Question

Decomposition of A is a first order reaction at T(K) and is given by \( A(g) \rightarrow B(g) + C(g) \).
In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in \( min^{-1} \)) of the reaction ? (\( \log 2 = 0.3 \))

• A. \( 6.9 \times 10^{-4} \)
• B. \( 6.9 \times 10^{-1} \)
• C. \( 6.9 \times 10^{-2} \)
• D. \( 6.9 \times 10^{-3} \)

Hint:

For a reaction \( A \rightarrow nB \), the total pressure relation is \( P_t = P_o + (n-1)x \). This helps you find the partial pressure of reactant quickly.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a first-order gas-phase reaction, the rate constant is determined by the natural logarithm of the ratio of initial partial pressure to the partial pressure at time \( t \).

Step 2: Key Formula or Approach:
\( k = \frac{2.303}{t} \log \frac{P_o}{P_A} \)

Step 3: Detailed Explanation:
Reaction: \( A(g) \rightarrow B(g) + C(g) \)
Initial pressure (t=0): \( P_o = 1 \text{ bar} \).
Pressure at time t: \( P_A = P_o - x \), \( P_B = x \), \( P_C = x \).
Total pressure \( P_t = (P_o - x) + x + x = P_o + x \).
Given \( P_t = 1.5 \text{ bar} \) and \( P_o = 1 \text{ bar} \) :
\( 1 + x = 1.5 \Rightarrow x = 0.5 \text{ bar} \).
Partial pressure of A at 100 min: \( P_A = 1 - 0.5 = 0.5 \text{ bar} \).
Now, calculate the rate constant \( k \) for \( t = 100 \text{ min} \) :
\[ k = \frac{2.303}{100} \log \frac{1.0}{0.5} = \frac{2.303}{100} \log 2 \]
\[ k = \frac{2.303 \times 0.3}{100} = \frac{0.6909}{100} \approx 6.9 \times 10^{-3} \text{ min}^{-1} \].

Step 4: Final Answer:
The rate constant is \( 6.9 \times 10^{-3} \text{ min}^{-1} \).