Question

Identify the options in which increase in rate of reaction will be maximum:
(I) $E_a$ = 40 Id/mole Temperature change 300 to 3 10 K
(II) $E_a$ =80 kJ/mole Temperature change 200 to 210 K
(III) $E_a$ = 60 Id/mole Temperature change 400 to 410 K

• A. II
• B. I
• C. III
• D. All four will experience same change.

Answer 1

The correct option is A II

If $K_1$ and $k_2$ are rates of reaction at temperature $T_1$ and $T_2$ then we have

$ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

This equation is derived as follows :

According to Arrhenius equation,

$\begin{array}{c}{k}_2=k_0{e}^{-Ea/R{T}_2}\\ k_1=k_0{e}^{-Ea/R{T}_1}\end{array}\}$Divide them and take loge on both sides

$\therefore ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{T_2-T_1}{T_1\times T_2}\right)$

Now in all 3 cases, $T_2-T_1={10}^{o}k$

$\therefore ln\left(\frac{k_2}{k_1}\right)\propto \frac{E_a}{T_1{T}_2}$

For (I) $ln\left(\frac{k_2}{k_1}\right)\propto \frac{40}{300\times 310}=4.3\times {10}^{-4}$

(II) $ln\left(\frac{k_2}{k_1}\right)\propto \frac{80}{200\times 210}=19\times {10}^{-4}$

(III) $ln\left(\frac{k_2}{k_1}\right)\propto \frac{60}{400\times 410}=3\times {10}^{-4}$

Therefore $\frac{k_2}{k_1}$ is maximum for (II)

Thus there is max increase in rate.