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 $t$)

Question

Aicharya · Accepted Answer

$\begin{array}{llll}\mathrm{A} ightarrow & \mathrm{B} & + & \mathrm{C} \ \mathrm{P}_{\mathrm{i}} & \mathrm{0} & & \mathrm{0} \ \mathrm{P}_{\mathrm{i}}-\mathrm{x} & \mathrm{x} & & \mathrm{x}\end{array}$ The formula used here is: $\begin{aligned} & k=\frac{2.303}{t} \log \frac{P_i}{P_t} \ & P_t=P_i+x \ & P_i-x=P_i-P_t+P_i \ & =2 P_i-P_t \ & k=\frac{2.303}{t} \times \log \frac{P_i}{\left(2 P_i-P_tight)} \end{aligned}$ Hence, the answer is option A. ,

Integrated rate law equation for a first order gas phase reaction is given by (where \(P_i\) is initial…

Solution