Question

Rate constant k varies with temperture by equation $logk\left(mi{n}^1\right)=log5\frac{2000kcal}{RT\times 2.303}.$ We can conclude that :

• A. The pre exponetial factor A is ${10}^5$.
• B. $E_a$ is 2000 kcal
• C. $E_a$ is 9.12 kcal
• D. The pre=exponential factor A is 5

Answer 1

Answer: (A), (B)

The pre exponetial factor A is ${10}^5$.
$E_a$ is 2000 kcal

Given equation is as follows:

$logk(mi{n}^{-1}=log5-\frac{2000}{RT\times 2.303}$

Comparing Arrhenius equation with the given equation, we get
$logk=logA-\frac{E_a}{2.3R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
Given equation :
$logk=log5-\frac{2000}{2.3RT}$
$\therefore logA=log5$ $\Rightarrow A={10}^5$

$E_a=2000kcal$

Hence, option A and B are correct.