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
activa tion energy (in kJ/ mol) for this process
is closest to__________.
(Given, R = 8.3 J mol–1 K–1, $\ln \left( {{3 \over 2}} \right) = 0.4$,
e–3 = 4.0)
Answer (integer)
3
Solution
Using Arrehenius equation
<br><br>K = A${e^{ - {{{E_a}} \over {RT}}}}$
<br><br>$$\ln \left( {{{{k_{400}}} \over {{k_{300}}}}} \right) = {{{E_a}} \over R}\left( {{1 \over {300}} - {1 \over {400}}} \right)$$
<br><br>$\Rightarrow$ $$\ln \left( {{{60} \over {40}}} \right) = {{{E_a}} \over R}\left( {{{100} \over {300 \times 400}}} \right)$$
<br><br>$\Rightarrow$ $\ln \left( {{3 \over 2}} \right) = {{{E_a}} \over {1200R}}$
<br><br>$\therefore$ E<sub>a</sub> = 0.4 × 1200 × 8.3 = 3984 J/mol
<br><br>E<sub>a</sub> = 3.984 kJ/mol
= 3.98 kJ/mol
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Chemical Kinetics are available — start with the harder ones if your accuracy is >70%.