Question

The reaction \(2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) is first order in \(\mathrm{H}_{2} \mathrm{O}_{2}\) and under certain conditions has a rate constant of 0.00752 \(\mathrm{s}^{-1}\) at \(20.0^{\circ} \mathrm{C} .\) A reaction vessel initially contains 150.0 \(\mathrm{mL}\) of 30.0\(\%\) \(\mathrm{H}_{2} \mathrm{O}_{2}\) by mass solution (the density of the solution is 1.11 \(\mathrm{g} / \mathrm{mL} )\) . The gaseous oxygen is collected over water at \(20.0^{\circ} \mathrm{C}\) as it forms. What volume of \(\mathrm{O}_{2}\) forms in 85.0 seconds at a barometric pressure of 742.5 \(\mathrm{mmHg}\) ? (The vapor pressure of water at this temperature is 17.5 \(\mathrm{mm} \mathrm{g} . )\)

Short answer

The volume of O2 gas produced in 85.0 seconds is calculated using the initial moles of H2O2, the first order rate equation, and the ideal gas law after correcting for the vapor pressure of water.

Step-by-step solution

Calculate the initial moles of H2O2

First, calculate the mass of the H2O2 solution by multiplying its volume by its density. Then, find the mass of H2O2 alone by considering the given percentage. Finally, calculate the moles of H2O2 using its molar mass (34.02 g/mol).

Determine the number of moles of O2 produced

Using the first order rate equation, find the concentration of H2O2 after 85.0 seconds. Then use the stoichiometry from the balanced equation to find the moles of O2 produced.

Correct the barometric pressure for water vapor

Subtract the vapor pressure of water from the barometric pressure to get the pressure of dry O2 gas.

Calculate the volume of O2 gas (using the ideal gas law)

The volume of O2 at the corrected pressure and given temperature is calculated by using the ideal gas law: PV = nRT.

Convert the volume of O2 to standard conditions

The volume of gases is often expressed at standard conditions (STP or SATP). This step is optional if the volume at the conditions provided is sufficient.

Key concepts

Rate Law

Understanding the rate law is crucial when studying the kinetics of chemical reactions, especially to determine the speed at which reactants turn into products. For a first-order reaction, the rate at which the substance reacts is directly proportional to its concentration.

The general form of the first-order rate law is given by the expression: ewline \( \text{Rate} = k[\text{A}] \)
where \( k \) is the rate constant and \( [\text{A}] \) is the concentration of reactant A. This type of reaction will have a constant half-life regardless of the initial concentration, which can be an important characteristic to identify the reaction order.
In the exercise example, the rate law helps us determine how much hydrogen peroxide \( (\mathrm{H}_{2} \mathrm{O}_{2}) \) will remain after a certain period by using the rate constant provided. By knowing the initial concentration of \( \mathrm{H}_{2} \mathrm{O}_{2} \), and the time, students can calculate the concentration at any given moment. This step is crucial for predicting the amount of product formed during the reaction.

Stoichiometry

Stoichiometry is the mathematical relationship between the reactants and products in a chemical reaction, based on the principle of conservation of mass. It allows us to predict the amounts of products formed from a given quantity of reactants and vice versa.

To employ stoichiometry in solving our example, it's important to write a balanced chemical equation. In the reaction \(2 \mathrm{H}_{2} \mathrm{O}_{2}(aq) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{O}_{2}(g)\), one mole of oxygen gas \( (\mathrm{O}_{2}) \) is produced for every two moles of hydrogen peroxide consumed. This 2:1 ratio is critical when calculating the moles of oxygen gas produced from the moles of hydrogen peroxide reacting over the given time frame.

Ideal Gas Law

The ideal gas law is a potent tool in comprehension of the behavior of gases under various conditions of temperature, volume, and pressure. The relationship is expressed by the equation:
\( PV = nRT \)
where \( P \) represents the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the absolute temperature.

In the context of our exercise, after calculating the moles of oxygen gas produced, we use the ideal gas law to find the volume that it occupies. It's essential first to correct for the vapor pressure of water since the oxygen is collected over water—this yields the partial pressure of the oxygen gas. With this pressure, the known temperature, and calculated moles, the ideal gas law provides the oxygen's volume. This equation reflects the directly proportional relationship between volume and the number of moles of a gas and temperature, and its inverse proportional relationship with pressure.