Question

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as \(\mathrm{Mn}\) or \(\mathbf{F} \mathrm{e}\)): $$2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)$$ What volume of pure \(\mathrm{O}_{2}(g),\) collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of 125 \(\mathrm{g}\) of a 50.0\(\% \mathrm{by}\) mass hydrogen peroxide solution? Ignore any water vapor that may be present.

Short answer

The volume of pure \(\mathrm{O}_2(g)\) generated by the decomposition of 125 g of a 50.0% by mass hydrogen peroxide solution, collected at \(27^{\circ} \mathrm{C}\) and 746 torr, is determined in three steps. First, we find the number of moles of \(\mathrm{H}_2\mathrm{O}_2\) in the 50% solution: \(n_{\mathrm{H}_2\mathrm{O}_2} = \frac{0.50 \times 125 \ g}{34.01\ \mathrm{g/mol}}\). Then, using the balanced chemical equation, we find the number of moles of generated \(\mathrm{O}_2\): \(n_{\mathrm{O}_2} = \frac{n_{\mathrm{H}_2\mathrm{O}_2}}{2}\). Finally, we use the Ideal Gas Law to find the volume of \(\mathrm{O}_2\): \(V = \frac{n_{\mathrm{O}_2} \cdot 0.0821 \ \mathrm{atm \cdot L/(mol\cdot K)} \cdot 300\ \mathrm{K}}{\frac{746}{760}\ \mathrm{atm}}\). Calculating these values, we find that the volume of pure \(\mathrm{O}_2(g)\) generated is approximately \(547.2 \ \mathrm{L}\).

Step-by-step solution

Calculate the moles of hydrogen peroxide in the solution

We are given a 50% by mass hydrogen peroxide solution, which means that 50% of the solution's mass is hydrogen peroxide. In the 125 g solution, we can find the mass of hydrogen peroxide by: Mass of \(\mathrm{H}_2\mathrm{O}_2 = 0.50 \times 125 \ g\) Now, to find the number of moles, we use the molar mass of \(\mathrm{H}_2\mathrm{O}_2\), which is \(34.01 \ \mathrm{g/mol}\): Number of moles = \(\frac{\text{Mass of }\mathrm{H}_2\mathrm{O}_2}{\text{Molar Mass of }\mathrm{H}_2\mathrm{O}_2}\)

Calculate the number of moles of O2 generated

Using the balanced chemical equation, we see that: \(2 \ \mathrm{H}_2\mathrm{O}_2 \longrightarrow \mathrm{O}_2\) Thus, for every 2 moles of \(\mathrm{H}_2\mathrm{O}_2\), we get 1 mole of \(\mathrm{O}_2\). So, we can find the number of moles of \(\mathrm{O}_2\) by dividing the number of moles of \(\mathrm{H}_2\mathrm{O}_2\) by 2.

Use the Ideal Gas Law to find the volume of O2

Now, we'll use the Ideal Gas Law (\(PV=nRT\)) to find the volume of \(\mathrm{O}_2(g)\). We're given temperature (T = \(27^{\circ}\mathrm{C} = 300\ \mathrm{K}\)) and pressure (P = \(746 \ \mathrm{torr} = \frac{746}{760} \ \mathrm{atm}\)). The gas constant (R) is \(0.0821 \ \mathrm{atm \cdot L/(mol\cdot K)}\). Substituting the known values in the Ideal Gas Law equation, we can find the volume (V) of \(\mathrm{O}_2(g)\): \(V = \frac{n \cdot R \cdot T}{P}\) Now let's plug in the values and perform the calculations.

Key concepts

decomposition reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances. In the case of hydrogen peroxide (\(\text{H}_2\text{O}_2\)), it naturally decomposes into water (\(\text{H}_2\text{O}\)) and oxygen gas (\(\text{O}_2\)). This process is accelerated by the presence of transition metal ions like manganese (Mn) or iron (Fe), which act as catalysts. The chemical equation representing this reaction is:

• \(2 \text{H}_2\text{O}_2 (aq) \rightarrow 2 \text{H}_2\text{O} (l) + \text{O}_2 (g)\)

Here, the catalyst speeds up the reaction but is not consumed by it. Understanding decomposition reactions helps in predicting the behavior of compounds, especially in the presence of catalysts.

molar mass calculation

Calculating molar mass is an essential step in many chemical reactions, involving converting a given mass of a substance to moles. Molar mass is the mass of one mole of a substance. For hydrogen peroxide (\(\text{H}_2\text{O}_2\)), its molar mass is calculated by adding the atomic masses of its constituent elements:

• Hydrogen (H): 1.01 g/mol \( \times 2 = 2.02 \text{ g/mol}\)
• Oxygen (O): 16.00 g/mol \( \times 2 = 32.00 \text{ g/mol}\)

The total molar mass of \(\text{H}_2\text{O}_2\) is \(34.01 \text{ g/mol}\). This molar mass is used to determine the number of moles in a specific mass of a substance, which is crucial for further calculations in reactions.

transition metal catalysis

Transition metal catalysis is a fascinating aspect of chemistry, involving metals from the d-block of the periodic table to accelerate chemical reactions. These metals, like Mn or Fe, are excellent catalysts due to their ability to change oxidation states easily, which assists in breaking and forming chemical bonds. When hydrogen peroxide decomposes, these metals speed up the reaction without being used up themselves. They offer a surface for the reaction to occur, making it proceed more quickly than it would without them. Catalysis by transition metals is important not only in laboratory reactions but also in industrial processes, leading to faster and more efficient production methods.

chemical equilibrium

Chemical equilibrium describes a state in which both the reactants and products in a chemical reaction are present in concentrations that have no further tendency to change with time. Understanding equilibrium is important in balancing reaction rates. For the decomposition of hydrogen peroxide, the process is often viewed as reaching equilibrium quickly, where reactants break down to form products dynamically. In cases where transition metals catalyze a reaction, equilibrium can shift toward the formation of products more efficiently. This knowledge is applied in creating conditions that favor the desired outcomes in various chemical industries. Although the decomposition reaction for \(\text{H}_2\text{O}_2\) releases gas, depending on the system's conditions, equilibrium concepts help manage and predict how much product can be expected under different conditions.