Question

The rate constant is doubled when temperature increases from 27${}^o$C to 37${}^o$C. Activation energy in kJ is:

• A. $34.2$
• B. $58.8$
• C. $100.8$
• D. $53.6$

Answer 1

The correct option is D $53.6$

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\left(\frac{k_2}{k_1}\right)$$=\frac{E_a}{2.303}\times \left(\frac{T_2–T_1}{T_1{T}_2}\right)$

We will get:

$log\left(\frac{2k}{k}\right)=\frac{E_a}{2.303\times 8.314}\left(\frac{310-300}{300\times 310}\right)$

$log\left(2\right)=\frac{E_a}{2.303\times 8.314}\left(\frac{10}{300\times 310}\right)$

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

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

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