Question

The rate of a reaction gets doubled when the temperature changes from $7^{∘}C$ to $17^{∘}C$. By what factor will it change for the temperature change from $17^{∘}C$ to $27^{∘}C$?

• A. $1.81$
• B. $1.71$
• C. $1.91$
• D. $1.76$

Answer 1

Answer: (C)

log$\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{1}{T_1}−\frac{1}{T_2}\right]$
log $2=\frac{E_a}{2.303R}\left[\frac{T_2−T_1}{T_1{T}_2}\right]$
$0.3=\frac{E_a}{2.303R}×\frac{10}{280×290}$
$Ea=0.3×2.303×280×29×R..\left(1\right)$
Now again
log $\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2−T_1}{T_1{T}_2}\right]$
log$\frac{k_2}{k_1}=\frac{0.3×2.303×280×29R}{2.303R}\left[\frac{10}{290×300}\right]=1.905$