The decomposition of sulphuryl chloride to `SO_(2) and Cl_(2)` is a first order reaction whose half life is 30 minutes . What percentage of the reactant will be decomposed in 2 hours

Explanation

To solve the problem of the decomposition of sulfuryl chloride (SO2Cl2) to sulfur dioxide (SO2) and chlorine gas (Cl2), we follow these steps:

### Step 1: Understand the Reaction Order and Given Data
The decomposition of sulfuryl chloride is a first-order reaction with a half-life (T_half) of 30 minutes. We need to find out the percentage of the reactant that will be decomposed in 2 hours.

### Step 2: Convert Time to Minutes
Since the half-life is given in minutes, we convert 2 hours into minutes:
\[ 2 \text{ hours} = 2 \times 60 = 120 \text{ minutes} \]

### Step 3: Calculate the Rate Constant (k)
For a first-order reaction, the rate constant (k) can be calculated using the half-life formula:
\[ k = \frac{0.693}{T_{half}} \]
Substituting the half-life:
\[ k = \frac{0.693}{30} \approx 0.0231 \text{ min}^{-1} \]

### Step 4: Use the First-Order Kinetics Equation
For a first-order reaction, the relationship between the initial concentration (A), the concentration after time (A - X), and time (t) is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{A}{A - X} \right) \]
Substituting the values we have:
\[ 0.0231 = \frac{2.303}{120} \log \left( \frac{A}{A - X} \right) \]

### Step 5: Solve for the Logarithmic Term
Rearranging the equation gives:
\[ \log \left( \frac{A}{A - X} \right) = 0.0231 \times 120 / 2.303 \]
Calculating the right side:
\[ \log \left( \frac{A}{A - X} \right) \approx 1.20 \]

### Step 6: Convert Logarithmic to Exponential Form
From the logarithmic equation, we can express it in exponential form:
\[ \frac{A}{A - X} = 10^{1.20} \approx 16 \]

### Step 7: Rearranging to Find the Decomposed Amount
Rearranging gives:
\[ A - X = \frac{A}{16} \]
Thus:
\[ X = A - \frac{A}{16} = A \left(1 - \frac{1}{16}\right) = A \left(\frac{15}{16}\right) \]

### Step 8: Calculate the Percentage Decomposed
The percentage decomposed can be calculated as:
\[ \text{Percentage decomposed} = \frac{X}{A} \times 100 = \frac{A \left(\frac{15}{16}\right)}{A} \times 100 = \frac{15}{16} \times 100 \approx 93.75\% \]

### Conclusion
Thus, the percentage of sulfuryl chloride that will be decomposed in 2 hours is approximately **93.75%**.

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