Question

Which of the following correctly represents integrated rate law equation for a first order reaction in gas phase:

• A. \( k = \frac{2.303}{t} \log \frac{P_i}{P_i - P} \)
• B. \( k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P} \)
• C. \( k = \frac{2.303}{2t} \log \frac{P_i}{P_i - P} \)
• D. \( k = \frac{2.303 P_i}{t} \log \frac{P_i}{2P_i - P} \)

Answer 1

The Correct Option is B

Consider a **first-order gas-phase decomposition reaction**: \[ A(g) \rightarrow B(g) + C(g) \]

Let the initial pressure of \( A \) be \( P_i \) and the pressure at time \( t \) be \( P \).

As the reaction proceeds, suppose the partial pressure of \( A \) that decomposes is \( x \).

Then, - Pressure of \( A \) left = \( P_i - x \) - Pressure of \( B \) formed = \( x \) - Pressure of \( C \) formed = \( x \)

Therefore, the total pressure at time \( t \) is: \[ P = (P_i - x) + x + x = P_i + x \] which gives \[ x = P - P_i \]

Hence, the pressure of \( A \) remaining is: \[ P_A = P_i - x = 2P_i - P \]

For a first-order reaction, the integrated rate law is: \[ k = \frac{2.303}{t} \log \frac{P_i}{P_A} \]

Substituting \( P_A = 2P_i - P \): \[ k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P} \]

Therefore, the correct integrated rate law for a first-order gas-phase reaction is: \[ \boxed{k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P}} \]