ABC26PH0942 · Chemical Kinetics

Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Enzyme Kinetics · Exam: NET JUNE 2017 · Marks: 2 · Difficulty: Hard

In a enzyme catalyzed reaction:
\[ \mathrm{E}+\mathrm{S} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}} \mathrm{ES} \xrightarrow{k_{2}} \mathrm{E}+\mathrm{P} \]
$\mathrm{k}_{2}=3.42 \times 10^{4} \mathrm{s}^{-1}$. If $[\mathrm{E}]_{0}=1.0 \times 10^{-2} \mathrm{mol} \mathrm{dm}^{-3}$, the magnitude of maximum velocity and turnover number using Michaelis-Menten kinetics are
(a)$3.42 \times 10^{2} \mathrm{mol} \mathrm{dm}^{-3} \mathrm{s}^{-1} ; 3.42 \times 10^{4} \mathrm{s}^{-1}$
(b)$3.42 \times 10^{6} \mathrm{moldm}^{-3} \mathrm{s}^{-1} ; 3.42 \times 10^{4} \mathrm{s}^{-1}$
(c)$3.42 \times 10^{4} \mathrm{mol} \mathrm{dm}^{-3} \mathrm{s}^{-1} ; 3.42 \times 10^{6} \mathrm{s}^{-1}$
(d)$3.42 \times 10^{4} \mathrm{mol} \mathrm{dm}^{-3} \mathrm{s}^{-1} ; 3.42 \times 10^{2} \mathrm{s}^{-1}$
Answer
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Explanation
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