Question

The equilibrium constant for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ is \(K_{c}=1.3 \times 10^{-2}\) at \(1000 \mathrm{~K} .\) (a) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) .\) (b) At this tempera- ture does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr?

Short answer

The equilibrium constant for the reverse reaction is approximately 76.92. At this temperature, the equilibrium favors the formation of NO and Br2.

Step-by-step solution

(a) Calculate the equilibrium constant for the reverse reaction

We are given the equilibrium constant for the following reaction: \[2 \operatorname{NO}(g) + \operatorname{Br}_2(g) \rightleftharpoons 2 \operatorname{NOBr}(g) \quad \text{with} \, K_c = 1.3 \times 10^{-2}\] Now, we have to find the equilibrium constant for the reverse reaction: \[2 \operatorname{NOBr}(g) \rightleftharpoons 2 \operatorname{NO}(g) + \operatorname{Br}_2(g)\] The relationship between the equilibrium constants of the forward and reverse reactions is given by: \[K_{c(reverse)} = \frac{1}{K_{c(forward)}}\] So, we can plug in the values and find out the equilibrium constant for the reverse reaction: \[K_{c(reverse)} = \frac{1}{1.3 \times 10^{-2}}\]

Calculate the equilibrium constant for the reverse reaction:

Let's find the equilibrium constant for the reverse reaction by dividing 1 by the given equilibrium constant of the forward reaction. \[K_{c(reverse)} = \frac{1}{1.3 \times 10^{-2}} \approx 76.92 \] The equilibrium constant for the reverse reaction is approximately 76.92.

(b) Determine the favored side at this temperature

We are given that the equilibrium for the forward reaction is \(K_c=1.3 \times 10^{-2}\) and for the reverse reaction is \(K_{c(reverse)} \approx 76.92\). When comparing the equilibrium constants, we can observe that: - If \(K_c > 1\), the reaction is product-favored, meaning that more products are formed at equilibrium. - If \(K_c < 1\), the reaction is reactant-favored, meaning that more reactants are present at equilibrium. In our case, the given reaction has an equilibrium constant of \(K_c = 1.3 \times 10^{-2}\) which is smaller than 1, and thus the reaction is reactant-favored. The reverse reaction has an equilibrium constant of 76.92 which is greater than 1, and thus, it is product-favored. Since the reverse reaction is product-favored, it means that more \(\mathrm{NO}\) and \(\mathrm{Br}_2\) are formed at equilibrium at this temperature.

Key concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant, denoted as \(K_c\), is a vital concept. It quantifies the ratio of concentrations of products to reactants at equilibrium for a given reaction. For a reaction given by \[aA + bB \rightleftharpoons cC + dD\] the equilibrium constant expression is: \[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\] where \([A]\), \([B]\), \([C]\), \([D]\) are the molar concentrations of the reactants and products. A high \(K_c\) value indicates a greater concentration of products at equilibrium, signifying a product-favored process. Conversely, a low \(K_c\) signals a predominance of reactants. The equilibrium constant varies with temperature, necessitating its evaluation at specific conditions.

Reaction Quotient

While the equilibrium constant \(K_c\) pertains to when a reaction is at equilibrium, the reaction quotient \(Q_c\) serves a similar function for non-equilibrium conditions. By comparing \(Q_c\) to \(K_c\), we can predict the direction a reaction will proceed to achieve equilibrium. For a reaction, \[aA + bB \rightleftharpoons cC + dD\] the reaction quotient \(Q_c\) is given by: \[Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\] When

• \(Q_c < K_c\): The reaction moves towards products, shifting right.
• \(Q_c > K_c\): The reaction shifts towards reactants, moving left.
• \(Q_c = K_c\): The system is at equilibrium.

This comparison offers a snapshot of progress and direction, helping chemists to manipulate conditions for desired objectives.

Product-favored Reaction

A product-favored reaction occurs when the equilibrium constant \(K_c\) for the chemical equation is greater than 1. This implies that at equilibrium, the concentration of products is higher than that of the reactants. Product-favored reactions are also referred to as product dominant, as more products form naturally when the system reaches equilibrium. Understanding whether a reaction is product-favored helps predict the outcome and manage reactions in industrial and laboratory settings. These reactions typically involve exothermic processes or favorable entropy changes that drive the system "forward" towards products.

Reverse Reaction

The reverse reaction is the reaction that proceeds in the opposite direction of the forward reaction. It's important in achieving a full understanding of a reaction's dynamics. When considering equilibrium, each forward reaction has a corresponding reverse reaction. The equilibrium constant for the reverse reaction \(K_{c(reverse)}\) can be found using the relationship: \[K_{c(reverse)} = \frac{1}{K_{c(forward)} }\] If the forward reaction is reactant-favored (\(K_c < 1\)), the reverse will generally be product-favored (\(K_{c(reverse)} > 1\)), indicating the preferred formation of the initial reactants when starting with products. In practical setups, considering reverse reactions can help in processes like purification or recycling of unreacted materials to enhance yield.