The iodine clock reaction is a classical chemical clock demonstration experiment to display chemical — Physical Chemistry — Kinetics Chemistry Question
Clock reaction
The iodine clock reaction is a classical chemical clock demonstration experiment to display chemical kinetics in action. In this reaction two clear solutions are mixed and after a short time delay, the colorless liquid suddenly turns to a shade of dark blue. The iodine clock reaction has several variations. One of them involves the reaction between peroxydisulfate(VI) and iodide ions:
Reaction A: S2O8 2– (aq) + 3 I– (aq) → 2 SO4 2– (aq) + I3 – (aq)
The I3 – ion formed in Reaction A reacts immediately with ascorbic acid (C6H8O6) present originally in the solution to form I– ion (Reaction B).
Reaction B: C6H8O6(aq) + I3 – (aq) → C6H6O6(aq) + 3 I– (aq) + 2 H+(aq)
When all the ascorbic acid present in the solution is consumed, the I3 – ion generated in Reaction A forms a blue colored complex with starch present in solution (Reaction C).
Reaction C: I3 – (aq) + starch → Blue-complex.
Thus, the time t elapsed between mixing the reactants and the appearance of the blue color depends on the amount of I3 – ion formed. Therefore 1/t can be used as a measure of reaction rate.
25.0 cm3 of (NH4)2S2O8 solution and 25.0 cm3 of KI solution are mixed at a temperature of 25 °C with 5.0 cm3 of C6H8O6 solution (0.020 mol dm-3) and 5.0 cm3 starch solution. The initial concentrations of (NH4)2S2O8 and KI are different. The elapsed time t for the appearance of blue color is measured. All the data are tabulated below.
[VISUAL]
Table:
| Experiment No | [(NH4)2S2O8]o (mol dm–3) | [KI]o (mol dm–3) | t (s) |
|---|---|---|---|
| 1 | 0.200 | 0.200 | 20.5 |
| 2 | 0.100 | 0.200 | 41.0 |
| 3 | 0.050 | 0.200 | 82.0 |
| 4 | 0.200 | 0.100 | 41.0 |
Find the rate law for Reaction A using the data given in Table.
Model Answer
From exp. 1 and 2: (k * (0.20)^x * (0.20)^y) / (k * (0.10)^x * (0.20)^y) = 41 / 20.5 ⇒ 2^x = 2 ⇒ x = 1
From exp. 1 and 4: (k * (0.20)^x * (0.20)^y) / (k * (0.20)^x * (0.10)^y) = 41 / 20.5 ⇒ 2^y = 2 ⇒ y = 1
Rate law for the reaction: Rate = k [S2O8 2–] [I–]
Using the data for the experiment 1, find the initial rate of Reaction A in mol dm–3 s– 1.
Model Answer
n(I3-) formed = n(ascorbic acid) = 0.020 mol dm-3 * 5.0 * 10^-3 dm3 = 0.10 mmol
n(I-) consumed = 3 * n(I3-) = 0.30 mmol
Total volume of reaction mixture = 25.0 + 25.0 + 5.0 + 5.0 = 60.0 cm3 = 0.0600 dm3
Initial rate of Reaction A = 1/3 * (delta[I-]/delta t) = 1/3 * ((0.30 mmol / 60 cm3) / 20.5 s) = 8.1 * 10^-5 mol dm-3 s-1
Calculate the rate constant for Reaction A at 25 °C.
Model Answer
Initial concentrations after mixing in Experiment 1:
[S2O8 2-] = 0.200 mol dm-3 * (25.0 cm3 / 60.0 cm3) = 0.0833 mol dm-3
[I-] = 0.200 mol dm-3 * (25.0 cm3 / 60.0 cm3) = 0.0833 mol dm-3
Rate = k [S2O8 2-] [I-]
8.1 * 10^-5 mol dm-3 s-1 = k * (0.0833 mol dm-3) * (0.0833 mol dm-3)
k = 1.16 * 10^-2 mol-1 dm3 s-1
The following mechanism is proposed for Reaction A:
I– (aq) + S2O8 2– (aq) → IS2O8 3– (aq)
IS2O8 3– (aq) → 2 SO4 2– (aq) + I+(aq)
I+(aq) + I– (aq) → I2(aq)
I2(aq) + I– (aq) → I3 – (aq)
Derive an equation for the rate of formation of I3 – (aq) assuming that the steady-state approximation can be applied to all intermediates. Is the given mechanism consistent with the rate law found in part 14.1?
Model Answer
Rate of formation of I3-: d[I3-]/dt = k4 [I2][I-]
Applying steady-state approximation to intermediates:
1) d[I2]/dt = k3 [I+][I-] - k4 [I2][I-] = 0 ⇒ k4 [I2][I-] = k3 [I+][I-]
2) d[I+]/dt = k2 [IS2O8 3-] - k3 [I+][I-] = 0 ⇒ k3 [I+][I-] = k2 [IS2O8 3-]
3) d[IS2O8 3-]/dt = k1 [I-][S2O8 2-] - k2 [IS2O8 3-] = 0 ⇒ k2 [IS2O8 3-] = k1 [I-][S2O8 2-]
Thus, the rate of formation of I3- is:
d[I3-]/dt = k1 [I-][S2O8 2-]
Yes, this mechanism is consistent with the rate law Rate = k [S2O8 2-] [I-] found in part 14.1.
Ascorbic acid is a weak diprotic acid. In order to find its first acid dissociation constant, Ka1, 50.0 cm3 of ascorbic acid solution (c = 0.100 mol dm-3) is titrated with NaOH solution with a concentration of 0.200 mol dm-3. After addition of 1.00 cm3 NaOH solution the pH value of the resulting solution is 2.86. Calculate acid dissociation constant Ka1 for ascorbic acid.
Model Answer
pH = 2.86 ⇒ [H3O+] = 1.38 * 10^-3 mol dm-3
n(ascorbic acid)initial = 50.0 cm3 * 0.100 mol dm-3 = 5.0 mmol
n(NaOH)added = 1.00 cm3 * 0.200 mol dm-3 = 0.20 mmol
n(ascorbic acid)left = 5.00 mmol - 0.20 mmol = 4.80 mmol
Using the exact equation for weak acid titration at low pH:
Ka1 = [H3O+] * (n(HA-) + [H3O+] * V_total) / (n(H2A) - [H3O+] * V_total)
where V_total = 51.0 cm3 ⇒ [H3O+] * V_total = 1.38 * 10^-3 mol dm-3 * 0.0510 dm3 = 0.0704 mmol
Ka1 = 1.38 * 10^-3 * (0.20 * 10^-3 + 0.0704 * 10^-3) / (4.80 * 10^-3 - 0.0704 * 10^-3) = 7.9 * 10^-5
Give the predominant species present in the solution at pH = 7.82 if Ka2 for ascorbic acid is 2.5 ⋅ 10– 12.
Model Answer
At pH = 7.82, which is significantly higher than pKa1 (4.10) and lower than pKa2 (11.60), the predominant species is the monoanion of ascorbic acid, HA-.