Looking for classes? Ksquare Career Institute, Bengaluru →
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
2000 K
B
$\dfrac{1000}{2.303}$ K
C
1000 K
D
$\dfrac{2000}{2.303}$ K
Detailed Solution
$k_1 = 10^{16}\,e^{-2000/T}$ and $k_2 = 10^{15}\,e^{-1000/T}$
The temperature at which $k_1 = k_2$ will be given by: $10^{16}\,e^{-2000/T} = 10^{15}\,e^{-1000/T}$
$\dfrac{e^{-2000/T}}{e^{-1000/T}} = \dfrac{10^{15}}{10^{16}}$
$e^{-1000/T} = 10^{-1}$
Taking natural logarithm: $\ln e^{-1000/T} = \ln 10^{-1}$
$2.303 \times \log e^{-1000/T} = 2.303 \times \log 10^{-1}$
$-\dfrac{1000}{T} = -2.303$
On solving, we get $T = \dfrac{1000}{2.303}$ K
The temperature at which $k_1 = k_2$ will be given by: $10^{16}\,e^{-2000/T} = 10^{15}\,e^{-1000/T}$
$\dfrac{e^{-2000/T}}{e^{-1000/T}} = \dfrac{10^{15}}{10^{16}}$
$e^{-1000/T} = 10^{-1}$
Taking natural logarithm: $\ln e^{-1000/T} = \ln 10^{-1}$
$2.303 \times \log e^{-1000/T} = 2.303 \times \log 10^{-1}$
$-\dfrac{1000}{T} = -2.303$
On solving, we get $T = \dfrac{1000}{2.303}$ K