JEE Main 202431 Jan, Morning shiftChemistryEasyChemical Kinetics62× since 2002Integrated rate law equation for a first order gas phase reaction is given by (where Pi\mathrm{P}_{\mathrm{i}}Pi is initial pressure and Pt\mathrm{P}_{\mathrm{t}}Pt is total pressure at time ttt)AOption A. k=2.303t×logPi(2Pi−Pt)k=\frac{2.303}{t} \times \log \frac{P_i}{\left(2 P_i-P_t\right)}k=t2.303×log(2Pi−Pt)PiBOption 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}}}k=t2.303×logPi(2Pi−Pt)COption C. k=2.303t×Pi(2Pi−Pt)k=\frac{2.303}{t} \times \frac{P_i}{\left(2 P_i-P_t\right)}k=t2.303×(2Pi−Pt)PiDOption D. k=2.303t×log2Pi(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)}k=t2.303×log(2Pi−Pt)2PiCheckSkip · show solution
