Problem 81 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2}+\mathrm{O}_{2}\) \(\mathrm{K}_{c_{1}}=2.5 \times 10^{30}\) \(\mathrm{NO}+\frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}\) \(\mathrm{K}_{c_{2}}=1.6\) find \(K_{c}\) for the reaction given below \(\frac{1}{2} \mathrm{~N}_{2}+\frac{1}{2} \mathrm{O}_{2}+\frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}\) (a) \(1.01 \times 10^{-15}\) (b) \(2.02 \times 10^{-15}\) (c) \(1.01 \times 10^{30}\) (d) \(2.02 \times 10^{15}\)

Step-by-Step Solution

Answer

The value of \(K_c\) for the given reaction is \(1.01 \times 10^{-15}\).

Step 1: Analyze Given Equations

The first reaction is \(2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2}+\mathrm{O}_{2}\) with an equilibrium constant \(K_{c_{1}} = 2.5 \times 10^{30}\). The second reaction is \(\mathrm{NO} + \frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}\) with \(K_{c_{2}} = 1.6\). We need to find the equilibrium constant for \(\frac{1}{2} \mathrm{~N}_{2} + \frac{1}{2} \mathrm{O}_{2} + \frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}\).

Step 2: Transform First Reaction (Divide by 2)

Divide both sides of the first reaction \(2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2} + \mathrm{O}_{2}\) by 2 to get \(\mathrm{NO} \rightleftharpoons \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2}\). The new equilibrium constant \(K'\) for this modified reaction is \(K' = \sqrt{K_{c_{1}}} = \sqrt{2.5 \times 10^{30}}\).

Step 3: Invert the Reaction

Since the desired reaction is the reverse of the modified reaction \(\mathrm{NO} \rightleftharpoons \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2}\), invert this reaction. The equilibrium constant for the reverse reaction \(\frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}\) is \(\frac{1}{\sqrt{K_{c_{1}}}}\).

Step 4: Combine Reactions

Now combine the result from Step 3 and the second given equation \(\mathrm{NO} + \frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}\). The overall equation becomes \(\frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2} + \frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}\).

Step 5: Calculate Overall Equilibrium Constant

The equilibrium constant for the overall reaction is the product of the constants for its component reactions: \[ K_{c} = \left( \frac{1}{\sqrt{K_{c_{1}}}} \right) \cdot K_{c_{2}} = \left( \frac{1}{\sqrt{2.5 \times 10^{30}}} \right) \cdot 1.6 \]. Solving gives \[ K_{c} = 1.01 \times 10^{-15} \].

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating state in a chemical reaction where the forward and reverse reactions occur at the same rate. This means that the concentrations of reactants and products remain constant over time. When we talk about a reaction reaching equilibrium, it doesn't imply that the reactants and products are in equal concentrations; rather, it signifies a state of balance. In our original exercise, for instance, the two reactions involve nitric oxide and nitrogen dioxide, which reach a condition where their rates of formation and decomposition are equal.

At equilibrium, the dynamic nature of the process doesn't stop—reactions continue to happen, but because they occur at the same rate in both directions, there is no net change in concentrations.

To quantify this state of balance, we use the equilibrium constant, denoted usually by the symbol \( K_c \) for concentration-based systems. The value of \( K_c \) indicates what's favored: a large \( K_c \) suggests products are favored, whereas a small \( K_c \) indicates reactants are favored. In the context of the given problem, the extremely large \( K_{c_1} \) value of \( 2.5 \times 10^{30} \) shows a strong favoring towards the product formation in the first reaction.

Reaction Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It's a crucial skill that allows chemists to predict the quantities of materials consumed and produced in a given reaction. In our problem, we see stoichiometry play a role when we transform and combine equations to form new reactions.

For example, the original reaction \( 2\text{NO} \rightleftharpoons \text{N}_2 + \text{O}_2 \) has coefficients that enable us to deduce the amounts of reactants needed to produce a specific quantity of products. By understanding stoichiometry, we are also able to manipulate reactions to produce new equilibrium equations, such as when we divided the entire first reaction by 2.

This operation is significant as it directly impacts the equilibrium constant. When a balanced chemical equation is multiplied or divided by a factor, it affects the equilibrium constant by raising it to the power of that factor or taking the corresponding root. This is evident in Step 2 where dividing by 2 leads us to determine \( K' = \sqrt{K_{c_1}} \). Instructors often emphasize the importance of keeping reactions balanced for this reason—balance maintains the integrity of stoichiometric calculations.

Equilibrium Constant Manipulations

Manipulating equilibrium constants (\( K_c \)) is a key technique to solve reactions that are combinations, reversals, or multiples of others. This skill is particularly useful in forming new reactions like the one in our exercise.

To solve the given problem, you first determine how the coefficients affect \( K_c \). With our first reaction tweaked by dividing through by two, its equilibrium constant has to be adjusted by taking the square root of the original \( K_{c_1} \). Essentially, this step is fundamental whenever any reaction is scaled by a particular factor.

Things become particularly interesting in Step 3, where you reverse a reaction. Reversing a chemical equation requires you to take the reciprocal of the equilibrium constant. Therefore, when we invert the reaction to create \( \frac{1}{2} \text{N}_2 + \frac{1}{2} \text{O}_2 \rightleftharpoons \text{NO} \), the constant becomes \( \frac{1}{\sqrt{K_{c_1}}} \). Lastly, when reactions are combined, the new equilibrium constant is the product of the constants of the individual reactions. This manipulation finally gives the desired \( K_c \)—a tiny value suggesting that, at equilibrium, the concentration of reactants is considerably higher than that of the product.

This entire exercise underscores the power of understanding equilibrium constant manipulations. It's a core method for calculating key reaction properties, even when new reaction pathways are introduced.