Question

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; K_{p}=5.0 \times 10^{12}\) (b) \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) ; K_{c}=5.8 \times 10^{-18}\)

Short answer

The short answer is: (a) The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) lies to the right, favoring the formation of products, as $$K_p = 5.0 \times 10^{12}$$ is significantly greater than 1. (b) The reaction \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) lies to the left, favoring the formation of reactants, as $$K_c = 5.8 \times 10^{-18}$$ is significantly less than 1.

Step-by-step solution

Reaction (a)

Given the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) with the equilibrium constant $$K_p = 5.0 \times 10^{12}$$. As $$K_p$$ is significantly greater than 1, we can say that this reaction lies to the right, favoring the formation of products \(($$\mathrm{NO}_{2}$$) over reactants (\)\mathrm{NO}\( and \)\mathrm{O}_{2}$).

Reaction (b)

For the reaction \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\), the given equilibrium constant is $$K_c = 5.8 \times 10^{-18}$$. Since $$K_c$$ is significantly less than 1, this reaction lies to the left, favoring the formation of reactants (\(\mathrm{HBr}\)) over the products (\(\mathrm{H}_{2}\) and \(\mathrm{Br}_{2}\)). The conclusions are: (a) The reaction lies to the right, favoring the formation of products. (b) The reaction lies to the left, favoring the formation of reactants.

Key concepts

Equilibrium Constant

The equilibrium constant is a critical concept in understanding chemical equilibrium. It tells us the extent to which a reaction proceeds and helps predict the relative concentrations of reactants and products at equilibrium.
In a balanced chemical reaction, the equilibrium constant is represented by either \( K_c \) or \( K_p \). For reactions using concentrations, \( K_c \) is used, while partial pressures use \( K_p \). Each constant offers valuable insights into the reaction's dynamics.
If the equilibrium constant is much larger than 1, it indicates that the reaction favors the formation of products. This means that at equilibrium, the concentration of products is greater than that of the reactants. Conversely, if the constant is much smaller than 1, it implies that the equilibrium favors the reactants.
Key points to remember include:

• \( K_c \) and \( K_p \) relate to concentrations and pressures, respectively.
• A large equilibrium constant means more products at equilibrium.
• A small equilibrium constant signifies a preference for reactants.
• The equilibrium position represents the ratio of product to reactant concentrations when the reaction is at rest.

Reaction Direction

The direction in which a chemical reaction proceeds towards equilibrium is crucial for understanding the behavior of the reaction under different conditions. By analyzing the equilibrium constant, we can determine whether a reaction moves to the right (favoring products) or to the left (favoring reactants).
In the exercise given, the reaction involving \( 2 \mathrm{NO}(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \) has an equilibrium constant \( K_p = 5.0 \times 10^{12} \). This large value indicates that the reaction proceeds to the right, heavily favoring products. In contrast, the reaction \( 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_2(g) + \mathrm{Br}_2(g) \) with \( K_c = 5.8 \times 10^{-18} \) goes to the left, favoring reactants.
The direction can be easily understood with these pointers:

• Large \( K \) values indicate a rightward direction with more products formed.
• Small \( K \) values suggest a leftward direction and a predominance of reactants.
• Reaction direction is essential for understanding chemical processes and predicting outcomes.

Product Formation

Product formation in a chemical reaction is largely influenced by the reaction's equilibrium state. When a reaction has a large equilibrium constant, it means that, at equilibrium, most of the reactant species have turned into products.
Taking from our exercise, in reaction (a), \( 2 \mathrm{NO}(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \), the high value of \( K_p \) reflects a large tendency to form \( \mathrm{NO}_2 \), the product. This is because the enormous \( K \) value shifts the balance heavily towards products.
Meanwhile, in reaction (b), the formation of products \( \mathrm{H}_2 \) and \( \mathrm{Br}_2 \) is less favored as \( K_c \) is very small. This indicates that under equilibrium, \( \mathrm{HBr} \) remains mostly unreacted, preferring to stay as a reactant.
Important points to recall on product formation:

• Product formation dominance relies on a high equilibrium constant.
• Small \( K \) values demonstrate limited product formation.
• Understanding product formation helps in optimizing reaction conditions for desired outputs.