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) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) .\) (c) Calculate \(K_{c}\) for \(\mathrm{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\).

Short answer

(a) At this temperature, the equilibrium favors NO and Br2 formation, as the equilibrium constant is less than 1. (b) The equilibrium constant for the reverse reaction, \(K_c^R\), is 76.92. (c) The equilibrium constant for the half reaction, \(K_c^{1/2}\), is 0.114.

Step-by-step solution

Determine favored side of the reaction

Given that equilibrium constant for the main reaction is \(K_c = 1.3 \times 10^{-2}\). Since \(K_c\) is less than 1, the equilibrium favors the reactant side, i.e., NO and Br2. Answer: Equilibrium favors NO and Br2 formation. (b) Calculate the equilibrium constant for the reverse reaction

Calculate equilibrium constant for the reverse reaction

We know that the equilibrium constant for the reverse reaction is the reciprocal of the forward reaction's equilibrium constant. So, for the reaction \(2 \text{NOBr} (g) \rightleftharpoons 2 \text{NO} (g) + \text{Br}_2 (g)\): \[ K_c^R = \frac{1}{K_c} = \frac{1}{1.3 \times 10^{-2}} = 76.92 \] Answer: The equilibrium constant for the reverse reaction, \(K_c^R\) is 76.92. (c) Calculate the equilibrium constant for the half reaction of NOBr dissociation

Calculate equilibrium constant for half reaction

The equilibrium constant for a half reaction is the square root of the full reaction's equilibrium constant. So, for the half reaction \(\text{NOBr} (g) \rightleftharpoons \text{NO} (g) + \frac{1}{2} \text{Br}_2 (g)\), we will find the square root of the given equilibrium constant: \[ K_c^{1/2} = \sqrt{1.3 \times 10^{-2}} = 0.114 \] Answer: The equilibrium constant for the half reaction, \(K_c^{1/2}\) is 0.114.

Key concepts

Equilibrium Constant

The equilibrium constant, often represented as \(K_c\), is a crucial number in chemistry that tells us about the ratio of the concentrations of products and reactants in a reaction at equilibrium. When a reversible chemical reaction reaches equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products remain constant.
For a general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant \(K_c\) is given by:\[K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]This formula shows how the concentrations of the chemical species at equilibrium determine the value of \(K_c\).

• If \(K_c \gt 1\), the equilibrium position favors the formation of products.
• If \(K_c \lt 1\), it favors the formation of reactants.

In the original exercise, the \(K_c = 1.3 \times 10^{-2}\), indicating that the equilibrium favors the reactants, namely NO and \(Br_2\).

Reversible Reactions

Reversible reactions are chemical reactions where the conversion of reactants to products and the conversion of products back to reactants occur simultaneously. These reactions do not go to completion but instead reach a state where both reactants and products are present in constant concentrations - the equilibrium state.
One of the key characteristics of reversible reactions is that they can be represented with a double arrow, \(\rightleftharpoons\), highlighting their ability to proceed in both directions. An important aspect of reversible reactions is that altering the conditions, such as temperature or pressure, can shift the equilibrium, favoring either the forward or reverse reaction.
In our exercise, both original reactions involve NOBr converting into NO and \(Br_2\) and vice versa. Recognizing that these reactions take place concurrently and depend on various factors such as the equilibrium constant aids in understanding their behavior.

Reaction Quotient

The reaction quotient, denoted as \(Q_c\), looks similar to the equilibrium constant, but it can be calculated at any point during a reaction, not just at equilibrium. Just like \(K_c\), it is determined using the expression:\[Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]What makes \(Q_c\) especially useful is that it helps predict how the direction of the reaction will shift to reach equilibrium.

• If \(Q_c = K_c\), the system is at equilibrium.
• If \(Q_c \gt K_c\), the system has more products than equilibrium, and the reverse reaction is favored to reach equilibrium.
• If \(Q_c \lt K_c\), the system has more reactants, and the forward reaction is favored.

Understanding the reaction quotient can help predict which way a reaction will shift when it's not yet at equilibrium, making it a valuable tool for chemists looking to control reaction conditions.