Question

\(A B_{3}(g)\) is dissociates as \(A B_{3}(g) \rightleftharpoons A B_{2}(g)+\frac{1}{2} B_{2}(g)\) When the initial pressure of \(A B_{2}\) is 800 torr and the total pressure developed at equilibrium is 900 torr. What fraction of \(A B_{3}(g)\) is dissociated? (a) \(10 \%\) (b) \(20 \%\) (c) \(25 \%\) (d) \(30 \%\)

Short answer

The fraction dissociated is 20%.

Step-by-step solution

Determine Initial Pressures

Set up the initial pressures for the reactant and products. Initially, we have AB3 at 800 torr and no AB2 or B2. So, P(AB3_initial) = 800 torr, P(AB2_initial) = 0 torr, and P(B2_initial) = 0 torr.

Determine Total Pressure Change

Calculate the change in pressure using the total pressure at equilibrium. Since the total pressure at equilibrium is 900 torr, the pressure change is 900 torr - 800 torr = 100 torr.

Find the Equilibrium Pressures

Write the equilibrium expression for the reaction. If x is the change in pressure for AB3, then the partial pressure of AB2 at equilibrium will be x since it is produced, and the pressure of B2 will be x/2 because of the stoichiometry of the reaction. This means the total pressure increase is x + x/2.

Solve for x

Now sum the partial pressures and equate to the overall pressure change. x + x/2 = 100 torr. This gives us a single variable equation to solve for x.

Calculate Fraction Dissociated

Once we know the value of x (change in pressure for AB3), we can find the fraction dissociated by dividing x by the initial pressure of AB3 and multiplying by 100 to get a percentage.

Key concepts

Understanding Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate, leading to a constant concentration of reactants and products. It's a dynamic process - even though the concentrations don't change over time, the molecules are still reacting with one another.

In the context of the textbook exercise, the dissociation of AB3 into AB2 and B2 represents such an equilibrium. Initially, AB3 starts to dissociate, but as AB2 and B2 form, some of them recombine to form AB3 again. At equilibrium, the rate at which AB3 dissociates equals the rate at which AB2 and B2 recombine, resulting in no net change in their concentrations, or in this case, their partial pressures.

Understanding the concept of equilibrium in chemical reactions is crucial for correctly solving many problems in chemistry. The ability to conceptualize the dynamic but stable nature of equilibrium is often the difference between fully grasping chemical processes and simply memorizing reactions.

Dissociation of Gases

Dissociation in chemistry generally refers to the process by which molecules break down into smaller components. For gases, dissociation occurs when the molecular bonds within a gaseous compound break under conditions such as temperature or pressure changes.

In our exercise, AB3 gas is dissociating to form AB2 and half a molecule of B2. This is important because the fraction of AB3 that dissociates at a given temperature and pressure determines the equilibrium composition of the reaction mixture. The stoichiometry - that is, the ratio in which AB3 dissociates to form AB2 and B2 - is vital for calculating the changes in partial pressures of the gases involved.

Remembering Stoichiometry

Stoichiometry influences the entire equilibrium system because it dictates how much each substance is formed or used up. In this problem, for every unit of AB3 that dissociates, we get one unit of AB2 and a half unit of B2 gas. This stoichiometric relationship is essential in calculating the partial pressures at equilibrium.

Partial Pressures and Equilibrium

Partial pressure is the pressure that a gas in a mixture would exert if it were alone in the container. It's an important concept in understanding gaseous equilibria because it helps us relate the amounts of gases present to the total pressure.

In our problem, we're dealing with a system where AB3 gas dissociates and creates a certain total pressure at equilibrium. To figure out how much has dissociated, we consider the partial pressures of the products formed. The total pressure in the system is the sum of the individual partial pressures of AB3, AB2, and B2.

Calculating Equilibrium Partial Pressures

Through the given total pressure at equilibrium and the stoichiometry of the reaction, we can deduce the partial pressures of the reaction products. Knowing that the pressure increase comes exclusively from the products of the dissociation, we can set up and solve equations that reflect the proportional relationships between the reactants and products, incorporating the change in pressure due to the dissociation process.