Question

What is the activation energy for a reaction whose rate constant doubles when temperature changes from ${30}^{\circ }C\text{to}{40}^{\circ }C$?

• A. $72.56{\text{KJ mol}}^{-1}$
• B. $45.45{\text{KJ mol}}^{-1}$
• C. $54.5{\text{KJ mol}}^{-1}$
• D. None of these

Answer 1

The correct option is C $54.5{\text{KJ mol}}^{-1}$

The temperature changes from ${30}^{o}C\text{to}{40}^{o}C$.
During this rise in temperature, the rate constant of reaction is doubled.

$\text{Let,}{k}_1=k$

$\text{then,}{k}_2=2k$

$T_1={30}^{\circ }C+273=303K$

$T_2={40}^{\circ }+273=313K$

According to Arrhenius the relationship between temperature and rate constant is given by,
$k=A{e}^{-E_a/RT}$

$\frac{k_2}{k_1}=\frac{A{e}^{-E_a/R{T}_{313}}}{A{e}^{-E_a/R{T}_{303}}}$

Taking ln on both sides,
$ln2=-\frac{E_a}{R{T}_{313}}+\frac{E_a}{R{T}_{303}}$
$ln2=-\frac{E_a}{R}[\frac{1}{313}-\frac{1}{303}]ln2=\frac{E_a}{R}\left[\frac{10}{313\times 303}\right]$

$E_a=\frac{ln2\times 8.314\times 313\times 303}{10}{E}_a=54476.8Jmo{l}^{-1}{E}_a\approx 54.5kJmo{l}^{-1}$