To solve the problem, we will follow these steps:
### Step 1: Write the dissociation reaction
The dissociation reaction of sulphuryl chloride `(SO2Cl2)` is given as:
\[ SO2Cl2(g) \rightleftharpoons SO2(g) + Cl2(g) \]
### Step 2: Set up the initial conditions
Assume we start with 1 mole of `SO2Cl2`. The initial amounts are:
- `SO2Cl2`: 1 mole
- `SO2`: 0 moles
- `Cl2`: 0 moles
### Step 3: Determine the change in moles
Given that `SO2Cl2` dissociates to the extent of 90%, we can calculate the amounts at equilibrium:
- Amount of `SO2Cl2` dissociated = \( 0.9 \times 1 \) mole = 0.9 moles
- Remaining `SO2Cl2` = \( 1 - 0.9 = 0.1 \) moles
- Amount of `SO2` produced = 0.9 moles
- Amount of `Cl2` produced = 0.9 moles
### Step 4: Calculate total moles at equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = 0.1 \, (\text{SO2Cl2}) + 0.9 \, (\text{SO2}) + 0.9 \, (\text{Cl2}) = 1.9 \, \text{moles} \]
### Step 5: Calculate partial pressures
Using the total pressure of 1 atm:
- Partial pressure of `SO2Cl2`:
\[ P_{SO2Cl2} = \frac{0.1}{1.9} \times 1 \, \text{atm} = 0.0526 \, \text{atm} \]
- Partial pressure of `SO2`:
\[ P_{SO2} = \frac{0.9}{1.9} \times 1 \, \text{atm} = 0.4748 \, \text{atm} \]
- Partial pressure of `Cl2`:
\[ P_{Cl2} = \frac{0.9}{1.9} \times 1 \, \text{atm} = 0.4748 \, \text{atm} \]
### Step 6: Calculate \( K_p \)
Using the formula for \( K_p \):
\[ K_p = \frac{P_{SO2} \cdot P_{Cl2}}{P_{SO2Cl2}} \]
Substituting the values:
\[ K_p = \frac{(0.4748)(0.4748)}{0.0526} = 4.2714 \]
### Step 7: Calculate \( \Delta G \)
Using the equation:
\[ \Delta G = \Delta G^0 - 2.303RT \log K_p \]
At equilibrium, \( \Delta G = 0 \), so:
\[ 0 = \Delta G^0 - 2.303RT \log K_p \]
Thus,
\[ \Delta G^0 = 2.303RT \log K_p \]
Substituting in the values:
- \( R = 8.314 \, \text{J/mol K} \)
- \( T = 375 \, \text{K} \)
- \( \log K_p = \log(4.2714) \approx 0.630 \)
Calculating \( \Delta G^0 \):
\[ \Delta G^0 = 2.303 \times 8.314 \times 375 \times 0.630 \]
\[ \Delta G^0 \approx 4.49 \, \text{kJ} \]
### Step 8: Determine the work done
The work done \( W \) is equal to \( -\Delta G^0 \):
\[ W \approx -4.49 \, \text{kJ} \]
### Final Answer
The work done in the process at the same temperature is approximately:
\[ W \approx -4.49 \, \text{kJ} \]
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