Question

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

• 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 $2=\frac{E_{a}}{2.303R}\left[\frac{T_{2}-T_{1}}{T_{1}T_{2}}\right]$
$0.3=\frac{E_{a}}{2.303R}\times\frac{10}{280\times290}$
$Ea=0.3\times2.303\times280\times29\times R ..\left(1\right)$
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$\frac{k_{2}}{k_{1}}=\frac{0.3\times2.303\times280\times29R}{2.303R}\left[\frac{10}{290\times300}\right]=1.905$