NEET UG 2019 — Chemistry Physical Chemistry

Key Idea Rate constants at two different temperatures is given by Arrhenius equation as follows: $$\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right], T_2>T_1$$logk1​k2​​=2.303REa​​[T1​1​−T2​1​],T2​>T1​ Given, $\mathrm{E}_{\mathrm{a}}=0, \mathrm{~T}_1=200 \mathrm{~K}, \mathrm{k}_1=1.6 \times 10^6 \mathrm{~s}^{-1}$Ea​=0, T1​=200 K,k1​=1.6×106 s−1 $$\mathrm{T}_2=400 \mathrm{~K}, \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$$T2​=400 K,R=8.314JK−1 mol−1 According to Arrhenius equation, $$\log \frac{\mathrm{k}_2}{\mathrm{k}_1}=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]$$logk1​k2​​=2.303REa​​[ T1​1​− T2​1​] On substituting the given values in above equation, we get $$\begin{aligned} & \log \frac{\mathrm{k}_2}{1.6 \times 10^6}=\frac{0}{2.303 \times 8.314}\left[\frac{\mathrm{l}}{200}-\frac{\mathrm{l}}{400}\right] \\ & \log \left(\frac{\mathrm{k}_2}{1.6 \times 10^6}\right)=0 \\ & \frac{\mathrm{k}_2}{1.6 \times 10^6}=10^0=1 \\ & \mathrm{k}_2=1.6 \times 10^6 \mathrm{~s}^{-1} \text { at } 400 \mathrm{~K} \end{aligned}$$​log1.6×106k2​​=2.303×8.3140​[200l​−400l​]log(1.6×106k2​​)=01.6×106k2​​=100=1k2​=1.6×106 s−1 at 400 K​ Thus, option (2) is correct.