Question

Hydrogen peroxide is unstable and decomposes readily: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ This reaction is accelerated by light, heat, or a catalyst. (a) Explain why hydrogen peroxide sold in drugstores comes in dark bottles. (b) The concentrations of aqueous hydrogen peroxide solutions are normally expressed as percent by mass. In the decomposition of hydrogen peroxide, how many liters of oxygen gas can be produced at STP from \(15.0 \mathrm{~g}\) of a 7.5 percent hydrogen peroxide solution?

Short answer

Hydrogen peroxide is stored in dark bottles to slow decomposition; 0.371 L of \( \mathrm{O_2} \) gas can be produced from the solution.

Step-by-step solution

Understand Why Dark Bottles Are Used

Hydrogen peroxide decomposes readily under the influence of light, heat, or catalysts. The dark bottles help to limit the exposure to light, which slows down the decomposition process and maintains the effectiveness of the hydrogen peroxide for a longer time.

Calculate the Mass of Hydrogen Peroxide

Given a solution that is 7.5% hydrogen peroxide, calculate the mass of hydrogen peroxide in 15.0 g of solution. Use the formula: \[ \text{Mass of } \mathrm{H_2O_2} = \frac{7.5}{100} \times 15.0 = 1.125 \text{ g} \] So, there are 1.125 g of hydrogen peroxide.

Convert Mass to Moles

Find the number of moles of hydrogen peroxide. The molar mass of \( \mathrm{H_2O_2} \) is approximately 34.02 g/mol. \[ \text{Moles of } \mathrm{H_2O_2} = \frac{1.125 \text{ g}}{34.02 \text{ g/mol}} = 0.0331 \text{ mol} \]

Determine Moles of Oxygen Gas Produced

The balanced chemical equation indicates that 2 moles of hydrogen peroxide produce 1 mole of oxygen gas. Therefore, calculate the moles of \( \mathrm{O_2} \) produced from 0.0331 moles of \( \mathrm{H_2O_2} \): \[ \text{Moles of } \mathrm{O_2} = \frac{1}{2} \times 0.0331 \text{ mol} = 0.01655 \text{ mol} \]

Calculate Volume of Oxygen Gas

Use the ideal gas law (at STP, 1 mole of gas occupies 22.4 L) to find the volume of \( \mathrm{O_2} \): \[ \text{Volume of } \mathrm{O_2} = 0.01655 \text{ mol} \times 22.4 \text{ L/mol} = 0.37072 \text{ L} \]

Round and Present Final Answer

Round the final volume to appropriate significant figures for the given data. The volume of \( \mathrm{O_2} \) produced at STP is 0.371 L.

Key concepts

Catalysts

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. Catalysts work by providing an alternative reaction pathway with a lower activation energy. This means less energy is required for the reaction to proceed, allowing it to happen faster.

In the decomposition of hydrogen peroxide, catalysts like manganese dioxide can be used to accelerate the reaction. This is especially useful in environments where heating or light exposure is not desirable. A key point to remember about catalysts is that they do not alter the overall energy balance (enthalpy change) of the reaction; they merely help reach the final products more quickly.

Many everyday reactions, from biological ones inside our bodies to large-scale industrial processes, benefit from the use of catalysts. In biotechnology, enzymes act as highly specific catalysts that facilitate vital reactions at body temperature.

Percent by Mass

Percent by mass is a way to express the concentration of a solution. It indicates how much solute is present in a given quantity of solution and is calculated as the mass of the solute divided by the total mass of the solution, then multiplied by 100.

For the hydrogen peroxide example, a 7.5 percent by mass solution contains 7.5 grams of hydrogen peroxide in every 100 grams of solution. If you have 15 grams of such a solution, you can find the mass of the hydrogen peroxide using the formula:

• Mass of hydrogen peroxide = (7.5/100) × 15 = 1.125 grams

This calculation is crucial for determining how much reactant is available to undergo decomposition, which impacts both the amount and volume of oxygen gas that can be produced.

Ideal Gas Law

The ideal gas law is a fundamental equation used to relate the properties of gases. It is given by the formula:

• \[ PV = nRT \]

where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature. For standard temperature and pressure (STP), these conditions are 0°C (273.15 K) and 1 atmosphere pressure.

Under STP, 1 mole of any ideal gas occupies 22.4 liters. This value often simplifies gas-related calculations in chemistry problems. In the decomposition of hydrogen peroxide, using this standard molar volume helps to compute the volume of oxygen gas produced based on the calculated moles from the reaction.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, often expressed in grams per mole (g/mol). It's crucial for converting between the mass of a substance and the amount in moles, which are used in stoichiometric calculations.

To calculate the molar mass of hydrogen peroxide \( (\mathrm{H_2O_2}) \), you sum the atomic masses of its constituent atoms:

• Hydrogen (H) has an atomic mass of approximately 1.01; oxygen (O) has an atomic mass of approximately 16.00.
• Molar mass of \( \mathrm{H_2O_2} = 2(1.01) + 2(16.00) = 34.02 \) g/mol

This calculation is fundamental in converting a mass measurement in grams to an amount in moles, which is indispensable in predicting the amounts of products formed in a chemical reaction, such as the decomposition of hydrogen peroxide into water and oxygen gas.