Passage 4: Hydrogen Peroxide

Hydrogen peroxide (H2O2) is a common component in fuels, antiseptics, and

bleaching agents due to its strong oxidizing properties. It decomposes into water

and O2 gas via the following relationship:

2H2O2 (aq) = 2H2O(l) + O2 (g) ΔH: -98kJ/mol (Reaction 1).

Researchers performed decomposition of H2O2 with potassium iodide (KI) at

various temperatures. Specifically, they measured the rate of oxygen gas evolution

at varying concentrations of H2O2 and KI as well as varying temperatures. Results

are summarized in Table 1:

The specific heat capacity of the hydrogen peroxide solution was measured to be

2.5 J/g°C and was slightly more dense than water at 1.45 g/cm3. Hydrogen peroxide

is quite environmentally friendly as an oxidizing agent because of this harmless

decomposition into water and oxygen gas.

At the end of the experiment, the researchers noticed that no potassium iodide

had been consumed over the course of their project, although its presence always

increased reaction rate relative to the same reaction run at the same temperature

and concentrations in its absence.

Imagine that the researchers performed this reaction in 2 separate reaction

vessels at the same temperature. Vessel 1 was open to the atmosphere,

but Vessel 2 had a lid over the solution, sealing it from the atmosphere.

Which vessel would facilitate the faster rate of reaction, and when would the

difference in rate be most significant?

A) Vessel 1; immediately upon the start of reaction

B) Vessel 2; immediately upon the start of reaction

C) Vessel 2; a long time after the start of reaction

D) Vessel 1; a long time after the start of reaction

Click to reveal answer

Correct answer: D. Henry’s Law of Solubility states that the

concentration of a gas dissolved in a liquid is directly proportional to the partial

pressure of the gas above the liquid. With oxygen gas (O2) being evolved as a

product of reaction, the partial pressure of O2 gas in Vessel 2 would increase (due

to it being a sealed container), and subsequently the amount of dissolved O2 gas

would also increase. This would slow the reaction due to Le Chatelier’s principle,

wherein an increase in products generally leads to an increase in the rate of reverse

reaction and a decrease in the rate of forward reaction. Additionally, this effect

would be more pronounced a long time after the start of reaction than immediately

after the start of reaction because dissolved O2 gas would increase with time in

Vessel 2.

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Day 25 MCAT Practice Question