To solve the problem, we will follow these steps:
### Step 1: Understand the Reaction
The reaction given is:
\[ A(g) \rightarrow 2B(g) + C(s) \]
### Step 2: Initial Conditions
We know that:
- The initial pressure of A, \( P_0 = 400 \, \text{mm Hg} \)
- The half-life of the reaction, \( t_{1/2} = 24 \, \text{min} \)
### Step 3: Determine the Rate Constant
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the given half-life:
\[ k = \frac{0.693}{24} \]
Calculating this gives:
\[ k \approx 0.028875 \, \text{min}^{-1} \]
### Step 4: Calculate the Time Elapsed
We need to find the pressure after 48 minutes. Since 48 minutes is twice the half-life (i.e., \( 48 = 2 \times 24 \)), we can use the half-life concept.
### Step 5: Determine the Amount of A Remaining
After one half-life (24 minutes), the amount of A remaining is:
\[ P_A = P_0 \cdot \left( \frac{1}{2} \right) = 400 \cdot \left( \frac{1}{2} \right) = 200 \, \text{mm Hg} \]
After another half-life (48 minutes), the amount of A remaining is:
\[ P_A = 200 \cdot \left( \frac{1}{2} \right) = 100 \, \text{mm Hg} \]
### Step 6: Calculate the Change in Pressure
The change in pressure due to the reaction can be calculated as follows:
- The pressure of A decreases by \( 400 - 100 = 300 \, \text{mm Hg} \).
- For every mole of A that reacts, 2 moles of B are produced. Therefore, the pressure contribution from B is:
\[ \Delta P_B = 2 \times (P_0 - P_A) = 2 \times 300 = 600 \, \text{mm Hg} \]
### Step 7: Calculate the Total Pressure
The total pressure \( P_{total} \) in the vessel after 48 minutes is:
\[ P_{total} = P_A + P_B \]
Where:
- \( P_A = 100 \, \text{mm Hg} \) (remaining pressure of A)
- \( P_B = 600 \, \text{mm Hg} \) (pressure contribution from B)
Thus:
\[ P_{total} = 100 + 600 = 700 \, \text{mm Hg} \]
### Final Answer
The pressure of the reaction mixture after 48 minutes is:
\[ \boxed{700 \, \text{mm Hg}} \]
