Question

A mixture in which the mole ratio of hydrogen to oxygen is 2: 1 is used to prepare water by the reaction $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ The total pressure in the container is 0.950 atm at \(25^{\circ} \mathrm{C}\) before the reaction. What is the final pressure in the container at \(125^{\circ} \mathrm{C}\) after the reaction, assuming an \(88.0 \%\) yield and no volume change?

Short answer

The final pressure in the container after the reaction is 1.299 atm.

Step-by-step solution

Calculate the number of moles of hydrogen and oxygen

From the information given: "mole ratio of hydrogen to oxygen is 2:1." Let the total number of moles, nT, equal the sum of the moles of hydrogen and oxygen before the reaction. So, $$\frac{n_{\mathrm{H_2}}}{n_{\mathrm{O_2}}} = \frac{2}{1}$$ $$n_{\mathrm{T}} = n_{\mathrm{H_2}} + n_{\mathrm{O_2}}$$

Find the moles of hydrogen and oxygen using ideal gas law

We know that the total pressure is 0.950 atm and temperature is \(25^{\circ} \mathrm{C}\). Converting to Kelvin: $$T = 25 + 273.15 = 298.15 \: \mathrm{K}$$ Using the ideal gas law, we have: $$PV = n_{\mathrm{T}}RT$$ Solving for \(n_{\mathrm{T}}\), we get: $$n_{\mathrm{T}} = \frac{PV}{RT}$$ Since we are considering no volume change during the process, cancel out the volume from both sides: $$n_{\mathrm{T}} = \frac{P}{R} \times T$$ Plug in the known values for pressure and temperature: $$n_{\mathrm{T}} = \frac{0.950 \: \mathrm{atm}}{0.0821 \: \mathrm{L \: atm/mol \: K}} \times 298.15 \: \mathrm{K}$$ $$n_{\mathrm{T}} = 34.73 \: \mathrm{moles}$$

Calculate moles of H2 and O2, and find the limiting reactant

Using the mole ratios \(\frac{n_{\mathrm{H_2}}}{n_{\mathrm{O_2}}} = \frac{2}{1}\) and \(n_{\mathrm{T}} = n_{\mathrm{H_2}} + n_{\mathrm{O_2}}\), we can find \(n_{\mathrm{H_2}}\) and \(n_{\mathrm{O_2}}\): \(n_{\mathrm{H_2}} = 2 \times n_{\mathrm{O_2}}\) So, \(n_{\mathrm{T}} = n_{\mathrm{H_2}} + n_{\mathrm{O_2}} = 3 \times n_{\mathrm{O_2}}\) $$n_{\mathrm{O_2}} = \frac{n_{\mathrm{T}}}{3} = \frac{34.73}{3} = 11.58 \: \mathrm{moles}$$ $$n_{\mathrm{H_2}} = 2 \times n_{\mathrm{O_2}} = 2 \times 11.58 = 23.16 \: \mathrm{moles}$$ Since O2 is the reactant with the smaller number of moles, it is the limiting reactant.

Calculate moles of water produced

Using the stoichiometry of the balanced equation (\(2\: \mathrm{H_2} + \mathrm{O_2} \rightarrow 2 \: \mathrm{H_2O}\)), we can find the moles of water produced. From an 88% yield, the moles of water produced is: $$n_{\mathrm{H_2O}} = 2 \times n_{\mathrm{O_2}} \times 0.88 = 2 \times 11.58 \times 0.88 = 20.42 \: \mathrm{moles}$$

Determine moles of unreacted H2

Moles of unreacted hydrogen are the initial moles of hydrogen minus the moles that reacted: $$n_{\mathrm{Unreacted \: H_2}} = n_{\mathrm{H_2}} - n_{\mathrm{H_2O}} = 23.16 - 20.42 = 2.74 \: \mathrm{moles}$$

