In order to decompose hydrogen peroxide (H2O2) with iodide ion as catalyst in neutral solution, the — Physical Chemistry — Kinetics Chemistry Question
Kinetics of the decomposition of hydrogen peroxide
In order to decompose hydrogen peroxide (H2O2) with iodide ion as catalyst in neutral solution, the 3 % H2O2 solution (which approximately corresponds to 30 g of H2O2 in 1 dm3 of solution) is mixed with a KI solution (0.1 mol dm-3) and water at different volumetric ratios; and the volume of the oxygen gas released V_O2 (cm3/min) is measured.
[VISUAL]
Determine the reaction order with respect to H2O2, and I–, respectively.
Model Answer
Chemical reaction: 2 H2O2 → 2 H2O + O2
The reaction rate is proportional to the volume of oxygen gas released in a unit of time.
In experiments 1, 2, and 3 when the volume of H2O2 solution doubles while keeping the same volume of KI solution, the reaction rate also doubles. Therefore, the rate is directly proportional to the concentration of H2O2. Hence, the reaction is the first-order with the respect to H2O2.
Similarly, from experiments 2, 4, and 5 the rate is directly proportional to the concentration of I–. Hence, the reaction is the first-order with the respect to I–.
Write down the chemical reaction, and determine the rate law.
Model Answer
Chemical reaction: 2 H2O2 → 2 H2O + O2
The rate law: v = k * c(H2O2) * c(I–)
Calculate the molarity of H2O2 at the beginning of the experiment 4 and after 4 min.
Model Answer
In the experiment #4, the solution of H2O2 is diluted three times; therefore, the concentration of H2O2 was reduced three times.
c0 = 10 g H2O2 / 1 dm3 = 10 / 34 = 0.294 mol dm-3.
Because the reaction proceeds slowly, the reaction rate (or the rate of releasing oxygen gas) is considered to be unchanged after of short period of time (4 min).
The volume of oxygen released after 4 min is equal to 4.25 × 4 = 17 cm3 O2.
Hence, n(O2) = pV / RT = (1)(17 * 10^-3) / (0.082)(298) = 0.695 * 10^-3 mol.
At the beginning, n(H2O2) = (0.294)(0.15) = 44.1 * 10^-3 mol.
After 4 min, n(H2O2) = 44.1 * 10^-3 – 2 * (0.695 * 10^-3) = 42.71 * 10^-3 mol.
Therefore, after 4 min c(H2O2) = 0.04271 / 0.15 = 0.285 mol dm-3.
The reaction mechanism involves a series of the following steps:
H2O2 + I– → k1 H2O + IO– (1)
IO− + H2O2 → k2 O2 + I− + H2O (2)
Do the two above steps have the same rate or different rates? Which step determines the overall rate of the oxygen release? Justify your answer.
Model Answer
The overall reaction: 2 H2O2 → 2 H2O + O2 (*)
v = - 1/2 d[H2O2]/dt
Consider three different cases:
i) If step (1) is slow and determines the overall rate, the rate of the overall reaction (*) will be the same as the rate of step (1):
v = - 1/2 d[H2O2]/dt = k1 [H2O2][I-]
which corresponds to the overall rate law as determined in section 13.2.
ii) If step (2) is slow, hence
v = - 1/2 d[H2O2]/dt = k2 [H2O2][IO-] (a)
Assume that the steady-state approximation is applied for IO-, we have
d[IO-]/dt = k1 [H2O2][I-] - k2 [IO-][H2O2] = 0 ⇒ [IO-] = (k1/k2) [I-] (b)
Replacing [IO-] from (b) in (a), we have:
v = - 1/2 d[H2O2]/dt = k1 [H2O2][I-]
which is also appropriate to the overall rate law.
iii) If the two steps have similar rates:
v = - 1/2 d[H2O2]/dt = 1/2 (k1 [H2O2][I-] + k2 [H2O2][IO-])
Let us assume that the concentration of IO- is in steady-state condition. Similar to the case ii), we have:
v = - 1/2 d[H2O2]/dt = k1 [H2O2][I-]
which corresponds to the overall rate law.
Among the three cases, case i) is the most appropriate to the overall rate law because no assumption is made. Besides, in case ii) the assumption of the steady-state IO- is not valid since step (2) is considered slow.