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

Question

In 2-chloro-3,4-dimethylhexane, how many chiral carbon atoms are present?

Answer 1

To determine the temperature at which the rate constants $k_1$ and $k_2$ are equal, we need to equate the two expressions for $k_1$ and $k_2$:

The given expressions for the rate constants are:

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

At the temperature where $k_1 = k_2$, we have:

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

Dividing both sides by $10^{15}$, we get:

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

Taking the natural logarithm on both sides, we have:

$\ln(10) + \ln(e^{-2000/T}) = \ln(e^{-1000/T})$

This simplifies to:

$\ln(10) - \frac{2000}{T} = -\frac{1000}{T}$

Rearranging the terms, we can solve for $T$:

$\ln(10) = \frac{2000}{T} - \frac{1000}{T}$

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

Thus, $T = \frac{1000}{\ln(10)}$

Using the approximation $\ln(10) = 2.303$, we substitute to find $T$:

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

Therefore, the correct answer is $\frac{1000}{2.303} K$.

Answer 2