Question

Rate constant k varies with temperature as given by equation:
$logk\left(mi{n}^{-1}\right)=5-\frac{2000K}{T}$. We can conclude:

• A. Pre-exponential factor A is 5.
• B. $E_a$ is 2000 kcal.
• C. Pre-exponential factor A is ${10}^5$
• D. $E_a$ is 9.152 kcal.

Answer 1

Answer: (C), (D)

The correct options are
C Pre-exponential factor A is ${10}^5$
D $E_a$ is 9.152 kcal.

$logk=logA=-\frac{E_a}{2.303RT}$
$\therefore logA=5$ and $\frac{-2000}{T}=-\frac{E_a}{2.303RT}$
On solving $E_a=9.152kcal$.