Problem 192 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) At temperature \(\mathrm{T}\) is \(4 \times 10^{-4} .\) The value of \(\mathrm{K}_{6}\) for the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons 1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is (a) \(4 \times 10^{-6}\) (b) \(2.5 \times 10^{2}\) (c) \(0.02\) (d) 50

Step-by-Step Solution

Answer

The value of \(K_6\) is 50, so the correct answer is (d).

Step 1: Identify Given Reaction and Equilibrium Constant

The problem provides the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g})\) with an equilibrium constant \(K_1\) of \(4 \times 10^{-4}\) at temperature \(T\).

Step 2: Write the Reverse Reaction for the Given Reaction

The reverse reaction for the first equation would be \(2\mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\). For the reverse reaction, the equilibrium constant \(K_2\) is the inverse of \(K_1\).

Step 3: Calculate the Equilibrium Constant for the Reverse Reaction

Using the formula for the equilibrium constant of the reverse reaction, we have \(K_2 = \frac{1}{K_1}\). Thus, \(K_2 = \frac{1}{4 \times 10^{-4}} = 2.5 \times 10^3\).

Step 4: Modify Reaction to Match the Given Reaction

The reaction given in the problem is \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\), which is half of the reverse reaction from Step 2. Therefore, the equilibrium constant for \(\mathrm{K}_6\) will be the square root of \(K_2\) because of the coefficients being halved.

Step 5: Calculate the Equilibrium Constant for the Final Reaction

To find \(K_6\) for the given reaction, take the square root of \(K_2\). Therefore, \(K_6 = \sqrt{2.5 \times 10^3} = 50\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to a state in a chemical reaction where the concentration of reactants and products no longer change with time. It occurs when the forward and backward reactions occur at the same rate, leading to a stable balance between the reactants and products.
The equilibrium constant, denoted as \(K\), is a crucial component in understanding chemical equilibrium. It provides a numerical value that indicates the ratio of the concentration of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced equation. A larger \(K\) value suggests that products are favored at equilibrium, while a smaller \(K\) value indicates a predominance of reactants.
Calculating and understanding equilibrium constants require knowledge of the reaction involved, including whether the reaction is being considered in its forward or reverse form. For reverse reactions, the equilibrium constant is simply the reciprocal of the constant for the forward reaction.

N2 and O2 Reaction

The reaction between nitrogen gas (\(\mathrm{N}_2\)) and oxygen gas (\(\mathrm{O}_2\)) to form nitrogen monoxide (\(\mathrm{NO}\)) is a classic example of chemical equilibrium. Represented by the equation \(\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g)\), this reaction helps illustrate how equilibrium principles apply.
In this reaction, nitrogen and oxygen gases react to form nitrogen monoxide in a balanced and specific stoichiometric ratio. The equilibrium constant \(K_1\) provided in the exercise is \(4 \times 10^{-4}\), indicating that at equilibrium, the amount of \(\mathrm{NO}(g)\) is lesser compared to the reactants \(\mathrm{N}_2(g)\) and \(\mathrm{O}_2(g)\). When calculating the equilibrium constant for the reverse reaction or adjusted stoichiometries, it is important to consider changes in reaction direction and individual coefficients to adjust the constant accordingly without error.

Stoichiometry

Stoichiometry in chemical reactions involves the quantitative relationship between reactants and products in a chemical reaction. It essentially provides a recipe that specifies the relative amounts of each substance involved.
For the reaction \(\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g)\), stoichiometry tells us:

• One mole of nitrogen gas reacts with one mole of oxygen gas to produce two moles of nitrogen monoxide.

When considering the changes in the stoichiometry of a reaction, such as halving or doubling coefficients as in the exercise, adjustments to the equilibrium constant must be made. For example, if every molecule in a reaction is halved, the effect on the equilibrium constant involves taking the square root of the original constant. This is because the powers of concentrations in the equilibrium expression directly relate to their stoichiometric coefficients.

Thermodynamics

Thermodynamics plays a crucial role in chemical equilibrium. It helps us understand the energy changes that occur during reactions and how these influence the position of equilibrium. This can be especially significant in reactions like the formation of \(\mathrm{NO}(g)\), which involve gases reacting.
The thermodynamics of a reaction can determine at which temperatures and pressures a reaction will have a certain \(K\) value. High temperatures could favor the formation of products in an endothermic reaction, where heat is absorbed, shifting equilibrium according to Le Chatelier's principle.
In a chemical equilibrium reaction like \(\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g)\), understanding the thermodynamics can help predict how changing conditions, such as temperature or pressure, might affect the equilibrium state.\(\Delta G\) (Gibbs free energy change) is related to the equilibrium constant through the equation \(\Delta G = -RT \ln K\), where a negative \(\Delta G\) suggests a spontaneous reaction under constant temperature and pressure.