JEE MAIN - Chemistry (2024 - 31st January Morning Shift - No. 11)
Integrated rate law equation for a first order gas phase reaction is given by (where $$\mathrm{P}_{\mathrm{i}}$$ is initial pressure and $$\mathrm{P}_{\mathrm{t}}$$ is total pressure at time $$t$$)
$$k=\frac{2.303}{t} \times \log \frac{P_i}{\left(2 P_i-P_t\right)}$$
$$\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}{\mathrm{P}_{\mathrm{i}}}$$
$$k=\frac{2.303}{t} \times \frac{P_i}{\left(2 P_i-P_t\right)}$$
$$\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{2 \mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}$$
Explanation
$$\begin{array}{llll} \mathrm{A} \rightarrow & \mathrm{B} & + & \mathrm{C} \\ \mathrm{P}_{\mathrm{i}} & 0 & & 0 \\ \mathrm{P}_{\mathrm{i}}-\mathrm{x} & \mathrm{x} & & \mathrm{x} \end{array}$$
$$\begin{aligned} & \mathrm{P_t=P_i+x} \\ & \mathrm{P_i-x=P_i-P_t+P_i} \\ & \mathrm{=2 P_i-P_t} \\ & \mathrm{K=\frac{2.303}{t} \log \frac{P_i}{2 P_i-P_t}} \end{aligned}$$
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