NEET UG 2013 — Chemistry Physical Chemistry

Given, initial temperature, $T_1=20+273=293 \mathrm{~K}$T1​=20+273=293 K Final temperature $$\begin{aligned} T_2 & =35+273 \\ & =308 \mathrm{~K} \\ R & =8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \end{aligned}$$T2​R​=35+273=308 K=8.314 J mol−1 K−1​ Since, rate becomes double on raising temperature, $\therefore r_2=2 r_1 \text { or } \frac{r_2}{r_1}=2$∴r2​=2r1​ or r1​r2​​=2 As rate constant, $k \propto r$k∝r $\therefore \frac{k_2}{k_1}=2$∴k1​k2​​=2 From Arrnhenius equation, we know that $$\begin{aligned} & \log \frac{k_2}{k_1}=-\frac{E_2}{2.303 R}\left[\frac{T_1-T_2}{T_1 T_2}\right] \\ & \log 2=-\frac{E_2}{2.303 \times 8.314}\left[\frac{293-308}{293 \times 308}\right] \\ & 0.3010=-\frac{E_2}{2.303 \times 8.314}\left[\frac{-15}{293 \times 308}\right] \end{aligned}$$​logk1​k2​​=−2.303RE2​​[T1​T2​T1​−T2​​]log2=−2.303×8.314E2​​[293×308293−308​]0.3010=−2.303×8.314E2​​[293×308−15​]​ $$\begin{aligned} \therefore E_2 & =\frac{0.3010 \times 2.303 \times 8.314 \times 293 \times 308}{15} \\ & =34673.48 \mathrm{~J} \mathrm{~mol}^{-1}=34.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$∴E2​​=150.3010×2.303×8.314×293×308​=34673.48 J mol−1=34.7 kJ mol−1​