Question

The rate constant of a reaction is $1.5 \times 10^{-3}$ at $25^{\circ} C$ and $2.1 \times 10^{-2}$ at $60^{\circ} C$. The activation energy is

• A. $\frac{35}{333} R \log _{e} \frac{2.1 \times 10^{-2}}{1.5 \times 10^{-2}}$
• B. $\frac{298 \times 333}{35} R \log _{e} \frac{21}{1.5}$
• C. $\frac{298 \times 333}{35} R \log _{e} 2.1$
• D. $\frac{298 \times 333}{35} R \log _{e} \frac{2.1}{1.5}$

Answer 1

Answer: (B)

$T_{1}=273+25=298 \,K$
$T_{2}=273+60=333\, K$
$\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.3 R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right) $
$\log _{e} \frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$
or $\log _{e} \frac{2.1 \times 10^{-2}}{1.5 \times 10^{-3}}=\frac{E_{a}}{R}\left(\frac{35}{333 \times 298}\right)$
$\therefore E_{a}=\frac{298 \times 333}{35} \times R \times \log _{e} \frac{21}{1.5}$