Q6 / 180 · COMEDK 2026 (9 May Shift 1) · Chemistry

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

  • A
    $\dfrac{2000}{2.303}\ K$
  • B
    $1000\ K$
  • Correct answer
    C
    $\dfrac{1000}{2.303}\ K$
  • D
    $2000\ K$

Solution

Given $k_1 = 10^{16} \cdot e^{-2000/T}$ and $k_2 = 10^{15} \cdot e^{-1000/T}$

Equating $k_1$ and $k_2$:

$10^{16} \cdot e^{-2000/T} = 10^{15} \cdot e^{-1000/T}$

$\dfrac{10^{16}}{10^{15}} = \dfrac{e^{-1000/T}}{e^{-2000/T}}$

$10 = e^{1000/T}$

Taking natural logarithm on both sides:

$\ln(10) = \dfrac{1000}{T}$

$2.303 \log_{10}(10) = \dfrac{1000}{T}$

$2.303 = \dfrac{1000}{T}$

$T = \dfrac{1000}{2.303}\ K$

Answer: $\dfrac{1000}{2.303}\ K$

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

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