Problem 48 - Introductory Chemistry Atoms First

Question

Consider the gas-state reaction \(2 \mathrm{~A}(g)+3 \mathrm{~B}(g) \rightleftarrows \mathrm{C}(g)+\mathrm{D}(g)\) (a) Write the equilibrium constant expression for the reaction. (b) Write the reaction in reverse. (c) Write the equilibrium constant expression for the reverse reaction that you wrote for part (b). (d) Compare your answers to \((\mathrm{a})\) and \((\mathrm{c})\). What conclusion can you draw from the comparison? (e) Suppose \(K_{\mathrm{eq}}\) for a reaction \(=10.0 .\) What will the value of \(K_{\text {eq }}\) be for the reverse reaction?

Step-by-Step Solution

Answer

(a) The equilibrium constant expression for the given reaction is: \(K = \frac{[C][D]}{[A]^2[B]^3}\) (b) The reverse reaction is: \(C(g) + D(g) \rightleftarrows 2A(g) + 3B(g)\) (c) The equilibrium constant expression for the reverse reaction is: \(K' = \frac{[A]^2 [B]^3}{[C][D]}\) (d) Since K and K' are reciprocals of each other, we can conclude that the equilibrium constant of the reverse reaction is the reciprocal of the equilibrium constant of the given reaction. (e) The value of the equilibrium constant for the reverse reaction is 0.1.

Step 1: Write the equilibrium constant expression for the given reaction

The reaction is given as: \[2A(g) + 3B(g) \rightleftarrows C(g) + D(g)\] For a general reaction: \[aA + bB \rightleftarrows cC + dD\] The equilibrium constant expression (K) can be written as: \[K = \frac{[C]^c [D]^d}{[A]^a [B]^b}\] For the given reaction, a = 2, b = 3, c = 1, and d = 1. So, the equilibrium constant expression for the given reaction is: \[K = \frac{[C][D]}{[A]^2[B]^3}\]

Step 2: Write the reverse reaction

The reverse reaction can be obtained by swapping the reactants and products of the given reaction: \[C(g) + D(g) \rightleftarrows 2A(g) + 3B(g)\]

Step 3: Write the equilibrium constant expression for the reverse reaction

For the reverse reaction, a = 1, b = 1, c = 2, and d = 3. So, the equilibrium constant expression for the reverse reaction is: \[K' = \frac{[A]^2 [B]^3}{[C][D]}\]

Step 4: Compare the expressions for equilibrium constants

When comparing the expressions for the equilibrium constants K and K', we can see that: \[K = \frac{[C][D]}{[A]^2[B]^3}\] \[K' = \frac{[A]^2 [B]^3}{[C][D]}\] Since K and K' are reciprocals of each other, we can conclude that the equilibrium constant of the reverse reaction is the reciprocal of the equilibrium constant of the given reaction.

Step 5: Find the value of the equilibrium constant for the reverse reaction

Given that the equilibrium constant for the forward reaction (K) is 10.0, we can find the equilibrium constant for the reverse reaction (K') by taking the reciprocal of K: \[K' = \frac{1}{K} = \frac{1}{10.0} = 0.1\] So, the value of the equilibrium constant for the reverse reaction is 0.1.

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs in reversible reactions when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. In the given exercise, where components A and B react to form C and D, equilibrium is achieved when these substances are converted between each other at a consistent rate.

At this point, although the reactants and products are still being produced and consumed, their concentrations remain constant. It's essential to understand that equilibrium does not mean the amounts of reactants and products are equal, but rather that their rates of formation are. The equilibrium constant expression quantitatively represents the ratio of product concentrations to reactant concentrations at equilibrium, raised to the power of their stoichiometric coefficients.

Reaction Quotient

The reaction quotient, Q, serves as a predictor to determine the direction in which a reaction will proceed to achieve equilibrium. It uses the same formula as the equilibrium constant expression, \(K\), but with current concentrations instead of equilibrium concentrations. When comparing Q and K:

• If \(Q < K\), the reaction will proceed forward to produce more products and reach equilibrium.
• If \(Q > K\), the reaction will shift backwards to produce more reactants and attain equilibrium.
• If \(Q = K\), the system is already at equilibrium and no shift in the reaction's direction will occur.

For students, utilizing Q can help predict the effects of changing concentrations on the position of equilibrium without having to wait for the reaction to actually reach equilibrium.

Le Chatelier's Principle

Le Chatelier's principle is a vital concept understanding how a system at equilibrium responds to external stresses. It states that if an external change, such as pressure, concentration, or temperature, is applied to a system at equilibrium, the system will adjust to counteract this change and re-establish equilibrium.

For example, adding more reactant to the system may cause the reaction to proceed in the forward direction to reduce the added reactants. Conversely, increasing the temperature in an exothermic reaction will favour the reverse reaction, as this would absorb the excess heat, maintaining the balance within the system. Learning to predict these shifts can dramatically improve problem-solving skills related to equilibrium reactions.

Reversible Reactions

Reversible reactions are those that can proceed in both the forward and reverse directions. They are often represented by a double-headed arrow (\(\rightleftarrows\)). In the provided exercise, the reaction between A, B, and C, D is reversible. This means that A and B can react to form C and D, but C and D can also react to form A and B.

Many chemical reactions are reversible, which can make achieving a complete reaction difficult. However, under certain conditions, such as high pressure or temperature, or with a catalyst, the reaction can be 'pushed' to favour one side, either towards the products or the reactants. Understanding reversible reactions is crucial for controlling chemical processes and for grasping the dynamic nature of chemical systems.