To solve the problem step by step, we will follow the details provided in the video transcript and the principles of first-order kinetics.
### Step 1: Understand the Reaction
The decomposition reaction is given as:
\[ A(g) \rightarrow 2B(g) + C(g) \]
### Step 2: Initial Conditions
At the start of the experiment (t = 0), the total pressure is given as:
\[ P_0 = 169 \, \text{mm} \]
At this point, the pressure of products B and C is 0, so:
- Pressure of A = 169 mm
- Pressure of B = 0 mm
- Pressure of C = 0 mm
### Step 3: Conditions After 9 Minutes
After 9 minutes, the total pressure is:
\[ P_t = 221 \, \text{mm} \]
Let \( x \) be the decrease in pressure of A after 9 minutes. According to the stoichiometry of the reaction:
- The pressure of B produced = \( 2x \)
- The pressure of C produced = \( x \)
### Step 4: Set Up the Equation
The total pressure after 9 minutes can be expressed as:
\[ P_t = P_0 - x + 2x + x \]
This simplifies to:
\[ 221 = 169 - x + 2x + x \]
\[ 221 = 169 + 2x \]
### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ 221 - 169 = 2x \]
\[ 52 = 2x \]
\[ x = 26 \, \text{mm} \]
### Step 6: Calculate Pressures at t = 9 minutes
Now we can find the pressures of A, B, and C after 9 minutes:
- Pressure of A = \( 169 - x = 169 - 26 = 143 \, \text{mm} \)
- Pressure of B = \( 2x = 2 \times 26 = 52 \, \text{mm} \)
- Pressure of C = \( x = 26 \, \text{mm} \)
### Step 7: Apply the First Order Kinetics Formula
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{T} \log \left( \frac{(P_B)^2 \cdot (P_C)}{P_A} \right) \]
Where:
- \( P_B = 52 \, \text{mm} \)
- \( P_C = 26 \, \text{mm} \)
- \( P_A = 143 \, \text{mm} \)
- \( T = 9 \, \text{minutes} \)
### Step 8: Substitute Values
Substituting the values into the formula:
\[ k = \frac{2.303}{9} \log \left( \frac{(52)^2 \cdot (26)}{143} \right) \]
### Step 9: Calculate the Logarithm
Calculating the inside of the logarithm:
\[ (52)^2 = 2704 \]
\[ 2704 \cdot 26 = 70204 \]
Now, calculate:
\[ \frac{70204}{143} \approx 491.6 \]
### Step 10: Final Calculation of \( k \)
Now we can calculate:
\[ k = \frac{2.303}{9} \log(491.6) \]
Using a calculator:
\[ \log(491.6) \approx 2.691 \]
Thus,
\[ k \approx \frac{2.303}{9} \times 2.691 \]
Calculating this gives:
\[ k \approx 0.256 \times 2.691 \approx 0.6804 \, \text{min}^{-1} \]
### Final Answer
The value of the rate constant \( k \) is approximately:
\[ k \approx 0.6804 \, \text{min}^{-1} \]
