Answer:
Option C
Explanation:
Key Idea: Arrhenius equation is given as:
$\log\frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]$
Given,
Activation energy of a reaction , EA =0
Rate constant , k1 =1.6 x 10-6 s-1
Temperature , T1 =280 K, T2 =300K
According to Arrhenius equation
$\log\frac{k_{2}}{1.6\times 10^{-6}}=\frac{0}{2.303 R}\left[\frac{1}{280}-\frac{1}{300}\right]$
$\log\frac{k_{2}}{1.6\times 10^{-6}}=0$
$\frac{k_{2}}{1.6\times 10^{-6}}=antilog 0$
$\frac{k_{2}}{1.6\times 10^{-6}}=1$
$\therefore$ $k_{2}=1.6\times 10^{-6}s^{-1}$