Question

A constant-volume tank contains a mixture of 1 mol of \(\mathrm{H}_{2}\) and \(0.5 \mathrm{mol}\) of \(\mathrm{O}_{2}\) at \(25^{\circ} \mathrm{C}\) and 1 atm. The contents of the tank are ignited, and the final temperature and pressure in the tank are \(2800 \mathrm{K}\) and 5 atm, respectively. If the combustion gases consist of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2},\) and \(\mathrm{O}_{2},\) determine \((a)\) the equilibrium composition of the product gases and ( \(b\) ) the amount of heat transfer from the combustion chamber. Is it realistic to assume that no \(\mathrm{OH}\) will be present in the equilibrium mixture?

Short answer

Answer: The equilibrium composition of the product gases after combustion is approximately 0.49 mol H2, 0.51 mol H2O, and 0.245 mol O2. The heat transfer during the combustion process is approximately -145.77 kJ.

Step-by-step solution

Calculate initial mixture composition

We are given that the initial tank contains 1 mol of H2 and 0.5 mol of O2. The total number of moles is: n_total_initial = n_H2_initial + n_O2_initial = 1 + 0.5 = 1.5 mol

Relate initial and final number of moles of each species

Let x be the fraction of complete combustion of H2. Then, n_H2_final = 1 - x n_H2O_final = x n_O2_final = 0.5 - 0.5x The total number of moles after combustion, n_total_final, will then be: n_total_final = n_H2_final + n_H2O_final + n_O2_final = 1 - x + x + 0.5 - 0.5x = 1.5 - 0.5x

Use ideal gas law to express final pressure and temperature in terms of x

We can write the ideal gas law for the initial state (1 atm and 25°C) and the final state (5 atm and 2800 K): PV_initial = n_total_initial RT_initial PV_final = n_total_final RT_final Since the volume is constant, V_initial = V_final. Therefore, we can write: P_initial / T_initial = n_total_initial R / (n_total_final R) Substitute the given values (P_initial = 1 atm, T_initial = 25°C + 273.15 K, P_final = 5 atm, T_final = 2800 K) and the expression for n_total_final: 1 / (25 + 273.15) = 1.5 / [(1.5 - 0.5x) * 2800] Simplify and solve for x: 1 / 298.15 = 1.5 / [4200 - 1400x] x ≈ 0.51

Equilibrium composition of product gases

Now we substitute the value of x obtained in step 3 into the expressions we found in step 2 for the final number of moles of each species: n_H2_final ≈ 1 - 0.51 = 0.49 mol n_H2O_final ≈ 0.51 mol n_O2_final ≈ 0.5 - 0.5 * 0.51 = 0.245 mol

Determine the heat transfer during combustion

For this step, we will need the standard enthalpy of formation values for H2O, H2, and O2 at 25°C. We have: ΔHf_H2O = -285.83 kJ/mol ΔHf_H2 = 0 kJ/mol (reference) ΔHf_O2 = 0 kJ/mol (reference) The heat transfer, Q, can be determined using the equation: Q = Σ n_final * ΔHf_final - Σ n_initial * ΔHf_initial Substitute the values obtained in step 4 and the enthalpy values: Q = [(0.245 * 0) + (0.49 * 0) + (0.51 * (-285.83))] - [(1 * 0) + (0.5 * 0)] Q ≈ -145.77 kJ

Realistic assumption of no OH presence

As there is no information suggesting that there should be any OH radicals in the equilibrium mixture, it is reasonable to assume no OH presence in this particular exercise. However, in a more detailed analysis, the possible presence or formation of OH radicals would need to be investigated based on the specific conditions and reactants involved. Final answers: (a) Equilibrium composition: H2 ≈ 0.49 mol, H2O ≈ 0.51 mol, O2 ≈ 0.245 mol (b) Amount of heat transfer: ≈ -145.77 kJ

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is a fundamental concept in chemistry and physics, connecting the pressure, volume, temperature, and amount of an ideal gas.
At a basic level, the ideal gas law is encapsulated in the equation:

\[ PV = nRT \]

where:

• \( P \) represents the pressure of the gas,
• \( V \) is the volume it occupies,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant, and
• \( T \) is the absolute temperature.

In the context of the exercise, applying the ideal gas law to both the initial and final states and factoring in that volume remains constant (\( V_{\text{initial}} = V_{\text{final}} \)), allows us to relate initial and final conditions and hence calculate the change in moles of gas during a chemical reaction, such as combustion. For the exercise to be improved upon, it's crucial that students understand each variable and how it relates to the others within the context of a closed system undergoing a reaction. Knowing that the number of moles may change due to the reaction and that \( R \) is constant is essential for solving for \( x \), our unknown representing the extent of the reaction. Simplifying the equation and solving for \( x \) encapsulates the practical application of the ideal gas law in finding equilibrium compositions of product gases in a chemical reaction.

Enthalpy of Formation

The enthalpy of formation, \( \Delta H_f \), measures the change in enthalpy when one mole of a compound is formed from its elements at standard conditions. It is a crucial concept in thermochemistry, serving as a reference to calculate the heat transfer during a reaction.

When dealing with combustion processes, we often reference the enthalpy of formation for each product and reactant. For example, in this exercise, we use the enthalpy of formation of water to determine the heat released. The equation used to calculate the heat transfer during a reaction is:

\[ Q = \Sigma n_{\text{final}} \times \Delta H_{f_{\text{final}}} - \Sigma n_{\text{initial}} \times \Delta H_{f_{\text{initial}}} \]

In essence, this equation represents the total change in enthalpy for a system, which in the case of an exothermic process such as combustion would be negative, indicating heat is released. For educational content, it's important to clarify the standard state conditions (298 K and 1 atm) and ensure that the students are comfortable converting units and understanding the significance of negative values for exothermic reactions and positive for endothermic ones. The exercise provides a strong foundation for understanding these concepts, but emphasizing how to interpret and use the enthalpy values enhances comprehension substantially.

Combustion Analysis

Combustion analysis involves determining the amounts of products formed when a substance combines with oxygen, a process that is central to energy generation and various chemical applications. This method typically involves measuring the masses of the reactants and products to deduce the composition of the original substance.

For the given exercise, students must consider the compositions of both the reactants (hydrogen and oxygen) and the products formed after the combustion reaction has reached equilibrium. The assumption of no intermediate species like OH being present simplifies the scenario into a stoichiometric problem where the balanced equation and conservation of mass help deduce the final amounts of \( H_2 \), \( O_2 \), and \( H_2O \).

As an educational improvement, it would be beneficial for students to understand that real combustion analysis includes considerations like incomplete combustion, limitations by equilibrium, and the possible presence of intermediate species. While these are not calculated in the textbook exercise, introducing these factors in a simplified way could prepare students for more complex scenarios. Additionally, tying back the role of the ideal gas law and the enthalpy of formation in interpreting the results of combustion analysis helps solidify the interconnectedness of these chemical concepts.