[14%] Iodide ion reacts with hydrogen peroxide in acidic solution according to the following equatio — Kinetics Chemistry Question
Problem Context
[14%] Iodide ion reacts with hydrogen peroxide in acidic solution according to the following equation:
2 I–(aq) + H2O2(aq) + 2 H+(aq) → I2(aq) + 2 H2O(l) (4)
The reaction is carried out at 18.8 °C in the presence of a mixture of CH3COOH and NaOH to regulate the pH, sodium thiosulfate (which reacts very rapidly with I2 to form tetrathionate ions), and starch. All the components except the hydrogen peroxide are premixed, and then the hydrogen peroxide solution is added and a stopwatch is started. The solution remains colorless until it suddenly turns blue, at which point the time t is recorded. The following data are obtained:
Run | 1.0 M CH3COOH, mL | 1.0 M NaOH, mL | 0.20 M KI, mL | 0.20 M H2O2, mL | 0.020 M Na2S2O3, mL | Distilled H2O, mL | t, s
A | 2.0 | 1.0 | 2.0 | 2.0 | 1.0 | 2.0 | 68.2
B | 4.0 | 1.0 | 2.0 | 2.0 | 1.0 | 0.0 | 68.9
C | 2.0 | 1.0 | 4.0 | 2.0 | 1.0 | 0.0 | 33.2
D | 2.0 | 1.0 | 2.0 | 4.0 | 1.0 | 0.0 | 32.9
Give a qualitative explanation for why the solution suddenly turns blue after a certain amount of time has elapsed.
Model Answer
I2 is produced in the reaction between H2O2 and I–. Initially, the I2 does not accumulate to a significant extent because it immediately reacts with the thiosulfate ion. Once the small amount of thiosulfate is consumed, then I2 begins to accumulate, and the solution turns blue because of the intensely blue complex formed between starch and iodine.
Calculate the initial [H+] in runs A and B and explain why [H+] will not change significantly over the course of the respective reactions. (The Ka of CH3COOH is 1.8 × 10-5.)
Model Answer
The NaOH will immediately react with the CH3COOH to form CH3COO–. Taking into account the 10-fold dilution, this means that in run A, [CH3COOH] = [CH3COO–] = 0.1 M. Since Ka = 1.8 × 10-5 = [H+][CH3COO-]/[CH3COOH]
[H+] = 1.8 × 10-5 M
In run B, with [CH3COOH] = 0.3 M and [CH3COO–] = 0.1 M, [H+] = 5.4 × 10-5 M. These solutions are buffers, with the H+ consumed in reaction 4 being replenished by the CH3COOH. Thus, at the point where the initially 0.0020 M S2O3 2- is consumed, the concentration of CH3COOH will fall to 0.098 M while [CH3COO–] rises to 0.102 M, and [H+] = 1.73 × 10-5 M, a change of only 4%.
The rate law for reaction (4) has the form Rate = k4[I–]m[H2O2]n[H+]p, where m, n, and p are integers. What are the values of m, n, and p under these experimental conditions? Briefly explain your reasoning.
Model Answer
In each case, [S2O32-]0 = 0.0020 M, so at the point where the blue color appears, ∆[H2O2] = –0.0010 M (since two moles of thiosulfate react per mole of I2 produced). Since rate = –∆[H2O2]/∆t, the time recorded is inversely proportional to the initial rate of the reaction. Comparing run B to run A, [H+] increases by a factor of 3 while all other concentrations are the same. The rate is essentially unchanged, so p = 0. Comparing run C to run A, [I–] increases by a factor of 2 while all other concentrations are the same. The rate roughly doubles, so m = 1. Comparing run D to run A, [H2O2] increases by a factor of 2 while all other concentrations are the same. The rate roughly doubles, so n = 1.
Rate = k4[I–][H2O2]
What is the value of the rate constant k4 for this reaction?
Model Answer
k4 = (–∆[H2O2]/∆t)/([I–][H2O2]). So:
Run A, k4 = (0.0010 M/68.2 s)/([0.040 M][0.040 M]) = 9.1 × 10-3 M-1 s-1
Run B, k4 = (0.0010 M/68.9 s)/([0.040 M][0.040 M]) = 9.2 × 10-3 M-1 s-1
Run C, k4 = (0.0010 M/33.2 s)/([0.080 M][0.040 M]) = 9.4 × 10-3 M-1 s-1
Run D, k4 = (0.0010 M/32.9 s)/([0.040 M][0.080 M]) = 9.5 × 10-3 M-1 s-1
Average = 9.3 × 10-3 M-1 s-1
The following mechanism for reaction (4) has been proposed:
H2O2(aq) + H+(aq) ⇄ H3O2+(aq) (fast, unfavorable)
H3O2+(aq) + I–(aq) → HOI(aq) + H2O(l) (slow)
HOI(aq) + H+(aq) ⇄ H2OI+(aq) (fast, unfavorable)
H2OI+(aq) + I–(aq) → I2(aq) + H2O(l) (fast)
Is this mechanism consistent with the given data? Justify your answer.
Model Answer
This mechanism predicts Rate = k2[I–][H3O2+] = k2K1[I–][H2O2][H+]. This is inconsistent with the observed data under these conditions, since it predicts first-order dependence on [H+] while a zero-order dependence is actually observed.
The experiments are repeated under the same conditions as Runs A and B above, except that phosphoric acid (H3PO4, Ka = 7.6 × 10-3) is substituted for acetic acid. The observed times for the solutions to turn blue with this substitution are 46.7 s and 31.5 s, respectively. Propose an interpretation for these observations.
Model Answer
H3PO4 is a much stronger acid than CH3COOH, with [H+] = 7.6 × 10-3 M and 2.3 × 10-2 M in the two new reactions. Even though there was no effect on the rate of reaction in the reactions run with acetic acid, now increasing [H+] increases the rate! This indicates that there is an acid-catalyzed pathway for the reaction (perhaps by the mechanism proposed in (e)) that has become significant at the lower pH of the reactions run in the presence of phosphoric acid.