Question

The activation energy of a certain reaction is $87$ $kJ$ ${mol}^{-1}$. What is the ratio of the rate constants for this reaction when the temperature is decreased from ${37}^{o}C$ to ${15}^{o}C$?

• A. $5.1/1$
• B. $8.3/1$
• C. $13.2/1$
• D. $24.4/1$

Answer 1

Answer: (C)

$13.2/1$

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

So,

$\frac{k_1}{k_2}=e^{-\frac{E_a}{R}(\frac{1}{t_1}-\frac{1}{T_2})}=e^{-\frac{E_a}{R}(\frac{1}{310}-\frac{1}{288})}=13.2$