Question

Rate constant $k$ varies with temperature by equation, $\log k\left(mi{n}^{-1}\right)=5-\frac{2000K}{T}$. 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 9.212 kcal

Answer 1

Answer: (C)

pre-exponential factor A is ${10}^5$

The Arrhenius expression is $k=A{e}^{-Ea/RT}.$
It can also be written as $logk=logA-\frac{E_a}{2.303RT}.$
But the given equation is $\log k\left(mi{n}^{-1}\right)=5-\frac{2000K}{T}.$
$logA=5$ or $A={10}^5.$
Also $\frac{E_a}{2.303RT}=\frac{2000K}{T}.$
${}_a=2.303\times 2\times 2000=9212cal=9.212kcal.$