A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 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, $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \ell \mathrm{n}\left(\frac{2}{3}\right)=0.4, \mathrm{e}^{-3}=4.0$)

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Chemical Kinetics

A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 K when the population of lactobacillus acidophilus in it doubles. The activation energy (in

kJ / mol

) for this process is closest to ____

(Given,

R = 8.3 J mol^{- 1} K^{- 1}, ℓ n (\frac{2}{3}) = 0.4, e^{- 3} = 4.0

)

Answer: 3.98

Solution

Since the population doubles after a fixed extent, the rate constant is inversely proportional to the time taken,

k \propto \frac{1}{t}

, so

\frac{k _{2}}{k _{1}} = \frac{t _{1}}{t _{2}} = \frac{60}{40} = \frac{3}{2}

. The Arrhenius relation

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

with

T_{1} = 300 K

and

T_{2} = 400 K

uses

\frac{1}{300} - \frac{1}{400} = \frac{1}{1200}

. Taking

ln \frac{3}{2} = 0.4

0.4 = \frac{E _{a}}{8.3} \cdot \frac{1}{1200}

, so

E_{a} = 0.4 \times 8.3 \times 1200 \approx 3984 J/mol \approx 3.98 kJ/mol

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Numerical answer

Practise more: all JEE Main PYQs on Chemical Kinetics →

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Chemical Kinetics

Solution

More JEE Main questions on Chemical Kinetics