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 \(\mathrm{K}_{\mathrm{c}}\) 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: (a) \(4 \times 10^{-6}\) (b) \(2.5 \times 10^{2}\) (c) \(0.02\) (d) 50

Short answer

The value of \(K_c\) for the reaction \(NO(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)\) is 50.

Step-by-step solution

Identify the given reaction and its equilibrium expression

We are given the reaction \(N_{2}(g)+O_{2}(g) \rightleftharpoons 2NO(g)\) with equilibrium constant \(K_c = 4 \times 10^{-4}\). The equilibrium expression for this reaction is \(K_c = \frac{[NO]^2}{[N_2][O_2]}\).

Define the reverse reaction's equilibrium expression

The reverse reaction \(2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)\) has its equilibrium expression described as \(K'_c = \frac{[N_2][O_2]}{[NO]^2}\). The equilibrium constant for the reverse reaction, \(K'_c\), is the reciprocal of the forward reaction's constant, so \(K'_c = \frac{1}{K_c} = \frac{1}{4 \times 10^{-4}} = 2.5 \times 10^{3}\).

Adjust the reaction to match the target equation

The target reaction is \(NO(g) \rightleftharpoons \frac{1}{2}N_{2}(g) + \frac{1}{2}O_{2}(g)\). This is half of the reverse reaction \(2NO(g) \rightleftharpoons N_2(g) + O_2(g)\), which entails finding the new equilibrium constant that corresponds to \(0.5\) times the reverse reaction.

Calculate the equilibrium constant for the target reaction

For a reaction change where the stoichiometry is halved, the equilibrium constant is taken to the power corresponding to the factor by which the reaction is scaled, which is \(K''_c = (2.5 \times 10^{3})^{0.5}\). Calculate the square root: \(K''_c = \sqrt{2.5 \times 10^{3}} \approx 50\).

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_c\), is a fundamental concept in chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants when a reaction is at equilibrium. For any given reaction, this constant provides insight into the extent to which reactants are converted into products.

For instance, consider the gas-phase reaction \(N_2(g) + O_2(g) \rightleftharpoons 2NO(g)\). The equilibrium expression for this process is \(K_c = \frac{[NO]^2}{[N_2][O_2]}\). This means that at equilibrium, the concentration of \(NO\) raised to the power of two is balanced by the concentrations of \(N_2\) and \(O_2\). An equilibrium constant of \(4 \times 10^{-4}\) for this reaction indicates that at equilibrium, the concentrations of reactants are favored over the products.

Understanding \(K_c\) is crucial for predicting the direction and extent of chemical reactions, especially in dynamic environments where conditions constantly change.

Gas-phase Reactions

Gas-phase reactions involve reactants and products that are primarily in the gaseous state. These reactions are crucial in various fields such as atmospheric chemistry, industrial processes, and even everyday phenomena like combustion.

One important characteristic of gas-phase reactions is that their equilibrium can be influenced by changes in pressure, temperature, and volume. The reaction \(N_2(g) + O_2(g) \rightleftharpoons 2NO(g)\) is a prime example of a gas-phase chemical process. It involves nitrogen and oxygen gases reacting to form nitric oxide gas.

In gas-phase reactions, the ideal gas law \(PV = nRT\) often comes into play. It relates pressure \(P\), volume \(V\), and temperature \(T\) of the gases to the number of moles \(n\), providing a comprehensive understanding of the reaction dynamics. These reactions are critical for understanding and controlling environmental emissions and designing high-efficiency industrial processes.

Stoichiometry

Stoichiometry refers to the calculation of relative quantities of reactants and products in chemical reactions. It provides a bridge between the macroscopic quantities used in processes and the microscopic interactions at the mole level.

In the context of equilibrium reactions like \(N_2(g) + O_2(g) \rightleftharpoons 2NO(g)\), stoichiometry helps us understand the mole ratios of reactants and products. This specific reaction shows a \(1:1:2\) stoichiometric relationship, meaning one mole of nitrogen reacts with one mole of oxygen to produce two moles of nitric oxide.

When a reaction is modified, such as halving the stoichiometry in \(NO(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g)\), the equilibrium constant must be adjusted accordingly. This involves raising the \(K_c\) to the power corresponding to the stoichiometric change, demonstrating stoichiometry's critical role in equilibrium calculations.

Le Chatelier's Principle

Le Chatelier's Principle is a cornerstone concept in understanding how equilibria respond to external changes. This principle states that if an external change is applied to a system at equilibrium, the system will adjust itself to partially counteract that change, re-establishing equilibrium.

Consider a scenario where the temperature or pressure is altered in a gas-phase reaction like \(N_2(g) + O_2(g) \rightleftharpoons 2NO(g)\). According to Le Chatelier's Principle, if the temperature is increased, the reaction may shift in a direction that absorbs heat, depending on whether it is exothermic or endothermic. Similarly, changes in pressure will favor the side of the reaction with fewer moles of gas.

This principle helps in predicting how changes in environmental conditions can influence the direction of reaction adjustments, making it vital for chemical industry applications where maintaining specific reaction conditions is essential for optimal yield.