Answer :
Sure! Let's address this step by step.
We are given the reaction:
[tex]\[ \text{SO}_3(g) \rightarrow \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \][/tex]
With the following parameters:
- [tex]\(\Delta H = 98.4\, \text{kJ/mol}\)[/tex] (Change in enthalpy)
- [tex]\(\Delta S = 0.09564\, \text{kJ/(mol·K)}\)[/tex] (Change in entropy)
Now, we need to determine the Gibbs free energy change ([tex]\(\Delta G\)[/tex]) at a specified temperature. For this calculation, let's use the standard temperature [tex]\(T = 298\, K\)[/tex].
To find [tex]\(\Delta G\)[/tex], we use the Gibbs free energy equation:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
Substituting in the given values:
[tex]\[ \Delta G = 98.4\, \text{kJ/mol} - (298\, \text{K} \times 0.09564\, \text{kJ/(mol·K)}) \][/tex]
Performing the multiplication inside the parentheses first:
[tex]\[ T \Delta S = 298\, \text{K} \times 0.09564\, \text{kJ/(mol·K)} = 28.50072\, \text{kJ/mol} \][/tex]
Now, subtract this result from [tex]\(\Delta H\)[/tex]:
[tex]\[ \Delta G = 98.4\, \text{kJ/mol} - 28.50072\, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 69.89928\, \text{kJ/mol} \][/tex]
Thus, the change in Gibbs free energy ([tex]\(\Delta G\)[/tex]) is:
[tex]\[ \Delta G = 69.89928\, \text{kJ/mol} \][/tex]
Summarizing our findings:
- [tex]\(\Delta H = 98.4\, \text{kJ/mol}\)[/tex]
- [tex]\(\Delta S = 0.09564\, \text{kJ/(mol·K)}\)[/tex]
- [tex]\(T = 298\, K\)[/tex]
- [tex]\(\Delta G = 69.89928\, \text{kJ/mol}\)[/tex]
These steps clearly illustrate how to determine the Gibbs free energy change given the enthalpy change, entropy change, and temperature of the reaction.
We are given the reaction:
[tex]\[ \text{SO}_3(g) \rightarrow \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \][/tex]
With the following parameters:
- [tex]\(\Delta H = 98.4\, \text{kJ/mol}\)[/tex] (Change in enthalpy)
- [tex]\(\Delta S = 0.09564\, \text{kJ/(mol·K)}\)[/tex] (Change in entropy)
Now, we need to determine the Gibbs free energy change ([tex]\(\Delta G\)[/tex]) at a specified temperature. For this calculation, let's use the standard temperature [tex]\(T = 298\, K\)[/tex].
To find [tex]\(\Delta G\)[/tex], we use the Gibbs free energy equation:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
Substituting in the given values:
[tex]\[ \Delta G = 98.4\, \text{kJ/mol} - (298\, \text{K} \times 0.09564\, \text{kJ/(mol·K)}) \][/tex]
Performing the multiplication inside the parentheses first:
[tex]\[ T \Delta S = 298\, \text{K} \times 0.09564\, \text{kJ/(mol·K)} = 28.50072\, \text{kJ/mol} \][/tex]
Now, subtract this result from [tex]\(\Delta H\)[/tex]:
[tex]\[ \Delta G = 98.4\, \text{kJ/mol} - 28.50072\, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 69.89928\, \text{kJ/mol} \][/tex]
Thus, the change in Gibbs free energy ([tex]\(\Delta G\)[/tex]) is:
[tex]\[ \Delta G = 69.89928\, \text{kJ/mol} \][/tex]
Summarizing our findings:
- [tex]\(\Delta H = 98.4\, \text{kJ/mol}\)[/tex]
- [tex]\(\Delta S = 0.09564\, \text{kJ/(mol·K)}\)[/tex]
- [tex]\(T = 298\, K\)[/tex]
- [tex]\(\Delta G = 69.89928\, \text{kJ/mol}\)[/tex]
These steps clearly illustrate how to determine the Gibbs free energy change given the enthalpy change, entropy change, and temperature of the reaction.