Calculate final pressure

Find the final moles of gas in the container, including unreacted hydrogen and water gas: $$n_{\mathrm{Final}} = n_{\mathrm{Unreacted \: H_2}} + n_{\mathrm{H_2O}} = 2.74 + 20.42 = 23.16 \: \mathrm{moles}$$ Convert the final temperature to Kelvin: $$T_{\mathrm{Final}} = 125 + 273.15 = 398.15 \: \mathrm{K}$$ Now, plug in our new values for n and T in the ideal gas law, knowing that the volume has not changed: $$P_{\mathrm{Final}} = \frac{n_{\mathrm{Final}} RT_{\mathrm{Final}}}{V}$$ Cancel volume from both sides as it remains constant: $$P_{\mathrm{Final}} = \frac{n_{\mathrm{Final}} R T_{\mathrm{Final}}}{P_{\mathrm{Initial}} / R T_{\mathrm{Initial}}} = \frac{23.16 \times 0.0821 \times 398.15}{0.950}$$ $$P_{\mathrm{Final}} = 1.299 \: \mathrm{atm}$$ The final pressure in the container at \(125^{\circ} \mathrm{C}\) after the reaction, assuming an 88.0% yield and no volume change, is 1.299 atm.

Key concepts

Mole Ratio

Understanding the mole ratio is crucial for solving many chemistry problems, especially those involving reactions. It tells you how much of one reactant is needed to react with a given amount of another. In the provided exercise, we have a classic example involving the synthesis of water, where the mole ratio between hydrogen and oxygen is a perfect 2:1, meaning two moles of hydrogen gas are needed for every one mole of oxygen gas for the reaction to proceed without any excess reactants. These whole-number ratios are derived from the balanced chemical equation and form the foundation of stoichiometric calculations.

For students who struggle with this concept, visual aids such as 'pictorial equations' using marbles or other simple counters to represent moles can be incredibly helpful for cementing understanding.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is typically expressed as PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is temperature in Kelvin. This law is used to predict the behavior of gases under different conditions and is instrumental in calculations involving gas reactions, as shown in our exercise. It's important to remember to always convert temperatures to Kelvin and that the value of R used must match the units of pressure and volume in your calculation.

To assist students in mastering the ideal gas law, interactive simulations that allow them to manipulate variables and observe results can demystify the relationships between these variables.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. By using the mole ratios from a balanced chemical equation, you can calculate how much product will be made from given amounts of reactants, or vice versa. It's essentially the 'recipe' part of chemistry that tells you how much of each 'ingredient' you need. This concept was applied in steps 3 and 4 of the exercise, where the stoichiometry of the water synthesis reaction was used to calculate the moles of water that would form from the reaction.

When helping students with stoichiometry, a step-by-step approach, breaking down each part of a chemical reaction, is very effective. Equations should always be balanced first to ensure the correct ratios are used.

Limiting Reactant

The limiting reactant is the reactant in a chemical reaction that is completely consumed first, limiting the amount of product that can be formed. Identifying the limiting reactant is vital because it determines the theoretical yield of the reaction. In our exercise, oxygen was the limiting reactant since it was present in smaller stoichiometric amounts compared to hydrogen when considering the 2:1 mole ratio.

To facilitate comprehension of this concept, using practice problems that allow students to calculate the amount of product that can be formed from given amounts of two or more reactants can sharpen their ability to identify the limiting reactant.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, described as the ratio of the actual yield to the theoretical yield, multiplied by 100%. The theoretical yield is the maximum amount of product that can be generated as predicted by stoichiometry, while the actual yield is the amount you actually get, which can be less due to practical losses. In the exercise, the reaction had an 88% yield, indicating that only 88% of the predicted water was produced. This aspect is crucial for practical chemistry, as it accounts for real-world inefficiencies.

To aid students with this topic, experiments demonstrating a reaction with measurable yields can highlight the differences between theoretical predictions and actual results, thus clarifying the percent yield concept.

Gas Laws

Gas laws describe how gases behave with changes in temperature, pressure, volume, and quantity. These laws include Boyle's Law, Charles's Law, Gay-Lussac's Law, and the Combined Gas Law, among others. They are simplified models for predicting the behavior of gases, useful for various applications from balloons to breathing mechanisms. In our exercise, the Combined Gas Law, which is a combination of Boyle's, Charles's, and Gay-Lussac's laws, was implicitly used to calculate the final pressure of the gas mixture in the container after the reaction and change in temperature.

When teaching gas laws, demonstrating each law with real-life examples and laboratory experiments can make these abstract concepts more tangible. Simple experiments, like observing the expansion of a balloon with temperature, can give students a visual and practical understanding of these laws.