Chemistry | MHT CET 2019 | Discussion Page

The activation energy of a reaction is zero, Its rate constant at 280 K is 1.6 x 10^-6 s^-1 , the rate constant at 300K is

$3.2 \times10^{-6}s^{-1}$

B) zero

$1.6 \times10^{-6}s^{-1}$

$1.6 \times10^{-5}s^{-1}$

Option C

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 , E_A =0

Rate constant , k₁ =1.6 x 10^-6 s^-1

Temperature , T₁ =280 K, T₂ =300K

According to Arrhenius equation

$\log\frac{k_{2}}{1.6\times 10^{-6}}=\frac{0}{2.303 R}\left[\frac{1}{280}-\frac{1}{300}\right]$

$\log\frac{k_{2}}{1.6\times 10^{-6}}=0$

$\frac{k_{2}}{1.6\times 10^{-6}}=antilog 0$

$\frac{k_{2}}{1.6\times 10^{-6}}=1$

$\therefore$ $k_{2}=1.6\times 10^{-6}s^{-1}$

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