Saturation Kinetics in the Iodide-Catalyzed Decomposition of H2O2
The rate of O2 evolution from H2O2 decomposition catalyzed by iodide is measured at 25 °C. At low [H2O2], the initial rate is first-order in H2O2; at high [H2O2], the rate becomes zero-order. Data ([H2O2] in M, v in 10^-5 M s^-1): 0.010, 1.0; 0.020, 1.8; 0.040, 2.9; 0.080, 3.6; 0.160, 3.8; 0.320, 3.9. The iodide concentration is constant at 0.010 M. The student proposes a Michaelis-Menten mechanism: H2O2 + I- ⇌ complex → products. The student wants to linearize the data to obtain Vmax and Km but is unsure whether to use a Lineweaver-Burk or Eadie-Hofstee plot, and whether the zero-order region indicates catalyst saturation or a change in mechanism.
1. Using the data, estimate Vmax and Km. Show your transformation and explain why the plot is linear.
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Use the Michaelis–Menten form v = Vmax[S]/(Km + [S]), which rearranges to 1/v = (Km/Vmax)(1/[S]) + 1/Vmax. With v in 10^-5 M s^-1, compute 1/[S] (M^-1) and 1/v (10^5 s M^-1): (100, 1.00), (50, 0.556), (25, 0.345), (12.5, 0.278), (6.25, 0.263), (3.125, 0.256). A Lineweaver–Burk plot of 1/v vs 1/[S] is linear. The intercept is about 0.25 × 10^5 s M^-1, so Vmax = 1/(0.25 × 10^5) = 4.0 × 10^-5 M s^-1. The slope from the first and last points is (1.00 - 0.256)/(100 - 3.125) = 0.00768 × 10^5 s. Then Km = slope × Vmax = 0.00768 × 10^5 × 4.0 × 10^-5 = 0.0307 M ≈ 0.031 M. The plot is linear because the mechanism follows a single saturable binding site; deviations would indicate cooperativity or a change in mechanism.
2. Why does the rate become zero-order at high [H2O2]? What does this imply about the catalyst?
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At high [H2O2], essentially all iodide is bound as the H2O2–I- complex. The rate-determining step is the decomposition of that complex, not the encounter with another H2O2. Increasing [H2O2] therefore cannot increase the rate further; the catalyst is saturated, and v = Vmax = kcat[I-]total. The zero-order region is a signature of Michaelis–Menten saturation, not a change to a different mechanism. The turnover frequency is Vmax/[I-] = 4.0 × 10^-5 / 0.010 = 4.0 × 10^-3 s^-1, meaning each iodide site turns over about once every 250 s under saturation conditions.
3. The student wonders whether an Eadie–Hofstee plot would be better. Compare the two linearizations and state which is more reliable when v has constant absolute error.
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Lineweaver–Burk plots 1/v against 1/[S]; the slope is Km/Vmax and the intercept is 1/Vmax. It is easy to use but distorts experimental error because reciprocals overemphasize the low-[S] points, where v is small and 1/v is large. Eadie–Hofstee plots v against v/[S]; the slope is -Km and the intercept is Vmax. It does not reciprocate v, so it is statistically preferable when the absolute error in v is constant. However, v appears on both axes, so the errors are correlated, and the plot can obscure deviations at high [S]. For these data both methods give Vmax ≈ 4.0 × 10^-5 M s^-1 and Km ≈ 0.03 M; the Eadie–Hofstee plot is the better choice for constant absolute error, while Lineweaver–Burk is better for rapid visualization.