To solve the problem, we need to determine the rate constant (k) for the decomposition of substance A based on the given pressures at the start and after 9 minutes. The reaction is as follows:
\[ A(g) \rightarrow 2B(g) + C(g) \]
### Step-by-Step Solution:
1. **Identify Initial Conditions**:
- The total pressure at the start of the experiment (P₀) is given as 169 mm.
2. **Determine Change in Pressure**:
- After 9 minutes, the total pressure (P) is 221 mm.
- The change in pressure (ΔP) can be calculated as:
\[
\Delta P = P - P₀ = 221 \, \text{mm} - 169 \, \text{mm} = 52 \, \text{mm}
\]
3. **Relate Change in Pressure to Decomposition**:
- According to the stoichiometry of the reaction, for every 1 mole of A that decomposes, 2 moles of B and 1 mole of C are produced. Therefore, if x mm of A decomposes, the pressure change can be expressed as:
\[
\Delta P = 2x + x = 3x
\]
- From the previous step, we know that \( \Delta P = 52 \, \text{mm} \). Thus:
\[
3x = 52 \implies x = \frac{52}{3} \approx 17.33 \, \text{mm}
\]
4. **Calculate Partial Pressures**:
- The pressure of A after 9 minutes:
\[
P_A = P₀ - x = 169 \, \text{mm} - 17.33 \, \text{mm} \approx 151.67 \, \text{mm}
\]
- The pressure of B produced:
\[
P_B = 2x = 2 \times 17.33 \approx 34.67 \, \text{mm}
\]
- The pressure of C produced:
\[
P_C = x \approx 17.33 \, \text{mm}
\]
5. **Use the First Order Kinetics Formula**:
- The formula for the rate constant (k) for a first-order reaction is given by:
\[
k = \frac{2.303}{t} \log\left(\frac{P_A}{P_0}\right)
\]
- Here, \( t = 9 \, \text{minutes} \), \( P_A \approx 151.67 \, \text{mm} \), and \( P_0 = 169 \, \text{mm} \).
6. **Calculate the Rate Constant**:
- Substitute the values into the formula:
\[
k = \frac{2.303}{9} \log\left(\frac{151.67}{169}\right)
\]
- Calculate the logarithm:
\[
\log\left(\frac{151.67}{169}\right) \approx \log(0.896) \approx -0.046
\]
- Now substituting back:
\[
k = \frac{2.303}{9} \times (-0.046) \approx -0.012 \, \text{min}^{-1}
\]
7. **Final Calculation**:
- Since we are looking for the positive value of k, we take the absolute value:
\[
k \approx 0.012 \, \text{min}^{-1}
\]
### Final Answer:
The value of the rate constant \( k \) is approximately **0.012 min⁻¹**.
