Verified Solution • 2025–26 Session moderate

QJEEM24Ched2b: JEE Main 2024 Chemistry - Class 12 JEEM24Ched2b Solution: Integrated rate law equation for a first order gas phase…

Integrated rate law equation for a first order gas phase reaction is given by (where Pi\mathrm{P}_{\mathrm{i}} is initial pressure and Pt\mathrm{P}_{\mathrm{t}} is total pressure at time tt )

Multiple Choice Options

  • A k=2.303t×log⁡Pi(2Pi−Pt)k=\frac{2.303}{t} \times \log \frac{P_i}{\left(2 P_i-P_t\right)}
  • B k=2.303t×log⁡(2Pi−Pt)Pi\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}}}
  • C k=2.303t×Pi(2Pi−Pt)k=\frac{2.303}{t} \times \frac{P_i}{\left(2 P_i-P_t\right)}
  • D k=2.303t×log⁡2Pi(2Pi−Pt)\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)}

Step-by-Step Verified Solution

A→amp;Bamp;+amp;CPiamp;0amp;amp;0Pi−xamp;xamp;amp;x\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}

amp;Pt=Pi+xamp;Pi−x=Pi−Pt+Piamp;=2Pi−Ptamp;K=2.303tlog⁡Pi2Pi−Pt\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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