Q44 / 180 · KCET 2023 · Chemistry

The rate constant $k_1$ and $k_2$ for two different reactions are $10^{16} \times e^{-2000 / T}$ and $10^{15} \times e^{-1000 / T}$ respectively. The temperature at which $k_1=k_2$ is

  • A
    $\frac{2000}{2.303} \mathrm{~K}$
  • B
    $2000 \mathrm{~K}$
  • Correct answer
    C
    $\frac{1000}{2.303} \mathrm{~K}$
  • D
    $1000 \mathrm{~K}$

Solution

$K_1=10^{16} \times e^{-\frac{2000}{T}}$ $K_2=10^{15} \times e^{-\frac{1000}{T}}$ Now, $K_1=K_2$ $\begin{aligned} 10^{16} \times e^{\frac{-2000}{T}} & =10^{15} \times e^{\frac{-1000}{T}} \\ 10 & =e^{\frac{1000}{T}}\end{aligned}$ Taking $\ln$ on both sides, \ln 10=\frac{1000}{T} \ln e $2.303 \log 10=\frac{1000}{T} \Rightarrow T=\frac{1000}{2.303} \mathrm{~K}$

Hence, the correct answer is option: 'C'.

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