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To determine the Gibbs free energy change ([tex]\(\Delta G_{rxn}\)[/tex]) for the reaction [tex]\(2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g)\)[/tex] and ascertain whether the reaction is spontaneous or nonspontaneous at [tex]\(300.0 \, \text{K}\)[/tex], we follow these steps:
### Step 1: Calculation of [tex]\(\Delta H_{rxn}\)[/tex]
We start by calculating the change in enthalpy ([tex]\(\Delta H_{rxn}\)[/tex]) for the reaction. This is given by:
[tex]\[ \Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) \][/tex]
For our reaction, it becomes:
[tex]\[ \Delta H_{rxn} = [2 \Delta H_f(\text{SO}_3(g))] - [2 \Delta H_f(\text{SO}_2(g)) + \Delta H_f(\text{O}_2(g))] \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{rxn} = [2(-396) \, \text{kJ/mol}] - [2(-297) \, \text{kJ/mol} + 0 \, \text{kJ/mol}] \][/tex]
[tex]\[ \Delta H_{rxn} = -792 \, \text{kJ} - (-594 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{rxn} = -792 \, \text{kJ} + 594 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{rxn} = -198 \, \text{kJ} \][/tex]
### Step 2: Calculation of [tex]\(\Delta S_{rxn}\)[/tex]
Next, we calculate the change in entropy ([tex]\(\Delta S_{rxn}\)[/tex]) for the reaction. This is given by:
[tex]\[ \Delta S_{rxn} = \sum S(\text{products}) - \sum S(\text{reactants}) \][/tex]
For our reaction, it becomes:
[tex]\[ \Delta S_{rxn} = [2 S(\text{SO}_3(g))] - [2 S(\text{SO}_2(g)) + S(\text{O}_2(g))] \][/tex]
Substituting the given values:
[tex]\[ \Delta S_{rxn} = [2(130.58) \, \text{J/(mol·K)}] - [2(191.50) \, \text{J/(mol·K)} + 205.00 \, \text{J/(mol·K)}] \][/tex]
[tex]\[ \Delta S_{rxn} = 261.16 \, \text{J/(mol·K)} - [383.00 \, \text{J/(mol·K)} + 205.00 \, \text{J/(mol·K)}] \][/tex]
[tex]\[ \Delta S_{rxn} = 261.16 \, \text{J/(mol·K)} - 588.00 \, \text{J/(mol·K)} \][/tex]
[tex]\[ \Delta S_{rxn} = -326.84 \, \text{J/(mol·K)} \][/tex]
### Step 3: Conversion of [tex]\(\Delta S_{rxn}\)[/tex] to kJ/(mol·K)
Since [tex]\(\Delta H_{rxn}\)[/tex] is in kJ, we should convert [tex]\(\Delta S_{rxn}\)[/tex] from J/(mol·K) to kJ/(mol·K):
[tex]\[ \Delta S_{rxn} = -326.84 \, \text{J/(mol·K)} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} \][/tex]
[tex]\[ \Delta S_{rxn} = -0.32684 \, \text{kJ/(mol·K)} \][/tex]
### Step 4: Calculation of [tex]\(\Delta G_{rxn}\)[/tex]
Now, we can calculate the Gibbs free energy change ([tex]\(\Delta G_{rxn}\)[/tex]) using the equation:
[tex]\[ \Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn} \][/tex]
Substituting the calculated values and the given temperature ([tex]\(T = 300.0 \, \text{K}\)[/tex]):
[tex]\[ \Delta G_{rxn} = -198 \, \text{kJ} - 300.0 \, \text{K} \times (-0.32684 \, \text{kJ/(mol·K)}) \][/tex]
[tex]\[ \Delta G_{rxn} = -198 \, \text{kJ} + 98.052 \, \text{kJ} \][/tex]
[tex]\[ \Delta G_{rxn} = -99.948 \, \text{kJ} \][/tex]
### Step 5: Determination of Spontaneity
Since [tex]\(\Delta G_{rxn}\)[/tex] is negative:
[tex]\[ \Delta G_{rxn} = -99.948 \, \text{kJ} < 0 \][/tex]
The reaction is spontaneous.
