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

Answer: (A)

$k = \frac{2.303}{t} \times \log \frac{P_i}{(2P_i - P_t)}$

Answer 2

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,

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