The mechanism of enzyme (E) catalysed reaction of a substrate (S) to yield product (P) is: E+S k₂k₁⇌[ES]… · ABC26PH0347
Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Hard
The mechanism of enzyme (E) catalysed reaction of a substrate (S) to yield product (P) is: \[ \mathrm{E}+\mathrm{S} \underset{\mathrm{k}_{2}}{\stackrel{\mathrm{k}_{1}}{\rightleftharpoons}}[\mathrm{ES}] \underset{\mathrm{k}_{-2}}{\stackrel{\mathrm{k}_{2}}{\rightleftharpoons}} \mathrm{E}+\mathrm{P} ; \quad \frac{-\mathrm{d}[\mathrm{S}]}{\mathrm{dt}}=\frac{\mathrm{k}_{1} \mathrm{k}_{2}[\mathrm{S}]_{0}+\mathrm{k}_{2} \mathrm{k}_{-2}[\mathrm{P}]}{\mathrm{k}_{1}[\mathrm{S}]_{0}+\mathrm{k}_{-2}[\mathrm{P}]+\mathrm{k}_{-1}+\mathrm{k}_{2}}[\mathrm{E}]_{0} \]
If a small amount of S is converted to P , the maximum rate for the reaction will be observed for (a)$\left(\mathrm{k}_{1}+\mathrm{k}_{2}\right) \gg \mathrm{k}_{1}[\mathrm{S}]_{0}$
(b)$\left(\mathrm{k}_{-1}+\mathrm{k}_{2}\right) \ll \mathrm{k}_{1}[\mathrm{S}]_{0}$
(c)$\left(\mathrm{k}_{1}+\mathrm{k}_{2}\right) \gg \mathrm{k}_{1}[\mathrm{S}]_{1}$
(d)$\mathrm{k}_{2} \ll \mathrm{k}_{1}$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
With little P present, −d[S]/dt = k₁k₂[S]₀[E]₀/(k₁[S]₀ + k₋₁ + k₂). This reaches its maximum k₂[E]₀ when (k₋₁ + k₂) ≪ k₁[S]₀ (b).
Study loop for Chemical Kinetics
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