Question

Ea for a certain biological reaction is 50 kJ/mol, by what factor ( how many times) will the rate of this reaction increase when body temperature increases from 37 C (normal ) to 40 C (fewer)?

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asked 2021-03-25
Ea for a certain biological reaction is 50 kJ/mol, by what factor ( how many times) will the rate of this reaction increase when body temperature increases from 37 C (normal ) to 40 C (fewer)?

Expert Answers (1)

2021-03-27
Arrhenius equation: k=Aexp(-Ea/RT)
where k is rate constant, A is pre-exponential factor, Ea is activation energy, R is molar gas constant and T is temperature in K.
\(\displaystyle{\ln{{\left({\frac{{{k}_{{2}}}}{{{k}_{{1}}}}}\right)}}}={\left({\frac{{-{E}{a}}}{{{R}}}}\right)}\times{\left({\frac{{{1}}}{{{T}_{{2}}}}}-{\frac{{{1}}}{{{T}_{{1}}}}}\right)}\)
\(\displaystyle{E}_{{a}}={50}\ {K}\frac{{j}}{{m}}{o}{l}={50000}\ \frac{{J}}{{m}}{o}{l}\)
\(\displaystyle{T}_{{1}}={37}{d}{e}{g}\ {C}={310}\ {K}\)
\(\displaystyle{T}_{{2}}={40}\ {d}{e}{g}\ {C}={313}\ {K}\)
\(\displaystyle{\ln{{\left({\frac{{{k}_{{2}}}}{{{k}_{{1}}}}}\right)}}}={\left({\frac{{-{500000}}}{{{8.314}}}}\times{\left({\frac{{{1}}}{{{313}}}}-{\frac{{{1}}}{{{310}}}}\right)}={0.18594}\right.}\)
\(\displaystyle{\frac{{{k}_{{2}}}}{{{k}_{{1}}}}}={\exp{{\left({0.18594}\right)}}}={1.2}\)
Since rate is proportional to rate constant
\(\displaystyle{\frac{{{R}{a}{t}{e}{\left({2}\right)}}}{{{R}{a}{t}{e}{\left({1}\right)}}}}={\frac{{{k}_{{2}}}}{{{k}_{{1}}}}}={1.2}\)
Thus the rate increases by a factor of 1.2 times
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