Chemistry - Chemical Kinetics Question with Solution | TestHub

Arrhenius equation:

$$\ln \frac{k^{\prime }}{k}=\frac{E_a}{R}\left(\frac{1}{T}-\frac{1}{T^{\prime }}\right)$$

Since $T^{\prime }>T$,

$$\left(\frac{1}{T}-\frac{1}{T^{\prime }}\right)>0$$

Therefore:

$$\frac{k_I^{\prime }}{k_I}>1$$

(A) True

For reaction II:

$E_a=2E_a$

Temperature increase also increases the rate constant:

$$\frac{k_{II}^{\prime }}{k_{II}}>1$$

(B) True

Because reaction II has double activation energy, its increase in rate constant is greater:

$$\frac{k_{II}^{\prime }}{k_{II}}>\frac{k_I^{\prime }}{k_I}$$

Hence:

$$\frac{k_I^{\prime }}{k_I}<\frac{k_{II}^{\prime }}{k_{II}}$$

(C) True

Option D is incorrect.

$$\boxed{CorrectAnswer:A,B,C}$$