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 \(\begin{array}{lllll}85.0 & \text { seconds at a barometric pressure of } & 742.5 & \mathrm{mmHg} ?\end{array}\) (The vapor pressure of water at this temperature is \(17.5 \mathrm{mmHg}\).)

Short answer

Calculate the initial moles of H2O2, find the remaining moles after 85 seconds using first-order kinetics, determine the moles of O2 produced, and use the ideal gas law with the corrected pressure to find the volume of O2.

Step-by-step solution

Calculate the initial amount of H2O2

First, calculate the mass of H2O2 in the 150.0 mL solution. Use the density to find the total mass of the solution and then use the percentage of H2O2 to find the mass of H2O2 present. Mass of solution = 150.0 mL * 1.11 g/mL = 166.5 g. Mass of H2O2 = 30.0% of 166.5 g = 0.30 * 166.5 g = 49.95 g.

Convert mass of H2O2 to moles

Convert the mass of H2O2 to moles using its molar mass (34.01 g/mol). Moles of H2O2 = 49.95 g / 34.01 g/mol = 1.468 moles.

Calculate the change in the amount of H2O2

Use the first-order rate law and the given rate constant to calculate the remaining moles of H2O2 after 85.0 seconds. The rate law for a first-order reaction is ln([A]0/[A]) = kt, where k is the rate constant and t is time. Rearrange to solve for [A], the concentration of H2O2 after time t: [A] = [A]0 * e^(-kt).

Find the moles of oxygen gas produced

Since the stoichiometry of the reaction is 2 moles of H2O2 producing 1 mole of O2, the change in moles of H2O2 will be twice the moles of O2 produced. Use the moles of H2O2 reacted to find the moles of O2 produced.

Apply Dalton's Law of Partial Pressures

Account for the vapor pressure of water by subtracting it from the barometric pressure to find the pressure of dry oxygen gas. Total pressure is 742.5 mmHg. The vapor pressure of water at 20.0°C is 17.5 mmHg. Pressure of dry oxygen gas = 742.5 mmHg - 17.5 mmHg = 725.0 mmHg.

Use the Ideal Gas Law to find the volume of O2

The ideal gas law is PV = nRT, where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvin. Convert the temperature to Kelvin and use the pressure of dry oxygen to find its volume.

Key concepts

First-Order Reaction

Understanding a first-order reaction is fundamental in the study of chemical kinetics. This type of reaction has a rate that is directly proportional to the concentration of one reactant. In mathematical terms, if we represent the reactant concentration as [A], the rate of the reaction can be expressed as Rate = k[A], where 'k' is the rate constant. A key characteristic of first-order reactions is that their half-life, the time it takes for half of the reactant to be consumed, is constant and independent of its initial concentration.

When solving problems involving first-order reactions, one often uses the integrated rate law, given by \( \text{ln} \frac{[A]_0}{[A]} = kt \), where \( [A]_0 \) is the initial concentration, [A] is the concentration at time 't', and 'k' is the rate constant. In practical terms, this equation allows us to determine how the concentration of a reactant changes over time, given the rate constant and the period of time.

Rate Constant

The rate constant, denoted as 'k', is a crucial parameter in chemical kinetics, providing a quantitative measure of how quickly a reaction proceeds. It is unique for each reaction and can be affected by factors like temperature and the presence of a catalyst. For a first-order reaction, the units of the rate constant are \( s^{-1} \), indicating the reaction rate per second.

Knowing the rate constant allows us to predict the speed of a reaction under specific conditions. This enables chemists to control and optimize reactions, whether it's speeding up a desired chemical process or slowing down a potentially hazardous one. In our original exercise, the rate constant is given, allowing us to calculate the concentration of remaining reactants at any given time.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's the 'recipe' for a reaction, telling us how much of each substance is needed to react completely with other substances, and what amount of product will be formed. The coefficients in a balanced chemical equation provide the molar ratio of reactants and products.

In our example reaction \(2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2\), the stoichiometry tells us that 2 moles of hydrogen peroxide produce 1 mole of oxygen gas. Hence, by determining the amount of hydrogen peroxide that has reacted, we can directly calculate the moles of oxygen gas produced. Accurate stoichiometric calculations are essential for predicting the outcomes of reactions and are the basis for the design of chemical processes.

Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures is a principle in chemistry that states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases. A gas's partial pressure is the pressure it would exert if it occupied the entire volume of the mixture alone.

This law becomes extremely relevant when collecting gases over water, as in our exercise. Since water vapor exerts its own pressure, we must account for it to determine the pressure of the gas of interest—oxygen, in this case. Subtracting the water vapor pressure from the total pressure gives us the dry oxygen gas's partial pressure. Applying Dalton's Law correctly is vital to accurately determine the amount of gaseous product in gas collection experiments.

Ideal Gas Law

The Ideal Gas Law is a well-established principle that describes the behavior of an ideal gas. Represented by the equation \(PV = nRT\), it relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T) of a gas. For the calculation, 'R' is a known constant (0.0821 L·atm/mol·K), and the temperature must be converted to Kelvin (K).

In the scenario provided in our exercise, we use the Ideal Gas Law to find the volume of oxygen gas produced from the known moles and temperature. Correct application of this law is crucial to determining the volume of gases in a variety of conditions, especially in laboratory settings where gas properties are to be measured or predicted.