### Conclusion
The Gibbs free energy change ([tex]\(\Delta G_{rxn}\)[/tex]) for the reaction [tex]\(2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g)\)[/tex] at [tex]\(300.0 \, \text{K}\)[/tex] is [tex]\(-100 \, \text{kJ}\)[/tex], and the reaction is spontaneous. Thus, the correct answer is:
[tex]\[ -100 \, \text{kJ/mol}, \text{spontaneous} \][/tex]
### Step 1: Calculation of [tex]\(\Delta H_{rxn}\)[/tex]
We start by calculating the change in enthalpy ([tex]\(\Delta H_{rxn}\)[/tex]) for the reaction. This is given by:
[tex]\[ \Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) \][/tex]
For our reaction, it becomes:
[tex]\[ \Delta H_{rxn} = [2 \Delta H_f(\text{SO}_3(g))] - [2 \Delta H_f(\text{SO}_2(g)) + \Delta H_f(\text{O}_2(g))] \][/tex]
Substituting the given values:
[tex]\[ \Delta H_{rxn} = [2(-396) \, \text{kJ/mol}] - [2(-297) \, \text{kJ/mol} + 0 \, \text{kJ/mol}] \][/tex]
[tex]\[ \Delta H_{rxn} = -792 \, \text{kJ} - (-594 \, \text{kJ}) \][/tex]
[tex]\[ \Delta H_{rxn} = -792 \, \text{kJ} + 594 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{rxn} = -198 \, \text{kJ} \][/tex]
### Step 2: Calculation of [tex]\(\Delta S_{rxn}\)[/tex]
Next, we calculate the change in entropy ([tex]\(\Delta S_{rxn}\)[/tex]) for the reaction. This is given by:
[tex]\[ \Delta S_{rxn} = \sum S(\text{products}) - \sum S(\text{reactants}) \][/tex]
For our reaction, it becomes:
[tex]\[ \Delta S_{rxn} = [2 S(\text{SO}_3(g))] - [2 S(\text{SO}_2(g)) + S(\text{O}_2(g))] \][/tex]
Substituting the given values:
[tex]\[ \Delta S_{rxn} = [2(130.58) \, \text{J/(mol·K)}] - [2(191.50) \, \text{J/(mol·K)} + 205.00 \, \text{J/(mol·K)}] \][/tex]
[tex]\[ \Delta S_{rxn} = 261.16 \, \text{J/(mol·K)} - [383.00 \, \text{J/(mol·K)} + 205.00 \, \text{J/(mol·K)}] \][/tex]
[tex]\[ \Delta S_{rxn} = 261.16 \, \text{J/(mol·K)} - 588.00 \, \text{J/(mol·K)} \][/tex]
[tex]\[ \Delta S_{rxn} = -326.84 \, \text{J/(mol·K)} \][/tex]
### Step 3: Conversion of [tex]\(\Delta S_{rxn}\)[/tex] to kJ/(mol·K)
Since [tex]\(\Delta H_{rxn}\)[/tex] is in kJ, we should convert [tex]\(\Delta S_{rxn}\)[/tex] from J/(mol·K) to kJ/(mol·K):
[tex]\[ \Delta S_{rxn} = -326.84 \, \text{J/(mol·K)} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} \][/tex]
[tex]\[ \Delta S_{rxn} = -0.32684 \, \text{kJ/(mol·K)} \][/tex]
### Step 4: Calculation of [tex]\(\Delta G_{rxn}\)[/tex]
Now, we can calculate the Gibbs free energy change ([tex]\(\Delta G_{rxn}\)[/tex]) using the equation:
[tex]\[ \Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn} \][/tex]
Substituting the calculated values and the given temperature ([tex]\(T = 300.0 \, \text{K}\)[/tex]):
[tex]\[ \Delta G_{rxn} = -198 \, \text{kJ} - 300.0 \, \text{K} \times (-0.32684 \, \text{kJ/(mol·K)}) \][/tex]
[tex]\[ \Delta G_{rxn} = -198 \, \text{kJ} + 98.052 \, \text{kJ} \][/tex]
[tex]\[ \Delta G_{rxn} = -99.948 \, \text{kJ} \][/tex]
### Step 5: Determination of Spontaneity
Since [tex]\(\Delta G_{rxn}\)[/tex] is negative:
[tex]\[ \Delta G_{rxn} = -99.948 \, \text{kJ} < 0 \][/tex]
The reaction is spontaneous.
### Conclusion
The Gibbs free energy change ([tex]\(\Delta G_{rxn}\)[/tex]) for the reaction [tex]\(2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g)\)[/tex] at [tex]\(300.0 \, \text{K}\)[/tex] is [tex]\(-100 \, \text{kJ}\)[/tex], and the reaction is spontaneous. Thus, the correct answer is:
[tex]\[ -100 \, \text{kJ/mol}, \text{spontaneous} \][/tex]
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