Step 1 : Understanding the Question:
This problem belongs to chemical kinetics, focusing specifically on a first-order gas-phase decomposition reaction. We are required to find the total pressure of the gas mixture after a given time interval, starting with a known initial pressure of pure reactant \( A \).
Step 2 : Key Formulas and Approach:
For a first-order reaction, the rate constant \( k \) is related to the half-life \( t_{1/2} \) by the equation:
\[ k = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{t_{1/2}} \]
The pressure of the remaining reactant over time is determined using the integrated rate law:
\[ P_A = P_0 e^{-kt} \]
Using stoichiometry, we can relate the remaining pressure of \( A \) to the pressures of the products and find the total pressure.
Step 3 : Detailed Explanation:
Step 1: Calculate the first-order rate constant \( k \). Given that \( t_{1/2} = 69.3\text{ s} \), we have:
\[ k = \frac{0.693}{69.3} = 0.01\text{ s}^{-1} \]
Step 2: Determine the remaining partial pressure of reactant \( A \) at \( t = 230\text{ s} \). Using the integrated rate law:
\[ P_A = 0.4 \times e^{-0.01 \times 230} = 0.4 \times e^{-2.3} \]
Using the approximation \( e^{-2.3} \approx 0.1 \), we find the partial pressure of \( A \) remaining:
\[ P_A = 0.4 \times 0.1 = 0.04\text{ atm} \]
Step 3: Establish the pressure stoichiometry of the reaction. Let \( x \) be the decrease in the partial pressure of \( A \):
\[ A(g) \rightarrow P(g) + Q(g) + R(g) \]
Initial pressures: \( P_0 \quad 0 \quad 0 \quad 0 \)
Pressures at time \( t \): \( (P_0 - x) \quad x \quad x \quad x \)
Since \( P_A = 0.4 - x = 0.04\text{ atm} \), it follows that \( x = 0.36\text{ atm} \).
Calculate the total pressure of the system, which is the sum of all individual partial pressures:
\[ P_T = (0.4 - x) + x + x + x = 0.4 + 2x \]
\[ P_T = 0.4 + 2(0.36) = 0.4 + 0.72 = 1.12\text{ atm} \]
Step 4 : Final Answer:
The total pressure of the system after 230 seconds is \( 1.12 \text{ atm} \), which corresponds to option (D).