# A sample of milk splits after

Question:

A sample of milk splits after $60 \mathrm{~min}$. at $300 \mathrm{~K}$ and after $40 \mathrm{~min}$. 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, $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \ln \left(\frac{2}{3}\right)=0.4, \mathrm{e}^{-3}=4.0$ )

Solution:

(3.98)

For a first order reaction, $k t=\ln \frac{[\mathrm{A}]}{\left[\mathrm{A}_{0}\right]}$

At $300 \mathrm{~K}, \mathrm{k}_{1} \times 60=\ln \frac{[\mathrm{A}]}{\left[\mathrm{A}_{0}\right]}$   ...(i)

At $400 \mathrm{~K}, \mathrm{k}_{2} \times 40=\ln \frac{[\mathrm{A}]}{\left[\mathrm{A}_{0}\right]}$    ....(ii)

From equation (1) and (2),

$\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{60}{40}$

$\ln \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}=\frac{\mathrm{E}_{a}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$

$\ln \left(\frac{60}{40}\right)=\frac{E_{a}}{8.3} \times \frac{100}{400 \times 300}$

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

$\Rightarrow \quad \mathrm{E}_{\mathrm{a}}=0.4 \times 8.3 \times 1200$

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

$\Rightarrow \quad \mathrm{E}_{\mathrm{a}}=3.984 \mathrm{~kJ} / \mathrm{mol}$

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