4. [11%] Ascorbic acid (Vitamin C, abbreviated HAsc) is a weak acid with pKa = 4.10. It is readily o — Kinetics Chemistry Question
Problem Context
- [11%] Ascorbic acid (Vitamin C, abbreviated HAsc) is a weak acid with pKa = 4.10. It is readily oxidized to form dehydroascorbic acid. One reagent that can carry out this oxidation is the ferricyanide ion, Fe(CN)6 3-.
Write a balanced net ionic equation for the reaction of Fe(CN)6 3- with HAsc in acidic solution.
Model Answer
C6H8O6(aq) + 2 Fe(CN)6 3–(aq) → C6H6O6(aq) + 2 Fe(CN)6 4–(aq) + 2 H+(aq)
The rate of this reaction can be followed by measuring [Fe(CN)6 3-] through its absorbance at 416 nm, where bright yellow Fe(CN)6 3- absorbs light strongly and all the other colorless components of the reaction do not. Two runs were conducted with [H+] = 0.06 M and an initial [Fe(CN)6 3-] = 5.22 × 10-4 M, but with different HAsc concentrations. Plots of ln([Fe(CN)6 3-]) as a function of time for the two runs are shown below.
What is the rate law for this reaction at [H+] = 0.06 M? Explain your reasoning.
Model Answer
The limiting reagent is Fe(CN)6 3-, with a large excess of both H+ and HAsc present (pseudo-first-order conditions). Since the natural logarithm of its concentration decays linearly with time, the reaction is first order in Fe(CN)6 3-. When the concentration of HAsc is increased by a factor of (0.0152 M/0.00507 M) = 3.00, the observed rate constant (= –slope) increases by a factor of (0.0132/0.00443) = 2.98. Thus the rate is linearly dependent on [HAsc]. ∴ Rate = k[HAsc][Fe(CN)6 3-]
What is the rate constant for this reaction at [H+] = 0.06 M? Make sure to include units.
Model Answer
Rate = k[HAsc] [Fe(CN)6 3-] = kobs[Fe(CN)6 3-], where kobs = –slope of the ln([Fe(CN)6 3-]) vs. time plot. So k[HAsc] = kobs, k = kobs/[HAsc]. From the first run, k = (0.00443 s-1)/(0.00507 M) = 0.874 M-1 s-1.
The behavior of the reaction varies with the acidity of the solution. Five runs were conducted at [HAsc] = 0.00507 M and an initial [Fe(CN)6 3-] = 5.22 × 10-4 M, but at varying [H+]. For each run, ln([Fe(CN)6 3-]) was plotted as a function of time and the slope of the plot was measured. The slope was multiplied by –1 and its variation as a function of [H+] is graphed below.
Propose a mechanistic explanation for the effect of [H+] on the reaction kinetics. Be sure to explain how your mechanistic explanation qualitatively predicts the appearance of the plot.
Model Answer
Evidently the reaction goes faster at low [H+], but even at high [H+], the rate levels off at a finite value rather than approaching zero. This is not consistent with a simple inverse order in H+, but rather a two-term rate law:
kobs = k1 + k2/[H+]
A plausible explanation is that either HAsc or its conjugate base can react with Fe(CN)6 3-, with Asc– reacting much more rapidly than HAsc. As [H+] decreases, the amount of Asc– increases and hence the rate increases. Even the highest pH studied, pH << pKa, so only a small fraction of the HAsc is in its conjugate base form, and [Asc–] is proportional to 1/[H+] under these conditions.