Question

The activation energy of a reaction at a given temperature is found to be $2.303RTJmo{l}^{-1}$ . The ratio of rate constant to the Arrhenius factor is

• A. $0.01$
• B. $0.1$
• C. $0.02$
• D. $0.001$

Answer 1

Answer: (B)

$k=A{e}^{-E_a/RT}$

$\frac{k}{A}=e^{-E_a/RT}$

$\ln \left(\frac{k}{A}\right)=\frac{-E_a}{RT}$

$2.303\log \left(\frac{k}{A}\right)=\frac{-E_a}{RT}$

$\log \left(\frac{k}{A}\right)=\frac{-E_a}{2.303RT}$

$\log \left(\frac{k}{A}\right)=\frac{-2.303RT}{2.303RT}$

$\log \left(\frac{k}{A}\right)=-1$

$\therefore \frac{k}{A}=0.1$