Question

The rate of a reaction gets doubled when the temperature changes from $7$ to $17$. By which factor will it change for the temperature change from $17$ to $27$? (as nearest integer)

Answer 1

$lo{g}_{10}\frac{k_2}{k_1}=\frac{2.303Ea}{R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

$\frac{Rlog2\times 280\times 290}{2.303\times 10}=Ea$

$log\frac{k_4}{k_3}=\frac{2.303\ast Rlog2\ast 280\ast 290}{2.303\ast 10\ast R}\left(\frac{10}{290\ast 300}\right)$

$lo{g}_{10}\left(\frac{k_4}{k_3}\right)=\frac{280}{300}log2$

$\therefore \frac{k_4}{k_3}=1.91$