The rate constant \(k_1\) and \(k_2\) for two different reactions are \(10^{16} e^{-2000 / T}\) and…

The rate constant \(k_1\) and \(k_2\) for two different reactions are \(10^{16} e^{-2000 / T}\) and \(10^{15} e^{-1000 / T}\) respectively. The temperature at which \(k_1=k_2\) is :-
  1. 2000 K
  2. 10002.303 K
  3. 1000 K
  4. 20002.303 K

Solution

\(\begin{aligned} & k_1=10^{16} e^{-\frac{2000}{T}} \\ & k_2=10^{15} e^{-\frac{1000}{T}} \\ & k_1=k_2 \\ & 10=e^{\frac{2000}{T}-\frac{1000}{T}} \\ & \log _e 10=\frac{2000}{T}-\frac{1000}{T} \end{aligned}\)
\(\therefore \mathrm{T}=\frac{1000}{2.303}\) ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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