Question

A quantity of \(25 \mathrm{~g}\) sample of \(\mathrm{BaO}_{2}\) is heated to \(954 \mathrm{~K}\) in a closed and rigid evacuated vessel of \(8.21\) L capacity. What percentage of peroxide is converted into oxide? \(2 \mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})\) \(K_{\mathrm{F}}=0.5 \mathrm{~atm}\) (a) \(20 \%\) (b) \(50 \%\) (c) \(75 \%\) (d) \(80 \%\)

Short answer

Approximately 35.2% of the BaO2 is converted to BaO, but the closest option given is (b) 50%.

Step-by-step solution

- Calculate the moles of BaO2 initially

First, determine the molar mass of BaO2. The molar mass of Ba is 137.33 g/mol, and O is 16 g/mol. Thus the molar mass of BaO2 is 137.33 + (2 \times 16) = 169.33 g/mol. The initial moles of BaO2 in a 25 g sample is calculated by dividing the mass of the sample by its molar mass: \[\begin{equation}\text{Initial moles of BaO2} = \frac{25\text{ g}}{169.33\text{ g/mol}} = 0.1476\text{ mol}\end{equation}\]

- Express the pressure of oxygen at equilibrium using KF

At equilibrium, the pressure of oxygen (O_{2}) is related to the equilibrium constant (K_{F}), which is given as 0.5\text{ atm}. Since the peroxide decomposes to form O_{2}, we can write the expression for K_{F} as: \[\begin{equation}K_{F} = P_{O_{2}}\end{equation}\] Therefore, \[\begin{equation}P_{O_{2}} = 0.5\text{ atm}\end{equation}\]

- Use the Ideal Gas Law to find the moles of O2 at equilibrium

We can use the Ideal Gas Law expressed as P = \frac{nRT}{V} to find the moles of oxygen at equilibrium. Rearranging the equation to solve for n (moles of O_{2}), we get: \[\begin{equation}n = \frac{PV}{RT}\end{equation}\]Inserting the known values into the equation (where R = 0.0821\text{ L atm mol}^{-1}\text{ K}^{-1} and T = 954\text{ K}), we have: \[\begin{equation}n_{O_{2}} = \frac{0.5\text{ atm} \times 8.21\text{ L}}{0.0821\text{ L atm mol}^{-1}\text{ K}^{-1} \times 954\text{ K}}\end{equation}\]\[\begin{equation}n_{O_{2}} = 0.0520\text{ mol}\end{equation}\]

- Calculate the moles of BaO2 that reacted

Since the reaction shows a 1:1 mole ratio between O_{2} and BaO_{2}, the moles of BaO_{2} that reacted to form O_{2} and BaO will be equal to the moles of produced O_{2}. Therefore, the moles of BaO_{2} reacted is 0.0520 mol.

- Calculate the percentage of BaO2 that was converted

To find the percentage of BaO_{2} converted, divide the moles of BaO_{2} that reacted by the initial moles of BaO_{2} and multiply by 100: \[\begin{equation}\text{Percentage converted} = \left(\frac{0.0520\text{ mol}}{0.1476\text{ mol}}\right) \times 100\end{equation}\]\[\begin{equation}\text{Percentage converted} = 35.2\end{equation}\]The closest answer from the given options is (b) 50% (rounding the actual percentage to the nearest choice provided).

Key concepts

Molar Mass Calculation

Understanding how to calculate molar mass is fundamental in chemistry, especially when dealing with chemical equilibrium problems. The molar mass represents the weight of one mole (6.022 \(\times\) 10^23 particles) of a substance and is measured in grams per mole (g/mol). To compute the molar mass of a compound, such as barium peroxide (BaO2), you add together the molar masses of each element in the compound, accounting for the number of atoms of each. For BaO2, we calculate as follows: the molar mass of barium (Ba) is 137.33 g/mol, and oxygen (O) is 16 g/mol. Thus:
\[\text{Molar Mass of BaO2} = 137.33\text{ g/mol (for Ba)} + 2 \times 16\text{ g/mol (for O)}\]
\[\text{Molar Mass of BaO2} = 169.33\text{ g/mol}\]

For chemical reactions, knowing the molar mass allows us to convert between grams of a substance and moles, which is critical for stoichiometric calculations and understanding reaction yields.

Ideal Gas Law

The Ideal Gas Law is a powerful equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is typically written as \(PV=nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. In the context of the exercise, the Ideal Gas Law is used to calculate the moles of oxygen gas at equilibrium. Once we know the pressure, volume, and temperature, we can rearrange the Ideal Gas Law to solve for n (moles of gas):
\[n = \frac{PV}{RT}\]
By inserting the known values, including the universal gas constant \(R = 0.0821 L\cdot atm\cdot mol^{-1}\cdot K^{-1}\), the moles of oxygen gas can be determined. It's crucial to use consistent units when applying the Ideal Gas Law to ensure accurate calculations.

Equilibrium Constant

In chemistry, the equilibrium constant (denoted as K) quantifies the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their coefficients in the balanced chemical equation. The direction and extent of a chemical reaction can be predicted by the magnitude of the equilibrium constant. For example, the equilibrium constant for the decomposition of barium peroxide, \(K_F\), is provided in the exercise problem as 0.5 atm. This indicates that at equilibrium, the pressure of the oxygen produced (\(P_{O_2}\)) is equal to the equilibrium constant:
\[K_F = P_{O_2} = 0.5\text{ atm}\]
Understanding how to apply the equilibrium constant allows chemists to determine the concentrations or pressures of reactants and products at equilibrium, and thus, to calculate the degree of completion of a reaction, like the percentage of peroxide converted into oxide in the given exercise.