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Sagot :
Certainly! Let's go through each part of the question step-by-step to provide a detailed solution.
i. Drawing a potential energy diagram and labeling the enthalpy of the reaction (2 points):
For the reaction [tex]\( SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) \)[/tex]:
1. On the potential energy diagram, the y-axis represents the potential energy, and the x-axis represents the progress of the reaction.
2. Since the reaction has a positive [tex]\( \Delta H \)[/tex] value ([tex]\( \Delta H = 98.4 \)[/tex] kJ/mol), it means the reaction is endothermic, meaning that the products have higher potential energy than the reactants.
3. Draw a curve that starts at the potential energy level of the reactants ([tex]\( SO_3 \)[/tex]) and ends at a higher potential energy level of the products ([tex]\( SO_2 \)[/tex] and [tex]\( \frac{1}{2} O_2 \)[/tex]).
4. Indicate [tex]\( \Delta H \)[/tex] (98.4 kJ/mol) as the difference in potential energy between the reactants and the products.
Here is a simplified version of what the potential energy diagram might look like:
```
Potential Energy
Products (SO_2 + 1/2 O_2)
/
/|
/ |
/ |
/ |
Reactants (SO_3) -------- / |
Reaction Progress
```
Label the vertical distance between the reactants and products as [tex]\( \Delta H = +98.4 \)[/tex] kJ/mol.
ii. Is the reaction endothermic or exothermic? Explain your answer (2 points):
The reaction [tex]\( SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) \)[/tex] is endothermic. This conclusion is reached because the enthalpy change ([tex]\( \Delta H \)[/tex]) for the reaction is positive ([tex]\( +98.4 \)[/tex] kJ/mol). In an endothermic reaction, energy is absorbed from the surroundings, resulting in products that have higher energy than the reactants.
iii. What is the Gibbs free energy of the reaction at 300 K? (2 points):
Gibbs free energy ([tex]\( \Delta G \)[/tex]) can be calculated using the formula:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
- Given:
- [tex]\( \Delta H = 98.4 \)[/tex] kJ/mol
- [tex]\( \Delta S = 0.09564 \)[/tex] kJ/(mol·K)
- Temperature ([tex]\( T \)[/tex]) = 300 K
Substitute the values into the formula:
[tex]\[ \Delta G = 98.4 \, \text{kJ/mol} - (300 \, \text{K} \times 0.09564 \, \text{kJ/(mol·K)}) \][/tex]
[tex]\[ \Delta G = 98.4 \, \text{kJ/mol} - 28.692 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 69.708 \, \text{kJ/mol} \][/tex]
Therefore, at 300 K, the Gibbs free energy of the reaction is [tex]\( 69.708 \)[/tex] kJ/mol.
i. Drawing a potential energy diagram and labeling the enthalpy of the reaction (2 points):
For the reaction [tex]\( SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) \)[/tex]:
1. On the potential energy diagram, the y-axis represents the potential energy, and the x-axis represents the progress of the reaction.
2. Since the reaction has a positive [tex]\( \Delta H \)[/tex] value ([tex]\( \Delta H = 98.4 \)[/tex] kJ/mol), it means the reaction is endothermic, meaning that the products have higher potential energy than the reactants.
3. Draw a curve that starts at the potential energy level of the reactants ([tex]\( SO_3 \)[/tex]) and ends at a higher potential energy level of the products ([tex]\( SO_2 \)[/tex] and [tex]\( \frac{1}{2} O_2 \)[/tex]).
4. Indicate [tex]\( \Delta H \)[/tex] (98.4 kJ/mol) as the difference in potential energy between the reactants and the products.
Here is a simplified version of what the potential energy diagram might look like:
```
Potential Energy
Products (SO_2 + 1/2 O_2)
/
/|
/ |
/ |
/ |
Reactants (SO_3) -------- / |
Reaction Progress
```
Label the vertical distance between the reactants and products as [tex]\( \Delta H = +98.4 \)[/tex] kJ/mol.
ii. Is the reaction endothermic or exothermic? Explain your answer (2 points):
The reaction [tex]\( SO_3(g) \rightarrow SO_2(g) + \frac{1}{2} O_2(g) \)[/tex] is endothermic. This conclusion is reached because the enthalpy change ([tex]\( \Delta H \)[/tex]) for the reaction is positive ([tex]\( +98.4 \)[/tex] kJ/mol). In an endothermic reaction, energy is absorbed from the surroundings, resulting in products that have higher energy than the reactants.
iii. What is the Gibbs free energy of the reaction at 300 K? (2 points):
Gibbs free energy ([tex]\( \Delta G \)[/tex]) can be calculated using the formula:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]
- Given:
- [tex]\( \Delta H = 98.4 \)[/tex] kJ/mol
- [tex]\( \Delta S = 0.09564 \)[/tex] kJ/(mol·K)
- Temperature ([tex]\( T \)[/tex]) = 300 K
Substitute the values into the formula:
[tex]\[ \Delta G = 98.4 \, \text{kJ/mol} - (300 \, \text{K} \times 0.09564 \, \text{kJ/(mol·K)}) \][/tex]
[tex]\[ \Delta G = 98.4 \, \text{kJ/mol} - 28.692 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta G = 69.708 \, \text{kJ/mol} \][/tex]
Therefore, at 300 K, the Gibbs free energy of the reaction is [tex]\( 69.708 \)[/tex] kJ/mol.
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