Problem 72 - A Complete Resource Book in Chemistry for JEE Main

Question

If equilibrium constants of reaction, \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\) is \(\mathrm{K}_{1}\), and \(1 / 2 \mathrm{~N}_{2}+1 / 2 \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}\) is \(\mathrm{K}_{2}\) then (a) \(\mathrm{K}_{1}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{1}=2 \mathrm{~K}_{2}\) (c) \(\mathrm{K}_{2}=\sqrt{\mathrm{K}_{1}}\) (d) \(\mathrm{K}_{1}=1 / 2 \mathrm{~K}_{2}\)

Step-by-Step Solution

Answer

The correct option is (c) \( \mathrm{K}_{2}=\sqrt{\mathrm{K}_{1}} \).

Step 1: Understand the Given Equations

We are given two equilibrium reactions: \( \mathrm{N}_2 + \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \) and \( \frac{1}{2} \mathrm{N}_2 + \frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{NO} \). The first reaction has an equilibrium constant \( K_1 \), while the second has \( K_2 \). Our task is to relate \( K_1 \) and \( K_2 \).

Step 2: Express Equilibrium Constants Algebraically

The equilibrium constant \( K_1 \) for the reaction \( \mathrm{N}_2 + \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \) is given as\[ \mathrm{K}_1 = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_2][\mathrm{O}_2]} \].The equilibrium constant \( K_2 \) for the reaction \( \frac{1}{2} \mathrm{N}_2 + \frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{NO} \) is given as\[ \mathrm{K}_2 = \frac{[\mathrm{NO}]}{([\mathrm{N}_2][\mathrm{O}_2])^{1/2}} \].

Step 3: Relate the Equilibrium Constants

The second reaction is essentially the first reaction divided by 2, i.e., each reactant and product is halved. The relation between these constants is that when the stoichiometry of a balanced equation is altered, specifically halved for all components, the new equilibrium constant is the square root of the original. Therefore, \( \mathrm{K}_2 = \sqrt{\mathrm{K}_1} \).

Step 4: Verify the Relation

We verify the relationship using the expressions for \( \mathrm{K}_1 \) and \( \mathrm{K}_2 \) derived earlier.From \( \mathrm{K}_2 = \frac{[\mathrm{NO}]}{([\mathrm{N}_2][\mathrm{O}_2])^{1/2}} \),\[ \mathrm{K}_2^2 = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_2][\mathrm{O}_2]} = \mathrm{K}_1 \].Thus, \( \mathrm{K}_2 = \sqrt{\mathrm{K}_1} \) is confirmed.

Key Concepts

Equilibrium Reaction

An equilibrium reaction occurs when the rate of the forward reaction equates to the rate of the reverse reaction. In this state, the concentrations of reactants and products remain constant over time. Consider the reaction: - \( \mathrm{N}_2 + \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \).- This represents a dynamic equilibrium where nitrogen and oxygen gases form nitrogen monoxide, and nitrogen monoxide decomposes back into nitrogen and oxygen. - Understanding this concept is crucial because it helps us calculate the equilibrium constants, which quantify the ratio of product concentrations to reactant concentrations at equilibrium. These constants are unique for every given temperature.- Equilibrium does not imply that the concentrations of reactants and products are equal, but rather that their ratios have stabilized according to the equilibrium constant at that particular temperature.

Stoichiometry

Stoichiometry pertains to the quantitative relationships between the amounts of reactants and products involved in a chemical reaction. It relies heavily on balanced chemical equations to give a clear representation of these relationships.- For the reaction \( \mathrm{N}_2 + \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \), the stoichiometry tells us that one molecule of nitrogen reacts with one molecule of oxygen to yield two molecules of nitrogen monoxide.- Another way to express this is through coefficients in a balanced equation, which represent the proportions of moles involved in the reaction. This balance is crucial for calculating equilibrium constants because changes in stoichiometry affect these calculations.- For example, the equilibrium reactions in the provided exercise showcase how altering the stoichiometry (by dividing all coefficients by two) alters the equilibrium constant.

Chemical Equilibrium

Chemical equilibrium is a state reached in a reversible reaction where the concentrations of reactants and products no longer change with time. This is not a state of no reactions but a balanced state where the forward and reverse reactions occur at the same rate. - During equilibrium, the system's observable properties, such as color, pressure, and concentration, remain unchanged. - The concept of chemical equilibrium is critical in understanding how changes in conditions like pressure, temperature, and concentration impact the position of equilibrium. - Le Chatelier’s Principle, which provides insights into how systems at equilibrium react to external changes, is often used to predict the effects of changing conditions.

Equilibrium Constant Calculation

Calculating equilibrium constants accurately is vital for predicting the composition of a system at equilibrium. These calculations derive from the balanced chemical equation.- The equilibrium constant \( K \) is determined by the expression involving the concentrations of the products and reactants. For the reaction \( \mathrm{N}_2 + \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} \), it is given by:- \[ K_1 = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_2][\mathrm{O}_2]} \]- For the reaction where stoichiometry is halved, such as \( \frac{1}{2} \mathrm{N}_2 + \frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{NO} \), the equilibrium constant is:- \[ K_2 = \frac{[\mathrm{NO}]}{([\mathrm{N}_2][\mathrm{O}_2])^{1/2}} \]- Notably, when the stoichiometry changes, the equilibrium constant changes accordingly. As illustrated, dividing the reaction coefficients by two results in an equilibrium constant that is the square root of the original.