Q87 · CSIR NET Chemistry, December 2011
Paper: CSIR NET December 2011 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Enzyme Kinetics · Marks: 2 · Difficulty: Hard
In the Michaelis-Menten mechanism for enzyme kinetics, the expression obtained is:
\[ \frac{\mathrm{v}}{[\mathrm{E}]_{0}[\mathrm{S}]}=1.4 \times 10^{12}-\frac{10^{4} \mathrm{v}}{[\mathrm{E}]_{0}} \]
The values of $\mathrm{k}_{3}\left(\mathrm{K}_{\text {ems }}, \mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}\right)$ and K (Michaelis constant, $\mathrm{mol} \mathrm{L}^{-1}$ ), respectively are(a)$1.4 \times 10^{12}, 10^{4}$
(b)$1.4 \times 10^{8}, 10^{4}$
(c)$1.4 \times 10^{8}, 10^{-4}$
(d)$1.4 \times 10^{12}, 10^{-4}$
Answer
Answer: C ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Eadie form: v/([E]₀[S]) = k₃/K − v/(K[E]₀). Comparing terms gives 1/K = 10⁴, so K = 10⁻⁴ mol L⁻¹, and k₃ = 1.4 × 10¹² × 10⁻⁴ = 1.4 × 10⁸ (c).
Study loop for Chemical Kinetics