Q52 / 200 · NEET 2008 (Screening) · Chemistry

The rate constants $k_1$ and $k_2$ for two different reactions 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
    $1000 \mathrm{~K}$
  • B
    $\frac{2000}{2.303} K$
  • C
    $2000 \mathrm{~K}$
  • Correct answer
    D
    $\frac{1000}{2.303} \mathrm{~K}$

Solution

Key Idea : The Arrhenius equation is represented as $k=A e^{-E_a / R T}$ In the given equations, first take log and then compare them. $\begin{aligned} & k_1=10^{16} e^{-2000 / T} \\ & k_2=10^{15} e^{-1000 / T} \end{aligned}$ On taking log, we get $\begin{aligned} & \log k_1=\log 10^{16}-\frac{2000}{2.303 T} \\ & \log k_2=\log 10^{15}-\frac{1000}{2.303 T} \\ & \because k_1=k_2 \end{aligned}$ Hence, from Eqs (i) and (ii) $T=\frac{1000}{2.303} \mathrm{~K}$

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

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