Question

Rtae constant k varies with temperature by equation, $logk\left(mi{n}^{-1}\right)=5-\frac{2000}{T\left(K\right)}$, we can conclude :

• A. Pre-exponential factor A is $5$
• B. $E_a$ is $2000\text{kcal}$
• C. Pre-exponential factor A is ${10}^5$
• D. $E_a$ is $9.212\text{kcal}$

Answer 1

Answer: (C), (D)

$E_a$ is $9.212\text{kcal}$

Given equation, $logk\left(mi{n}^{-1}\right)=5-\frac{2000}{T\left(K\right)}$
On comparing given equation with: $logk=logA-\frac{E_a}{2.303RT}$
$\Rightarrow logA=5$
Taking antilog
$\Rightarrow A={10}^5$
And $\frac{E_a}{2.303RT}=\frac{2000}{T\left(K\right)}$
$\Rightarrow \frac{E_a}{2.303R}=2000$
$\Rightarrow E_a=2000\times 2.303R$
$\Rightarrow E_a=2000\times 2.303\times 2$
$\Rightarrow E_a=9.212\text{k cal}$