The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \cdot e^{-2000 / T}$ and…

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
  1. $1000 \mathrm{~K}$
  2. $\frac{2000}{2.303} K$
  3. $2000 \mathrm{~K}$
  4. $\frac{1000}{2.303} \mathrm{~K}$

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)
$T=\frac{1000}{2.303} \mathrm{~K}$

Asked in: NEET 2008 (Screening)

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