Question

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

• A. $\frac{2000}{2.303}K$
• B. $2000K$
• C. $\frac{1000}{2.303}K$
• D. $1000K$

Answer 1

Answer: (C)

$\frac{1000}{2.303}K$

k1 = ${10}^{16}{e}^{-2000/T}$
k2 = ${10}^{15}{e}^{-2000/T}$

By taking log on both of them
$⟹\log k1=\log {10}^{16}-\frac{2000}{2.303}t$
$⟹\log k2=\log {10}^{15}-\frac{1000}{2.303}t$

Now, k1 = k2, Hence from equations $T=\frac{1000}{2.303}k$