The following data were obtained during the first order thermal decomposition of `SO_(2)CI_(2)` at a constant volume.
`SO_(2)CI_(2(g)) to SO_(2(g)) + CI_(2(g))`
`{:("Experiment","Time"//s^(-1),"Total pressure"//"atm"),(1,0,0.5),(2,100,0.6):}`
What is the rate of reaction when total pressure is 0.65 atm?

Options

A. `0.35 atm s^(-1)`
B. `2.235 xx 10^(-3) atm s^(-1)`
C. `7.8 xx 10^(-4) atm s^(-1)`
D. `1.55 xx 10^(-4) atm s^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction given is the thermal decomposition of \( SO_2Cl_2 \) into \( SO_2 \) and \( Cl_2 \):
\[ SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g) \]
This is a first-order reaction.

### Step 2: Analyze the Data
We have the following data from the experiments:
- Experiment 1: Time = 0 s, Total Pressure = 0.5 atm
- Experiment 2: Time = 100 s, Total Pressure = 0.6 atm

### Step 3: Determine Initial Pressure
The initial pressure \( P_0 \) can be determined from the first experiment, which is:
\[ P_0 = 0.5 \, \text{atm} \]

### Step 4: Calculate the Rate Constant \( k \)
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{P_0}{2P_0 - P_t} \right) \]
Where:
- \( P_t \) is the total pressure at time \( t \).
- For the second experiment, \( t = 100 \, \text{s} \) and \( P_t = 0.6 \, \text{atm} \).

Substituting the values:
\[ k = \frac{2.303}{100} \log \left( \frac{0.5}{2 \times 0.5 - 0.6} \right) \]
Calculating the denominator:
\[ 2 \times 0.5 - 0.6 = 1 - 0.6 = 0.4 \]
Now substituting:
\[ k = \frac{2.303}{100} \log \left( \frac{0.5}{0.4} \right) \]
Calculating the logarithm:
\[ \log \left( \frac{0.5}{0.4} \right) = \log(1.25) \approx 0.09691 \]
Now substituting back:
\[ k = \frac{2.303}{100} \times 0.09691 \approx 0.00223 \, \text{s}^{-1} \]

### Step 5: Calculate the Rate of Reaction at \( P_t = 0.65 \, \text{atm} \)
Using the relationship for the rate of reaction:
\[ \text{Rate} = k \cdot P_{SO_2Cl_2} \]
Where:
\[ P_{SO_2Cl_2} = 2P_0 - P_t \]
Substituting \( P_t = 0.65 \, \text{atm} \):
\[ P_{SO_2Cl_2} = 2 \times 0.5 - 0.65 = 1 - 0.65 = 0.35 \, \text{atm} \]
Now substituting \( k \) and \( P_{SO_2Cl_2} \) into the rate equation:
\[ \text{Rate} = 0.00223 \cdot 0.35 \approx 0.0007805 \, \text{atm/s} \]
Converting to scientific notation:
\[ \text{Rate} \approx 7.805 \times 10^{-4} \, \text{atm/s} \]

### Final Answer
The rate of reaction when the total pressure is 0.65 atm is approximately:
\[ \text{Rate} \approx 7.81 \times 10^{-4} \, \text{atm/s} \]

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