To find the rate constant \( k \) for the first-order reaction \( A(g) \rightarrow 2B(g) + C(g) \), we can follow these steps:
### Step 1: Understand the Reaction
The reaction given is a first-order reaction where one mole of \( A \) produces two moles of \( B \) and one mole of \( C \).
### Step 2: Set Up Initial Conditions
The initial pressure of \( A \) is given as:
\[
P_A = 90 \, \text{mm Hg}
\]
At time \( t = 0 \), the pressure of products \( B \) and \( C \) is \( 0 \, \text{mm Hg} \).
### Step 3: Determine the Pressure at Time \( t \)
After \( 10 \) minutes, the total pressure is:
\[
P_{\text{total}} = 180 \, \text{mm Hg}
\]
### Step 4: Relate Pressure Changes to Reaction Progress
Let \( x \) be the change in pressure of \( A \) that has reacted. The pressure of \( A \) at time \( t \) will be:
\[
P_A - x = 90 - x
\]
The pressure of \( B \) formed will be \( 2x \) and the pressure of \( C \) formed will be \( x \). Therefore, the total pressure at time \( t \) can be expressed as:
\[
P_A - x + 2x + x = P_A + 2x = 180
\]
Substituting \( P_A = 90 \):
\[
90 + 2x = 180
\]
### Step 5: Solve for \( x \)
Rearranging the equation:
\[
2x = 180 - 90
\]
\[
2x = 90 \implies x = 45
\]
### Step 6: Calculate Remaining Pressure of \( A \)
Now, the pressure of \( A \) remaining after \( 10 \) minutes is:
\[
P_A - x = 90 - 45 = 45 \, \text{mm Hg}
\]
### Step 7: Use the First-Order Rate Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[
k = \frac{2.303}{t} \log \left( \frac{P_{A,0}}{P_{A,t}} \right)
\]
Where:
- \( P_{A,0} = 90 \, \text{mm Hg} \) (initial pressure)
- \( P_{A,t} = 45 \, \text{mm Hg} \) (pressure after time \( t \))
- \( t = 10 \, \text{minutes} = 600 \, \text{seconds} \)
### Step 8: Substitute Values into the Equation
Substituting the values:
\[
k = \frac{2.303}{600} \log \left( \frac{90}{45} \right)
\]
Calculating the logarithm:
\[
\log \left( \frac{90}{45} \right) = \log(2) \approx 0.301
\]
Thus,
\[
k = \frac{2.303}{600} \times 0.301
\]
### Step 9: Calculate \( k \)
Calculating \( k \):
\[
k \approx \frac{2.303 \times 0.301}{600} \approx \frac{0.693}{600} \approx 1.155 \times 10^{-3} \, \text{s}^{-1}
\]
### Final Result
The rate constant \( k \) is approximately:
\[
k \approx 1.15 \times 10^{-3} \, \text{s}^{-1}
\]
