JEE MAIN - Chemistry (2020 - 9th January Evening Slot - No. 8)
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)
(Given, R = 8.3 J mol–1 K–1, $$\ln \left( {{3 \over 2}} \right) = 0.4$$, e–3 = 4.0)
Answer
3.98TO3.99
Explanation
Using Arrehenius equation
K = A$${e^{ - {{{E_a}} \over {RT}}}}$$
$$\ln \left( {{{{k_{400}}} \over {{k_{300}}}}} \right) = {{{E_a}} \over R}\left( {{1 \over {300}} - {1 \over {400}}} \right)$$
$$ \Rightarrow $$ $$\ln \left( {{{60} \over {40}}} \right) = {{{E_a}} \over R}\left( {{{100} \over {300 \times 400}}} \right)$$
$$ \Rightarrow $$ $$\ln \left( {{3 \over 2}} \right) = {{{E_a}} \over {1200R}}$$
$$ \therefore $$ Ea = 0.4 × 1200 × 8.3 = 3984 J/mol
Ea = 3.984 kJ/mol = 3.98 kJ/mol
K = A$${e^{ - {{{E_a}} \over {RT}}}}$$
$$\ln \left( {{{{k_{400}}} \over {{k_{300}}}}} \right) = {{{E_a}} \over R}\left( {{1 \over {300}} - {1 \over {400}}} \right)$$
$$ \Rightarrow $$ $$\ln \left( {{{60} \over {40}}} \right) = {{{E_a}} \over R}\left( {{{100} \over {300 \times 400}}} \right)$$
$$ \Rightarrow $$ $$\ln \left( {{3 \over 2}} \right) = {{{E_a}} \over {1200R}}$$
$$ \therefore $$ Ea = 0.4 × 1200 × 8.3 = 3984 J/mol
Ea = 3.984 kJ/mol = 3.98 kJ/mol
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