Q19 · CSIR-NET Chemistry, June 2018

Paper: CSIR-NET June 2018 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Enzyme Kinetics · Marks: 2 · Difficulty: Medium

For an enzyme-substrate reaction,
\[ \mathrm{E+S} \underset{k_{-1}}{\overset{k_{1}}{\rightleftharpoons}} \mathrm{ES}, \qquad \mathrm{ES} \xrightarrow{k_{2}} \mathrm{E+P}, \]
the slope and intercept of plot between $\frac{1}{r}$ and $\frac{1}{[S]}$ are $10^{-2}\,\mathrm{s}$ and $10^{2}\,\mathrm{M}^{-1}\mathrm{s}$, respectively. If $E_{0}=10^{-6}\,\mathrm{M}$ and $\frac{k_{-1}}{k_{2}}=1000$, the value of $k_{1}$ will be close to (in units of $\mathrm{M}^{-1}\mathrm{s}^{-1}$) [$r$ is the rate of reaction and $E_{0}$ is initial concentration of enzyme
(a)$1 \times 10^{11}$
(b)$1 \times 10^{4}$
(c)$1 \times 10^{8}$
(d)$1 \times 10^{6}$
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
1/r = (K_M/k₂E₀)(1/[S]) + 1/(k₂E₀). Intercept 10² gives k₂E₀ = 10⁻², so k₂ = 10⁴ s⁻¹. Slope 10⁻² gives K_M = 10⁻⁴ M. K_M = (k₋₁ + k₂)/k₁ = 1001k₂/k₁, so k₁ ≈ 1001 × 10⁴/10⁻⁴ ≈ 1 × 10¹¹ M⁻¹ s⁻¹ (a).

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