Chemical Kinetics - Result Question 53
####54. 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) $1000 K$
(b) $\frac{2000}{2.303} K$
(c) $2000 K$
(d) $\frac{1000}{2.303} K$
[2008]
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Solution:
- (d) Given, $k_1=10^{16} \cdot e^{-\frac{2000}{T}}$
and $k_2=10^{15} \cdot e^{-\frac{1000}{T}}$
When $k_1$ and $k_2$ are equal at any temperature $T$, we have
$ \begin{aligned} & 10^{16} \cdot e^{-\frac{2000}{T}}=10^{15} \cdot e^{-\frac{1000}{T}} \\ & \text{ or } 10 \times 10^{15} \cdot e^{-\frac{2000}{T}}=10^{15} \cdot e^{-\frac{1000}{T}} \\ & \text{ or } 10 . e^{-\frac{2000}{T}}=e^{-\frac{1000}{T}} \\ & \text{ or } \ln 10-\frac{2000}{T}=-\frac{1000}{T} \end{aligned} $
or $\ln 10=\frac{2000}{T}-\frac{1000}{T}$
or $2.303 \log 10=\frac{1000}{T}$
or $2.303 \times 1 \times T=1000 \quad[\therefore \log 10=1]$
or $T=\frac{1000}{2.303} K$
