4AB › Chemistry PYQ › CSIR-NET › December 2016 › Q85Q85 · CSIR-NET Chemistry, December 2016 Paper: CSIR-NET December 2016 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Enzyme Kinetics · 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 Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation V_max = k₂[E]₀ = 3.42×10⁴ × 10⁻² = 3.42×10² mol dm⁻³ s⁻¹; turnover number = k₂ = 3.42×10⁴ s⁻¹.
← Q84 · All questions in this paper · Chemical Kinetics chapter list · Q86 → Study loop for Chemical Kinetics
Open in whiteboard · Browse this chapter in the app