] using the formula:
[tex]\[
\Delta G = \Delta H - T \Delta S
\][/tex]
Substitute the values into the equation:
[tex]\[
\Delta G = -890.3 \, \text{kJ/mol} - 298 \, \text{K} \times (-0.242 \, \text{kJ/(mol⋅K)})
\][/tex]
3. Perform the multiplication first:
[tex]\[
298 \, \text{K} \times (-0.242 \, \text{kJ/(mol⋅K)}) = -72.116 \, \text{kJ/mol}
\][/tex]
4. Substitute this value back into the equation:
[tex]\[
\Delta G = -890.3 \, \text{kJ/mol} + 72.116 \, \text{kJ/mol}
\][/tex]
5. Finally, sum the values:
[tex]\[
\Delta G = -818.184 \, \text{kJ/mol}
\][/tex]
So, the Gibbs free energy change [tex]\(\Delta G\)[/tex] is [tex]\(-818.184 \, \text{kJ/mol}\)[/tex].
