Question

A sample of milk splits after $60\min \mathrm{at}300\mathrm{K}$ and after $40\min \mathrm{at}400\mathrm{K}$ when the population of lactobacillus acidophilus in it doubles. The activation energy (in $\mathrm{KJ}/\mathrm{mol}$) for this process is closest to (given, $\mathrm{R}=8.3\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$ , $\ln$ $\frac{2}{3}$$=0.4,{\mathrm{e}}^{-3}=4.0$) -

Answer 1

1. For a first-order reaction, we can write - $\ln \frac{K_2}{K_1}=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$ where ${\mathrm{E}}_{\mathrm{a}}$ $=$ activation energy, $\mathrm{R}$ $=$ gas constant, ${\mathrm{T}}_1$ $=$ lower temperature, ${\mathrm{K}}_1$ $=$ rate constant at temperature ${\mathrm{T}}_1$ , ${\mathrm{T}}_2$ $=$ higher temperature, ${\mathrm{K}}_2$ $=$ rate constant at temperature ${\mathrm{T}}_2$ .
2. Values are given to us - $\mathrm{R}=8.3\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$ , ${\mathrm{T}}_1=300\mathrm{K}$ , ${\mathrm{T}}_2=400\mathrm{K}$
3. Calculation of rate constants - full life of a first order reaction - $\mathrm{T}$ $=$ $\frac{1}{\mathrm{K}}$

${\mathrm{K}}_1$ $=$ $\frac{1}{60}$s^-1

${\mathrm{K}}_2$ $=$ $\frac{1}{40}$s^-1

4. Calculation of activation energy -

$\ln \frac{60}{40}=\frac{E_a}{8.3}\left[\frac{1}{300}-\frac{1}{400}\right]$

$\ln \frac{3}{2}=\frac{E_a}{8.3}\times \frac{100}{400\times 300}$

${\mathrm{E}}_{\mathrm{a}}=\ln \frac{3}{2}\times 8.3\times 1200$

${\mathrm{E}}_{\mathrm{a}}=3984$ $\mathrm{J}/\mathrm{mol}$

${\mathrm{E}}_{\mathrm{a}}=3.984$ $\mathrm{KJ}/\mathrm{mol}$

The activation energy (in $\mathrm{KJ}/\mathrm{mol}$) for this process is ${\mathrm{E}}_{\mathrm{a}}=3.984$ $\mathrm{KJ}/\mathrm{mol}$.