Question

\mathrm{~A} 0.10-\mathrm{mol}\( sample of \)\mathrm{H}_{2}(\mathrm{~g})\( and a \)0.10-\mathrm{mol}\( sample of \)\mathrm{Br}_{2}(\mathrm{~g})\( are placed into a \)2.0\( - L container. The reaction \)\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons\( \)2 \mathrm{HBr}(\mathrm{g})\( is then allowed to come to equilibrium. A \)0.20\(-mol sample of HBr is placed into a second 2.0-L sealed container at the same temperature and allowed to reach equilibrium with \)\mathrm{H}_{2}\( and \)\mathrm{Br}_{2}\(. Which of the following will be different in the two containers at equilibrium? Which will be the same? (a) Amount of \)\mathrm{Br}_{2}\(; (b) concentration of \)\mathrm{H}_{2}\(; (c) the ratio \)[\mathrm{HBr}] /\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]\(; (d) the ratio \)[\mathrm{HBr}] /\left[\mathrm{Br}_{2}\right]\(; (e) the ratio \)\left[\mathrm{HBr}^{2} /\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]\right.$; (f) the total pressure in the container. Explain each of your answers.

Short answer

Different in the two containers at equilibrium: (a) Amount of \(Br_2\), (b) concentration of \(H_2\), (f) total pressure. The same in both containers at equilibrium: (c) the ratio \([HBr]/[H_2][Br_2]\), (d) the ratio \([HBr]/[Br_2]\), (e) the ratio \([HBr]^2/[H_2][Br_2]\).

Step-by-step solution

Identify the Equilibrium Expression

For the reaction \(H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)\), write the equilibrium expression. The equilibrium constant \(K_c\) is expressed in terms of the concentrations of the products and reactants raised to the power of their stoichiometric coefficients: \(K_c = \frac{[HBr]^2}{[H_2][Br_2]}\).

Analyze the First Container

Determine the changes in concentration from initial to equilibrium for the first container. Initial concentrations are \([H_2] = [Br_2] = 0.10\,mol/L\) and \([HBr] = 0\,mol/L\). At equilibrium, some \(H_2\) and \(Br_2\) would have reacted to form \(HBr\). Assume \(x\, mol/L\) of \(H_2\) and \(Br_2\) react to form \(2x\, mol/L\) of \(HBr\).

Analyze the Second Container

Do the same for the second container, where initial concentrations are \([HBr] = 0.20\,mol/L\) and \([H_2] = [Br_2] = 0\,mol/L\). Similarly, assume \(y\, mol/L\) of \(HBr\) dissociates to form \(y\, mol/L\) of \(H_2\) and \(Br_2\) each.

Compare Equilibrium States

Establish that at equilibrium the value of \(K_c\) remains constant for a given temperature. This implies the ratios \(\frac{[HBr]^2}{[H_2][Br_2]}\) will be the same in both containers regardless of the initial conditions.

Examine Differences (a) and (b)

The amount of \(Br_2\) and the concentration of \(H_2\) will generally be different in the two containers at equilibrium due to different starting conditions (more \(HBr\) in the second container initially suggests less \(Br_2\) and \(H_2\) at equilibrium there than in the first).

Examine Ratios (c), (d), and (e)

The ratio \([HBr]/[H_2][Br_2]\) (part c), \([HBr]/[Br_2]\) (part d), and \([HBr]^2/[H_2][Br_2]\) (part e) will all be the same in both containers because they are defined by the equilibrium constant \(K_c\) which depends only on temperature.

Consider Total Pressure (f)

The total pressure in each container will depend on the total number of moles of gas present. Since the amount of gas at equilibrium could be different, the total pressures might also be different.

Key concepts

Equilibrium Constant

The equilibrium constant, represented as K_c, plays a pivotal role in the study of chemical reactions that are reversible—a common scenario in chemistry where the reactants and products interconvert. This constant is deeply rooted in the law of mass action, which states that for a reversible reaction at equilibrium and at a constant temperature, a certain ratio of the concentration of products to reactants raised to their stoichiometric coefficients remains constant.

In the example provided, the reaction in question is H₂(g) + Br₂(g) \rightleftharpoons 2HBr(g). The equilibrium constant for this particular reaction is calculated with the formula \(K_c = \frac{[HBr]^2}{[H_2][Br_2]}\). What's important to note is that the equilibrium constant is only influenced by temperature. Changes in concentrations or pressure do not affect the value of K_c, although they will shift the position of equilibrium to re-establish the constant ratio given by K_c.

In practical terms, understanding K_c enables chemists to predict the extent of the reaction and the concentrations of each species at equilibrium. If K_c is much greater than 1, the reaction favors products; whereas if it is much less than 1, reactants are favored.

Reaction Quotient

The reaction quotient, noted as Q, is the cousin of the equilibrium constant and is used when the system is not at equilibrium. It has the same form as the equilibrium expression but uses the initial concentrations of the reactants and products, before the system reaches equilibrium. For instance, in the provided exercise, if we wanted to know the direction in which the reaction will proceed initially, we'd calculate Q using the initial concentrations of H₂, Br₂, and HBr.

The value of Q can tell us if the reaction will move forward or backward to reach equilibrium. If Q < K_c, the forward reaction is favored, meaning more products will be formed, and if Q > K_c, the reaction will proceed in the reverse direction, creating more reactants. Comparing Q and K_c provides insight into the initial shift the reaction must make to achieve equilibrium. This concept reinforces why the ratios mentioned in the exercise—(c), (d), and (e)—will be equal in both containers when they are at equilibrium, regardless of their initial compositions.

Le Chatelier's Principle

Le Chatelier's principle is a cornerstone in chemical equilibrium, providing a qualitative tool to predict how a system at equilibrium responds to changes in concentration, pressure, or temperature. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change. Now let's apply it to the exercise scenario.

If the concentration of one of the reactants is increased, the principle suggests that the equilibrium will shift to produce more products, reducing the concentration of added reactants. Conversely, if the concentration of a product increases, the reaction will favor the formation of reactants. For example, in step 2 and step 3 of the solution, changing the initial amounts of H₂, Br₂, and HBr leads to different equilibrium concentrations in each container. This reflects Le Chatelier's principle in action.

The principle also explains how changes in total pressure (part f) affect the system. Since an increase in pressure favors the side with fewer moles of gas, and a decrease has the opposite effect, the total pressure in the two containers might differ at equilibrium due to the different gas moles present.