Discussion Forum



1)

The rate of reaction  doubles when its temperature changes from 300K  to 310 K. Activation energy of such a reaction  will be (R=8.314 JK-1 mol-1  and log 2= 0.301)


A) $53.6kJ mol^{-1}$

B) $48.6kJ mol^{-1}$

C) $58.5kJ mol^{-1}$

D) $60.5kJ mol^{-1}$

Answer:

Option A

Explanation:

From Arrhenius equation,

  $\log \frac{{k_{2}}}{k_{1}}=\frac{-E_{a}}{2.303 R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$

  Given,   $\frac{{k_{2}}}{k_{1}}=2;T_{2}=310 K$

          T1=300 K

 On putting values, 

   $\Rightarrow\log 2=\frac{-E_{a}}{2.303\times8.314}\left(\frac{1}{310}-\frac{1}{300}\right)$

   $  \Rightarrow$         $ E_{a}=53598.6 J/mol$

   =53.6 kJ/mol