Problem 75 - Chemistry: Principles and Reactions

Question

Consider a vessel with a movable piston. A reaction takes place in the vessel at constant pressure and a temperature of \(200 \mathrm{~K}\). When reaction is complete, the pressure remains the same and the volume and temperature double. Which of the following balanced equations best describes the reaction? (a) \(\mathrm{A}+\mathrm{B}_{2} \longrightarrow \mathrm{AB}_{2}\) (b) \(\mathrm{A}_{2}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}\) (c) \(2 \mathrm{AB}+\mathrm{B}_{2} \longrightarrow 2 \mathrm{AB}_{2}\) (d) \(2 \mathrm{AB}_{2} \longrightarrow \mathrm{A}_{2}+2 \mathrm{~B}_{2}\)

Step-by-Step Solution

Answer

Question: Determine the balanced equation that best describes the reaction given the following information: The reaction takes place between gases at a constant pressure. The volume of the gaseous mixture doubles, and the absolute temperature of the mixture doubles as well. Answer: (d) \(2 \mathrm{AB}_{2} \longrightarrow \mathrm{A}_{2}+2 \mathrm{B}_{2}\)

Step 1: Formulate the relationships for initial and final amounts of moles in each reaction option

For each reaction option, we can establish an equation that relates the initial total moles (n_i) to the final total moles (n_f). (a) A + B_2 -> AB_2: n_i = n_A + n_B2 n_f = n_AB2 (b) A_2 + B_2 -> 2AB: n_i = n_A2 + n_B2 n_f = 2n_AB (c) 2AB + B_2 -> 2AB_2: n_i = 2n_AB + n_B2 n_f = 2n_AB2 (d) 2AB_2 -> A_2 + 2B_2: n_i = 2n_AB2 n_f = n_A2 + 2n_B2

Step 2: Identify the relationship between volume and moles for the final conditions

The final conditions are given as the volume doubling and the temperature doubling as well. From the ideal gas law, PV = nRT, we know the relationship of P, V, n, and T. Since pressure remains constant, we can focus on the relationship between n, and V, keeping the T doubling in mind: Initial: n_i * R * T_i = P * V_i Final: n_f * R * 2T_i = P * 2V_i Dividing the initial by final equations, we find: n_f = 2n_i

Step 3: Find the reaction with the correct relationship between initial and final moles

Now, let's compare the relationship n_f = 2n_i to the equations established in Step 1 to find the reaction best describing the situation: (a) 2(n_A + n_B2) = n_AB2 - This does not satisfy the condition. (b) 2(n_A2 + n_B2) = 2n_AB - Canceling a factor of 2 in both sides leaves n_A2 + n_B2 = n_AB, which does not satisfy the condition. (c) 2(2n_AB + n_B2) = 2n_AB2 - Canceling a factor of 2 in both sides leaves 2n_AB + n_B2 = n_AB2, which does not satisfy the condition. (d) 2(2n_AB2) = n_A2 + 2n_B2 - This satisfies the relationship n_f = 2n_i. So, the balanced equation that best describes the reaction is: (d) \(2 \mathrm{AB}_{2} \longrightarrow \mathrm{A}_{2}+2 \mathrm{B}_{2}\)

Key Concepts

Understanding the Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure (P), volume (V), temperature (T), and the amount of gas in moles (n) in a simple yet powerful relationship: PV = nRT. Here, R is the ideal gas constant. The law provides a means to predict one property of a gas if the other three are known.

The law is particularly useful in stoichiometry because it helps us understand how gases will behave under different conditions of temperature and pressure. For example, in our exercise, we keep pressure constant and scrutinize the effects of volume and temperature on the amount of gas, which allows us to deduce changes in moles. The ideal gas law underpins many chemical calculations and is essential for understanding gas behavior during reactions like the one we're examining.

In the exercise solution, the law is applied after determining the stoichiometric relationship between reactants and products, helping us to conclude that the volume and temperature doubling (while pressure remains constant) implies the moles of gas must double too, leading us to the correct equation that represents the reaction.

The Mole Concept in Stoichiometry

The mole concept is a bridge between the microscopic world of atoms and molecules, and the macroscopic world we can measure. One mole refers to Avogadro’s number (approximately 6.022 x 10²³) of particles, be they atoms, molecules, ions, or others. It becomes hugely valuable when dealing with chemical equations and reactions.

In stoichiometry, the mole concept allows us to calculate amounts of reactants needed or products formed. The exercise at hand required an understanding of how moles are involved in the balancing of chemical equations, by comparing the total mole count before and after the reaction. This comparison gives insights into the changes occurring in a system and whether the equation aligns with the physical changes such as volume and temperature alterations, as mentioned in the ideal gas law.

The step by step solution showcases the process of equating the moles of reactants and products to find the equation that fits the new conditions post-reaction. Through this approach, we engage the mole concept to unravel the right chemical reaction that matches the observed changes.

Chemical Equilibrium Clarified

Chemical equilibrium occurs in a reversible chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction, leading to constant concentrations of the reactants and products. It's dynamic because the reactions continue to occur, but there's no net change in concentration of each engaged species.

In relation to our textbook problem, understanding chemical equilibrium is necessary even though the equilibrium concept doesn't directly apply because we are considering a complete reaction. However, the notion of balancing equations—the core of stoichiometry—is similar to achieving equilibrium in the sense that both concepts involve proportions and ratios of reactants and products.

The exercise improvement advice emphasizes ensuring a thorough explanation of the equilibrium concepts, even if in the context of a problem that doesn't involve a reaction at equilibrium. It's important because it prepares students to think about the balance and relative quantities of chemicals in all reactions, which is the essence of stoichiometry and helpful when they encounter true equilibrium problems.