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

Question

The equilibrium constant for the reaction, [2004] \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) At temperature \(\mathrm{T}\) is \(4 \times 10^{-4} .\) The value of \(\mathrm{K}_{\text {cor the }}\) 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 equilibrium constant is 50.

Step 1: Understand the Given Data

We have an equilibrium constant for the reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g})\) at temperature \(T\) which is \(4 \times 10^{-4}\). Our goal is to find the equilibrium constant for the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\).

Step 2: Write the Reverse Reaction

The original reaction is \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g})\). The reverse reaction is \(2\mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g})\). The equilibrium constant for this reverse reaction is the reciprocal of the given one, so it is \(\frac{1}{4 \times 10^{-4}} = 2.5 \times 10^3\).

Step 3: Adjust for Stoichiometry

We are given a reaction: \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\), which is half of the reverse reaction. The equilibrium constant is the square root of that for the reaction \(2\mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g})\). Thus, \(K = \sqrt{2.5 \times 10^3} = 50\).

Step 4: Conclusion

The equilibrium constant for the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\) at the same temperature is 50. Therefore, the correct answer is option (d) 50.

Key Concepts

Reversible Reactions

Reversible reactions are essential in understanding chemical processes because they involve reactants and products that can convert into each other. Such reactions are symbolized by a double arrow (\(\rightleftharpoons\)), indicating that both the forward reaction (reactants to products) and the reverse reaction (products to reactants) can occur. This duality means that over time, these reactions can reach a point where the rates of the forward and reverse reactions are equal.

• In reversible reactions, the position of equilibrium can vary based on different conditions such as temperature, pressure, and concentration changes.
• These reactions are dynamic, constantly proceeding in both directions even at equilibrium.
• Because of their reversible nature, the conversion to products is never complete, unlike irreversible reactions.
• The concentration of reactants and products remains constant over time when a reaction is at equilibrium.

Understanding reversible reactions is crucial in predicting how systems respond to changes, which is particularly relevant in industrial processes and biological systems.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the forward and reverse reactions occur at the same rate. At this point, the concentrations of reactants and products remain unchanged. This is not an indication that the reactions have stopped, but rather they are occurring simultaneously at equal rates, maintaining a balance.

• The equilibrium constant (\(K\)) is a numerical value that represents the ratio of product concentrations to reactant concentrations at equilibrium.
• This constant depends on temperature, meaning that changes in temperature can affect the equilibrium position.
• An equilibrium constant greater than 1 indicates a greater concentration of products compared to reactants, favoring the forward reaction.
• Conversely, a \(K\) value less than 1 suggests that reactants are favored.

Chemical equilibrium doesn't imply equal concentrations of reactants and products, but rather stable proportions as dictated by the \(K\) value.

Stoichiometry Adjustments

Stoichiometry adjustments involve modifying the coefficients in a chemical reaction to reflect changes in the quantities of substances involved. In the context of equilibrium, it's essential to understand how these changes affect equilibrium constants. Changes in the stoichiometry of a reaction can require recalculating the equilibrium constant to maintain the correct relation between products and reactants.

• The stoichiometric coefficient indicates the number of moles of each substance participating in the reaction.
• When a reaction's coefficients are multiplied by a common factor, the equilibrium constant does not change. However, if the equation is reversed or altered such as halving coefficients, the numerical value of \(K\) must be adjusted accordingly.
• For instance, if the balanced equation is halved, the equilibrium constant is adjusted by taking the square root of the original \(K\) value.
• This adjustment ensures consistency in predicting and measuring chemical reactions in varied conditions.

Understanding stoichiometry adjustments helps in ensuring accurate calculations in chemical reactions, especially when conditions or assumptions change.