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 temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr?

Short answer

(a) The equilibrium constant for the reverse reaction is \(K_{c,reverse} \approx 76.92\). (b) At this temperature, the equilibrium favors the reactants, NO and Br2, as \(K_{c,forward} < 1\).

Step-by-step solution

Part (a): Calculate \(K_c\) for reverse reaction

To find the equilibrium constant for the reverse reaction, we need to use the following relationship: \(K_{c,reverse} = \frac{1}{K_{c,forward}}\) Here, the forward reaction's equilibrium constant is given as \(K_{c,forward}=1.3 \times 10^{-2}\). Now, let's calculate the equilibrium constant for the reverse reaction: \(K_{c,reverse} = \frac{1}{1.3 \times 10^{-2}}\)

Part (a): Result

By solving the equation above, we get, \(K_{c,reverse} \approx 76.92\)

Part (b): Determine the favored side of the reaction

Based on the value of \(K_{c}\), we can determine which side of the reaction is favored when the system reaches equilibrium. If \(K_{c} > 1\), then the products are favored, and if \(K_{c}<1\), then the reactants are favored. In the given reaction, we have, \(K_{c,forward} = 1.3 \times 10^{-2}\) Since \(K_{c,forward} < 1\), the equilibrium favors the reactants which are NO and Br2. The value of \(K_{c,reverse}\) is not required for this analysis as it already indicates a greater affinity towards products of the reverse reaction, which are also reactants of the forward reaction.

Key concepts

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), is a crucial concept in chemical equilibrium. It provides a quantitative measure of the concentrations of products and reactants at equilibrium for a given chemical reaction. This constant is specific to a particular reaction at a specific temperature. The equilibrium constant expression depends on the stoichiometry of the balanced chemical equation and is calculated using the equilibrium concentrations of the reactants and products. For the general reaction \[ aA + bB \rightleftharpoons cC + dD \]\(K_c\) is expressed as:\[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]where \([A], [B], [C], [D]\) are the molar concentrations of the reactants and products.
Another important aspect of \(K_c\) is its reciprocity in reversible reactions. If we know \(K_c\) of a reaction in one direction, the \(K_c\) for the reverse reaction is simply the reciprocal. For instance, if for the forward reaction \(K_c = 1.3 \times 10^{-2}\), the equilibrium constant for the reverse reaction will be:\[K_{c,reverse} = \frac{1}{1.3 \times 10^{-2}} \approx 76.92\] Understanding this principle helps in evaluating reaction dynamics under different conditions.

Reversible Reactions

Reversible reactions are those chemical reactions where the conversion of reactants to products and the conversion of products back to reactants occur simultaneously. Over time, the system reaches a state of equilibrium where the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products remain constant.
These reactions are denoted by a double arrow \(\rightleftharpoons\), indicating that both the forward and reverse processes are significant. Reversible reactions are common in chemical systems and are influenced by factors such as temperature, pressure, and concentration.

• The direction in which the reaction initially moves depends on the initial concentrations of the reactants and products and the reaction's equilibrium constant.
• Changes in these conditions can shift the equilibrium position, either favoring the formation of more products or more reactants.

This dual nature ensures that reversible reactions are dynamic and can adapt to changes within their environment, a defining feature that distinguishes them from irreversible reactions.

Reaction Favorability

The concept of reaction favorability is pivotal in understanding which direction a reaction will predominantly proceed to achieve equilibrium. It is closely linked to the value of the equilibrium constant \(K_c\). Whether the reaction favors the formation of products or the retention of reactants at equilibrium can be evaluated by the magnitude of \(K_c\):

• If \(K_c > 1\), products are favored, meaning at equilibrium, the concentration of products is greater.
• If \(K_c < 1\), reactants are favored, indicating a higher concentration of reactants at equilibrium.

In the given problem, the forward reaction has \(K_c = 1.3 \times 10^{-2}\), which is much less than 1. This implies that at equilibrium, the reactants (NO and \(Br_2\)) are favored over the product \(NOBr\). Thus, most of the mixture will comprise \(NO\) and \(Br_2\), indicating a system balance skewed towards the reactants when temperature conditions are maintained at 1000 K.