Question

The equilibrium constant for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{NOBr}(g) $$ is \(K_{c}=1.3 \times 10^{-2}\) at \(1000 \mathrm{~K}\). (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor \(\mathrm{NOBr}\) ? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\). (c) Calculate \(K_{c}\) for \(\operatorname{NOBr}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\).

Short answer

(a) The equilibrium favors the reactants, NO and Br₂, since \(K_c = 1.3 \times 10^{-2}\) which is less than 1. (b) The equilibrium constant for the reverse reaction is \(K_c' = 76.92\). (c) The equilibrium constant for the rewritten reaction is \(K_c'' = 0.114\).

Step-by-step solution

(a) Determining the favored side of the reaction

To find out whether the equilibrium favors reactants NO and Br₂ or the product NOBr, we need to take a look at the given equilibrium constant Kc. If Kc is greater than 1, it means the equilibrium favors the products, and if Kc is less than 1, it means the equilibrium favors the reactants. In this case, the given equilibrium constant is \(K_c = 1.3 \times 10^{-2}\), which is less than 1. So, the equilibrium favors the reactants, NO and Br₂.

(b) Calculating the equilibrium constant for the reverse reaction

The reverse reaction can be written as follows: $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Br}_2(g) $$ To find the equilibrium constant for the reverse reaction, we simply need to take the inverse of the given equilibrium constant for the initial reaction. In this case, the initial equilibrium constant is \(K_c = 1.3 \times 10^{-2}\). Therefore, the equilibrium constant for the reverse reaction is: $$ K_c' = \frac{1}{K_c} = \frac{1}{1.3 \times 10^{-2}} = 76.92 $$

(c) Calculating the equilibrium constant for the rewritten reaction

The rewritten reaction can be expressed as follows: $$ \mathrm{NOBr}(g) \longrightarrow \mathrm{NO}(g) + \frac{1}{2} \mathrm{Br}_2(g) $$ To find the equilibrium constant for this rewritten reaction, we need to take the square root of the given equilibrium constant. The reason for taking the square root is because the stoichiometric coefficients of the initial reaction are doubled in this new reaction. In this case, the initial equilibrium constant is \(K_c = 1.3 \times 10^{-2}\). Therefore, the equilibrium constant for the rewritten reaction is: $$ K_c'' = \sqrt{K_c} = \sqrt{1.3 \times 10^{-2}} = 0.114 $$

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs in a chemical reaction when the rates of the forward and reverse reactions are equal, leading to a constant concentration of the reactants and products over time. It's a dynamic state, not static; the reactions continue to occur, but with no net change because they balance each other out. Equilibrium does not mean that the reactants and products are in equal concentrations, but rather that their concentrations have stabilized in a ratio that will remain constant over time, depending on the conditions of the system.

For the given exercise, the equilibrium constant, represented as \(K_c\), is a numerical value that provides information on the concentration ratio of the products to the reactants at equilibrium. When \(K_c\) is less than one, like \(1.3 \times 10^{-2}\) in our example, it indicates that the reactants, in this case, NO(g) and Br₂(g), are favored at equilibrium at 1000 K. This key insight helps students predict the direction in which a given reaction mixture will proceed to reach equilibrium.

Le Chatelier's Principle

Le Chatelier's principle is a valuable tool in predicting the behavior of a chemical system at equilibrium when it's subjected to changes. According to this principle, if a system at equilibrium experiences a change in concentration, temperature, volume, or pressure, the system will adjust itself to counteract the effect of the applied change and a new equilibrium will be established.

For instance, if there's an increase in the concentration of one of the reactants, the equilibrium will shift to favor the formation of products to reduce the added reactants. Conversely, an increase in product concentration would see the system favor the formation of reactants. In the case of the exercise, if conditions were to change, applying Le Chatelier's principle would help in determining the new equilibrium concentrations.

Reaction Quotient

The reaction quotient, \(Q_c\), is a measure that indicates the direction in which a reaction mixture will proceed to reach equilibrium. It's calculated using the same expression as the equilibrium constant \(K_c\), but with the current concentrations of the reactants and products, not just those at equilibrium.

Comparing \(Q_c\) to the equilibrium constant \(K_c\) helps predict the direction of the reaction:

• If \(Q_c < K_c\), the reaction will proceed forward to produce more products.
• If \(Q_c > K_c\), the reaction will proceed in the reverse direction to produce more reactants.
• If \(Q_c = K_c\), the reaction is at equilibrium and no net change is expected.

This concept is integral in the step-by-step solutions provided to students. By understanding it, they can better grasp the dynamics of the equilibrium process and calculate the direction and extent of a reaction's shift when it's not at equilibrium.