Question

Integrated rate law equation for a first order gas phase reaction is given by (where \(P_i\)​ is initial pressure and \(P_t\)​ is total pressure at time t)

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

Answer 1

The Correct Option is A

Consider the reaction:

\[ A \rightarrow B + C \]

Initial pressures:

\[ P_i \quad 0 \quad 0 \]

After reaction:

\[ P_i - x \quad x \quad x \]

Total pressure at time \(t\):

\[ P_t = P_i + x \]

Therefore:

\[ P_i - x = P_i - P_t + P_i \] \[ = 2P_i - P_t \]

Hence,

\[ k = \frac{2.303}{t}\times \log \frac{P_i}{2P_i - P_t} \]