Question

A plot of $\frac{1}{T}$ vs. $logk$ for a reaction gives the slope $(-1\times {10}^{4}K)$. The energy of activation for the reaction is :
(Given $R=8.314J{K}^{-1}{mol}^{-1}$)

• A. $1.202kJ{mol}^{-1}$
• B. $83.14kJ{mol}^{-1}$
• C. $8314J{mol}^{-1}$
  
• D. $191.47kJmo{l}^{-1}$

Answer 1

The correct option is D $191.47kJmo{l}^{-1}$

$logk=logA-\frac{E_a}{2.303RT}$

Slope $=\frac{-E_a}{2.303R}=-1\times {10}^4$

$\Rightarrow E_a=2.303\times 8.314\times {10}^4=191.47kJmo{l}^{-1}$