Question

The rate of a particular reaction doubles when temperature change from ${27}^0$C to ${37}^0$C. Calculate the energy of the activation of such reaction.

• A. 53582 KJ
• B. 53.58 KJ
• C. 29.86 KJ
• D. none of these

Answer 1

Answer: (B)

53.58 KJ

We are given that:

When $T_1=27+273=300K$

Let $K_1=k$

When $T_2=37+273=310K$

$K_2=2k$

Substituting these values the equation:

$log\frac{K_2}{K_1}$

= $\frac{Ea}{2.303R}\times \frac{(T_2–T_1)}{T_1{T}_2}$

We will get:

$log\frac{2k}{k}$ $=\frac{E_a}{2.303\times 8.314}\times \frac{310–300}{300\times 310}$

$lo{g}_2=\frac{E_a}{2.303\times 8.314}\times \frac{10}{300\times 310}$

$E_a=53598.6Jmo{l}^{-1}$

$E_a=53.6kJmo{l}^{-1}$

Hence, the energy of activation of the reaction is $53.6kJmo{l}^{-1}$