Question

A \(20.0\) -\(\mathrm{L}\) stainless steel container at \(25^{\circ} \mathrm{C}\) was charged with \(2.00\) atm of hydrogen gas and \(3.00\) atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at \(25^{\circ} \mathrm{C} ?\) If the exact same experiment were performed, but the temperature was \(125^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C},\) what would be the pressure in the tank?

Short answer

The final pressure in the container at \(25^\circ C\) is 4.03 atm, and the final pressure in the container at \(125^\circ C\) is 5.29 atm.

Step-by-step solution

Write down the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between hydrogen gas and oxygen gas to produce water is: \[2 \, \text{H}_{2}(g) + \text{O}_{2}(g) \rightarrow 2 \, \text{H}_{2}\text{O}(g)\]

Calculate the initial moles of H2 and O2 based on the initial pressures and the volume

First, we need to determine the moles of hydrogen gas and oxygen gas present in the container using the Ideal Gas Law: \[PV = nRT\] Where: \(P\) is the pressure in atm \(V\) is the volume in liters \(n\) is the number of moles \(R\) is the gas constant (0.0821 L⋅atm/(mol⋅K)) \(T\) is the temperature in Kelvin We have been given the initial pressures of Hydrogen gas (2 atm) and Oxygen gas (3 atm), and we are also given the volume of the container (20 L). The temperature is given as \(25^\circ C \) which is \(298 K\). For hydrogen gas: \(P_{H_2} = 2 \, \text{atm}\) \(V_{H_2} = 20 \, \text{L}\) \(T_{H_2} = 298 \, \text{K}\) \(n_{H_2} = \frac{P_{H_2} V_{H_2}} {RT_{H_2}} = \frac{2 \cdot 20}{0.0821 \cdot 298} = 1.63 \, \text{mol}\) For oxygen gas: \(P_{O_2} = 3 \, \text{atm}\) \(V_{O_2} = 20 \, \text{L}\) \(T_{O_2} = 298 \, \text{K}\) \(n_{O_2} = \frac{P_{O_2} V_{O_2}} {RT_{O_2}} = \frac{3 \cdot 20}{0.0821 \cdot 298} = 2.45 \, \text{mol}\)

Determine the limiting reactant

Based on the balanced chemical equation, we can see that 2 moles of hydrogen gas react with 1 mole of oxygen gas. So, we can calculate the required moles of oxygen gas to react completely with the given moles of hydrogen gas: \(n_{O_2} = \frac{n_{H_2}}{2} = \frac{1.63}{2} = 0.82 \, \text{mol}\) Since we have 2.45 moles of oxygen gas but only need 0.82 moles to react completely with the hydrogen gas, hydrogen gas is the limiting reactant.

Determine moles of water produced and final moles of gases

The stoichiometry of the balanced chemical equation shows that the moles of water produced are equal to the moles of hydrogen gas consumed: \(n_{H_2\text{O}} = n_{H_2} = 1.63 \, \text{mol}\) The moles of oxygen gas remaining after the reaction are: \(n_{O_2 \text{ final}} = n_{O_2} - 0.82 = 2.45 - 0.82 = 1.63 \, \text{mol}\)

Calculate the final pressure at 25°C

The final moles of gases inside the container are the sum of the moles of water produced and the moles of oxygen gas remaining: \(n_\text{final} = n_{H_2\text{O}} + n_{O_2 \text{ final}} = 1.63 + 1.63 = 3.26 \, \text{mol}\) In order to find the final pressure, we can use the Ideal Gas Law again: \(P_\text{final} = \frac{n_\text{final} RT}{V} = \frac{3.26 \cdot 0.0821 \cdot 298}{20} = 4.03 \, \text{atm}\) So the final pressure in the container at \(25^\circ C\) is 4.03 atm.

Calculate the final pressure at 125°C

Now we need to determine the final pressure if the temperature was raised to \(125^\circ C\) after the reaction. First, convert the temperature to Kelvin: \(T = 125 + 273.15 = 398.15 \, \text{K}\) Using the Ideal Gas Law again for the final pressure: \(P_\text{final} = \frac{n_\text{final} RT}{V} = \frac{3.26 \cdot 0.0821 \cdot 398.15}{20} = 5.29 \, \text{atm}\) So the final pressure in the container at \(125^\circ C\) is 5.29 atm.

Key concepts

Ideal Gas Law

The Ideal Gas Law is pivotal in understanding how gases behave under different conditions of temperature, volume, and pressure. It is articulated by the equation
\[ PV = nRT \]
where P represents the pressure, V denotes the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin. In the given problem, we use this law to establish a relationship between the known quantities (P, V, T) and the sought variable, which is the number of moles (n) of the reactant gases hydrogen and oxygen. By rearranging the law,
\( n = \frac{PV}{RT} \)
we can calculate the initial moles before the reaction occurs. Understanding this equation provides invaluable insight into predicting how gases will respond when subjected to changes in their environment, such as temperature shifts or volume alterations in a closed system.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, thus dictating the maximum amount of product that can be formed. Identifying the limiting reactant is crucial as it influences how the other reactants are used and what volume of products are produced. For the provided exercise, we compare the stoichiometric ratio of hydrogen and oxygen in the balanced equation with the initially calculated moles to determine which one limits the reaction. Here, hydrogen gas emerges as the limiting reactant since it will run out before the oxygen gas, based on their respective mole ratios needed for the reaction to proceed. More simply put, there isn't enough hydrogen gas to react with all of the oxygen gas available, halting the reaction as soon as the hydrogen is expended.

Gas Stoichiometry

Gas stoichiometry is the quantitative analysis of reactants and products in a gaseous state during a chemical reaction. In this context, the stoichiometry lays out how different volumes of gases interact based on the balanced chemical equation. Adhering to the evidence that the volumes of gases at the same temperature and pressure contain the same number of molecules (Avogadro's law), gas stoichiometry enables us to predict the volume of gases formed as products or needed as reactants. For the equation used in our calculation, the stoichiometry dictates that two moles of hydrogen will produce two moles of water vapor, thereby allowing us to calculate the moles of water formed once the reaction is complete. These stoichiometric relationships are indispensable for chemists to ensure that processes are economically viable and to predict the yields of reactions.

Equilibrium Pressure Calculation

After the limiting reactant has been used up, the reaction ceases and the system achieves a state of dynamic equilibrium. Calculating the equilibrium pressure involves understanding the partial pressures of the remaining gases post-reaction within the closed container. In the given scenario, we consider both the products and any excess reactants. Since our system has a constant volume and amount of gas, we use the Ideal Gas Law again to find the total pressure after the reaction. The total pressure, also known as the equilibrium pressure, is the sum of the partial pressures of all gases present. This exercise exemplifies the application of the Ideal Gas Law to find equilibrium pressure, highlighting its importance not only in initial reaction setup but also in analyzing outcomes post-reaction. In real-life applications, such calculations are essential for designing safety systems for chemical storage and reactions.