Question

Rate constant $(K)$ varies with temperature as given by equation
$\log K\left({\min }^{-1}\right)=5-\frac{2000}{T}$
Consider the following about this equation
I. Pre exponential factor is $10^5$.
II. $E_a$ is $9.212kcal$.
III. Variation of $\log K$ with $\frac{1}{T}$ is linear.
Select the correct statement.

• A. I, II and III
• B. Both I and II
• C. Both II and III
• D. Both I and III

Answer 1

Answer: (B)

Given,
$\log k\left({\min }^{-1}\right)=5-\frac{2000}{T}$ ...(i)
$\because {\log }_{10}k={\log }_{10}A-\frac{E_a}{2.303RT}$ ...(ii)
On comparing equations (i) and (ii)
I. $\log A=5$ or $A=10^5$
Thus, (i) is true
II. $\frac{E_a}{2.303R}=2000$
$E_a=2.30\times 2000\times 0.002kcal$
$\therefore E_a=9.212kcal$
Thus (ii) is true.
III. From Eq. (ii) it is clear that, $\log k\propto \frac{1}{T}$
$\therefore$ Thus (III) is also false.