Chemistry - Chemical Kinetics Question with Solution | TestHub

Explaining the Question:

Find the activation energy of the reaction.

Concept:

This question is based on first-order kinetics.

Solution:

Given temperatures are $T_1=27°C=300K$ and $T_2=47°C=320K$.

For 50% completion at $t_1=30min$:

The rate constant $K_1$ for a first-order reaction is given by:

$K_t=\ln \left(\frac{A_0}{A_t}\right)$

$A_t=\frac{A_0}{2}$

50% completion at $t_1=30min$:

$K_{}\times 30=\ln \frac{A_0}{\frac{A_0}{2}}$

$K_1=\frac{\ln 2}{30}mi{n}^{-1}$

For 50% completion at $t_2=10min$:

$K_2{t}_2=\ln 2$

$K_2\times 10=\ln 2$

$K_2=\frac{\ln 2}{10}mi{n}^{-1}$

$$\frac{K_1}{K_2}=\frac{10}{30}=\frac{1}{3}$$

Using the Arrhenius equation at two temperatures:

$$\ln \left(\frac{K_2}{K_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$$

$$\ln \left(\frac{\frac{\ln 2}{10}}{\frac{\ln 2}{30}}\right)=\frac{E_a}{8.314Jmo{l}^{-1}{K}^{-1}}\left(\frac{1}{300K}-\frac{1}{320K}\right)$$

$$\ln (3)=\frac{E_a}{8.314}\left(\frac{320-300}{300\times 320}\right)$$

$$1.0986=\frac{E_a}{8.314}\left(\frac{20}{96000}\right)$$

$$E_a=\frac{1.0986\times 8.314\times 96000}{20}$$

$$E_a=43899.42Jmo{l}^{-1}$$

$$E_a=43.899kJmo{l}^{-1}$$