The expression for the equilibirum constant (Keq) for the enzyme catalyzed reaction given below, is E+S k₂k₁⇌… · ABC26PH0404
Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium
The expression for the equilibirum constant $\left(\mathrm{K}_{\mathrm{eq}}\right)$ for the enzyme catalyzed reaction given below, is \[ \mathrm{E}+\mathrm{S} \underset{\mathrm{k}_{2}}{\stackrel{\mathrm{k}_{1}}{\rightleftarrows}} \mathrm{ES} \underset{\mathrm{k}_{4}}{\stackrel{\mathrm{k}_{3}}{\rightleftarrows}} \mathrm{P}+\mathrm{E} \]
(a)$\frac{\mathrm{k}_{1} \mathrm{k}_{3}}{\mathrm{k}_{2} \mathrm{k}_{4}}$
(b)$\frac{\mathrm{k}_{1} \mathrm{k}_{2}}{\mathrm{k}_{3} \mathrm{k}_{4}}$
(c)$\frac{\mathrm{k}_{2} \mathrm{k}_{3}}{\mathrm{k}_{1} \mathrm{k}_{4}}$
(d)$\frac{\mathrm{k}_{1} \mathrm{k}_{4}}{\mathrm{k}_{2} \mathrm{k}_{3}}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
At equilibrium both steps are balanced: K = (k₁/k₂)(k₃/k₄) = k₁k₃/(k₂k₄) (a).
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