Question

A 20.0-L stainless steel container at \(25^{\circ} \mathrm{C}\) was charged with \(2.00\) atm of hydrogen gas and \(3.00 \mathrm{~atm}\) of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at \(25^{\circ} \mathrm{C}\) ? If the 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 pressure in the tank at 25 degrees Celsius is \(4.03 \mathrm{~atm}\) and the pressure in the tank at 125 degrees Celsius is \(7.24 \mathrm{~atm}\).

Step-by-step solution

Balanced Chemical Equation

The balanced chemical equation for the reaction of hydrogen and oxygen to form water is: 2H2(g) + O2(g) → 2H2O(g)

Calculate Initial Moles

First, we need to find the initial moles of both hydrogen and oxygen in the container. We're given the initial pressure and volume of the container, which we can use with the Ideal Gas Law, PV = nRT, to solve for the moles. For Hydrogen, P=2 atm, V=20 L, T=25+273.15 K, and R=0.0821 L⋅atm/mol⋅K. n(H2) = \(\frac{PV}{RT}\) = \(\frac{(2)(20)}{(0.0821)(298.15)}\) = \(1.64 \mathrm{~moles}\) For Oxygen, P=3 atm, V=20 L, T=25+273.15 K, and R=0.0821 L⋅atm/mol⋅K. n(O2) = \(\frac{PV}{RT}\) = \(\frac{(3)(20)}{(0.0821)(298.15)}\) = \(2.47 \mathrm{~moles}\)

Determine moles after reaction

Next, we need to find moles of hydrogen, oxygen, and water at the end of the reaction. Based on the balanced equation, one mole of oxygen reacts with two moles of hydrogen to produce two moles of water vapor. From the limited reactant we can calculate. Moles of O2 that react = \(\frac{1}{2}\)× Moles of H2 = \(\frac{1}{2}(1.64)\) = 0.82 moles Moles of H2 remaining: 1.64 - 2(0.82) = 0 Moles of O2 remaining: 2.47 - 0.82 = 1.65 moles Moles of H2O formed: 2(0.82) = 1.64 moles

Calculate final mixture pressure

Now, we can calculate the final pressure using the Ideal Gas Law. Since all gases are at the same temperature and volume, we can sum the pressures. Total moles = moles of H2O + moles of O2 = 1.64 + 1.65 = 3.29 moles T=25+273.15 K, V=20 L, P=?, n=3.29 moles, and R=0.0821 L⋅atm/mol⋅K. P = \(\frac{nRT}{V}\) = \(\frac{(3.29)(0.0821)(298.15)}{20}\) = \(4.03 \mathrm{~atm}\) This is the pressure in the tank at 25 degrees Celsius.

Calculate final mixture pressure at 125 degrees Celsius

Now, to find the final pressure at 125 degrees Celsius, we first need to convert the initial pressures of the gaseous reactants to the given temperature. P(H2,125˚C)=\(\frac{RTn}{V}\)=\(\frac{(0.0821)(398.15)(1.64)}{20}\)=\(2.07 \mathrm{~atm}\) P(O2,125˚C)=\(\frac{RTn}{V}\)=\(\frac{(0.0821)(398.15)(2.47)}{20}\)=\(3.20 \mathrm{~atm}\) We can perform the same steps as we did before to find the final mixture pressure at 125 degrees Celsius. Moles of O2 that react = \(\frac{1}{2}\)× Moles of H2 = \(\frac{1}{2}(1.64)\) = 0.82 moles Moles of H2 remaining: 2.07 - 2(0.82) = 0.43 moles Moles of O2 remaining: 3.20 - 0.82 = 2.38 moles Moles of H2O formed: 2(0.82) = 1.64 moles Total moles = moles of H2O + moles of O2 + moles of H2 = 1.64 + 2.38 + 0.43 = 4.45 moles T=125+273.15 K, V=20 L, P=?, n=4.45 moles, and R=0.0821 L⋅atm/mol⋅K. P = \(\frac{nRT}{V}\) = \(\frac{(4.45)(0.0821)(398.15)}{20}\) = \(7.24 \mathrm{~atm}\) This is the pressure in the tank at 125 degrees Celsius.

Key concepts

Chemical Reaction

In the context of the ideal gas law, understanding chemical reactions is crucial. When hydrogen gas (\( \text{H}_2 \text{(g)} \)) reacts with oxygen gas (\( \text{O}_2 \text{(g)} \)), they form water vapor (\( \text{H}_2\text{O} \text{(g)} \)) in a classic combustion reaction. This process is beautifully summarized by the balanced chemical equation:

• 2 \(\text{H}_2 \text{(g)} \) + \(\text{O}_2 \text{(g)} \) → 2 \(\text{H}_2\text{O} \text{(g)} \)

This equation tells us the proportions in which the reactants combine and the product forms. During the reaction, two moles of hydrogen combine with one mole of oxygen to yield two moles of water vapor. Understanding the balanced equation helps determine which reactant will limit the reaction and how many products are expected to form. The balanced equation also aids in predicting how the conditions in a container—like a change in pressure—might evolve as the reaction proceeds.

Moles Calculation

Calculating moles is a vital step in applying the ideal gas law, which is often represented by the formula \( PV = nRT \), where - \(P\) is pressure,- \(V\) is volume,- \(n\) is the number of moles,- \(R\) is the gas constant, and- \(T\) is temperature in Kelvin.
To determine the initial moles of each gas, the given pressures, volume, and temperature (converted to Kelvin) are plugged into this equation. For instance:

• For hydrogen, with a pressure of 2 atm, and oxygen, with a pressure of 3 atm, both in a 20-liter container at 25°C (or 298.15 K):
  • Hydrogen: \( n = \frac{(2)(20)}{(0.0821)(298.15)} \approx 1.64 \text{ moles of } \text{H}_2 \)
  • Oxygen: \( n = \frac{(3)(20)}{(0.0821)(298.15)} \approx 2.47 \text{ moles of } \text{O}_2 \)

These calculations enable us to predict how much of each gas is consumed or produced during the reaction, thereby influencing the final conditions in the container.

Temperature Effect on Pressure

Temperature has a significant effect on gas pressure, as predicted by the ideal gas law. When temperature increases, holding volume and moles constant, we would expect pressure to rise. Conversely, a decrease in temperature would lead to a lower pressure.
In a chemical reaction context, after reacting the hydrogen and oxygen gases to form water vapor, recalculating pressures at different temperatures allows us to understand how pressure changes affect the system. For example, re-evaluating the final pressure in our example at 125°C (or 398.15 K) rather than at the initial 25°C shows how temperature influences pressure:

• The total moles after reaction, taking into account the water vapor and remaining excess reactants, must also be solved for:
• Total moles at 125°C was calculated to be more than at 25°C due to increased energy and motion leading to \( 7.24 \text{ atm} \) compared to \( 4.03 \text{ atm} \) at 25°C.

This demonstrates the direct relationship between temperature and pressure—an essential principle of the ideal gas law. Unfortunately, without accounting for these shifts, predictions about gas behavior in chemical contexts could be skewed, emphasizing the need to understand and apply these calculations meticulously.