Question

At a particular temperature, \(8.1\) moles of \(\mathrm{NO}_{2}\) gas is placed in a 3.0-L container. Over time the \(\mathrm{NO}_{2}\) decomposes to NO and \(\mathrm{O}_{2}\) : $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be \(1.4 \mathrm{~mol} / \mathrm{L}\). Calculate the value of \(K\) for this reaction.

Short answer

The equilibrium constant \(K\) for this reaction is \(0.81\).

Step-by-step solution

Calculate the initial concentration of NO2

Given the number of moles of NO2 and the volume of the container, we can calculate the initial concentration (in \(\mathrm{mol} / \mathrm{L}\)) of NO2: \([\mathrm{NO}_{2}]_{initial} = \frac{8.1 \text{ moles}}{3.0 \text{ L}} = 2.7 \text{ M}\)

Write the balanced chemical equation and the expression for K

The balanced chemical reaction is given in the problem: $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ To calculate the equilibrium constant K, we need to express K in terms of the equilibrium concentrations of each substance: $$K = \frac{[\mathrm{NO}]^{2}[\mathrm{O}_{2}]}{[\mathrm{NO}_{2}]^{2}}$$

Set up an ICE table (Initial, Change, Equilibrium)

Here is an ICE (Initial, Change, Equilibrium) table for the reaction: | | \(\mathrm{NO}_{2}\) | \(\mathrm{NO}\) | \(\mathrm{O}_{2}\) | | ------- | ------- | ------- | ------- | | Initial | 2.7 M | 0 M | 0 M | | Change | -2x | +2x | +x | | Equilibrium | \(2.7 - 2x\) | \(1.4\) | \(x\) | Note that \([\mathrm{NO}]\) at equilibrium is given as \(1.4 \mathrm{~mol} / \mathrm{L}\). Since \(2 \text{ moles}\) of NO are formed for every mole of O2, we can write the change in concentration of NO as +2x and that of O2 as +x.

Calculate the value of K

Now, we can plug the equilibrium concentrations of the components into the K expression: $$K = \frac{[\mathrm{NO}]^{2}[\mathrm{O}_{2}]}{[\mathrm{NO}_{2}]^{2}} = \frac{(1.4)^{2}(x)}{(2.7-2x)^{2}}$$ We know that \([\mathrm{NO}]_{equilibrium}=1.4\mathrm{~mol} / \mathrm{L}\). Since the stoichiometry of the reaction states that 2 moles of NO are produced for each mole of O2, we can find the change in the concentration (x) by dividing the equilibrium concentration of NO by 2: $$x = \frac{1.4 \mathrm{~mol} / \mathrm{L}}{2} = 0.7 \mathrm{~mol} / \mathrm{L}$$ Now we can plug the value of x into the K expression: $$K = \frac{(1.4)^{2}(0.7)}{(2.7-2(0.7))^{2}} = \frac{1.96 \times 0.7}{(2.7 - 1.4)^{2}}$$ Now we can simplify and calculate K: $$K = \frac{1.372}{1.3^{2}} = \frac{1.372}{1.69} = 0.81$$ Therefore, the equilibrium constant \(K\) for this reaction is \(0.81\).

Key concepts

Chemical Equilibrium

Chemical equilibrium represents a state in a chemical reaction where the rates of the forward reaction (reactants converting to products) and the reverse reaction (products converting back to reactants) are equal. At this point, the concentration of reactants and products remain constant over time. It's important to note, however, that being in a state of equilibrium does not mean that the reactants and products are in equal concentrations, but rather that their concentrations are stable because the processes of formation and decomposition are happening at the same rate.

When considering a reversible reaction such as:
\[2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{O}_{2}(g)\]
The concept of equilibrium constant (K) comes into play. This is a numerical value that represents the ratio of the concentration of products to reactants, each raised to the power of their stoichiometric coefficients in the balanced equation. The equilibrium constant is a reflection of the position of equilibrium and is influenced by temperature.

ICE Table

An ICE table, which stands for Initial, Change, and Equilibrium, is a systematic way to organize the concentration of reactants and products over the course of a reaction heading towards equilibrium. It helps in solving equilibrium problems by tracking the changes in concentration as the reaction progresses. The 'Initial' row lists the initial concentrations before the reaction starts. The 'Change' row denotes the changes that occur during the reaction, often expressed in terms of a variable, like 'x,' to indicate the amount consumed or produced. The 'Equilibrium' row contains the expressions that represent the concentrations of the reactants and products after the reaction has reached equilibrium.

The use of an ICE table is crucial when dealing with reactions where not all the concentrations are known at equilibrium. It allows us to use the stoichiometry of the reaction and the initial conditions to calculate the unknowns.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of each substance in the mixture at the point of chemical equilibrium. Knowing the equilibrium concentrations of the reactants and products is essential for calculating the equilibrium constant, K. In the explained exercise, we were given the equilibrium concentration of \(\mathrm{NO}\) which is \((1.4 \mathrm{~mol} / \mathrm{L}\). The concentration of \((\mathrm{O}_{2}\) and the final concentration of \((\mathrm{NO}_{2}\) need to be calculated taking into account the reaction stoichiometry and the shift in concentrations as the reaction achieves equilibrium. These concentrations are what we eventually plug into the equilibrium expression to solve for the equilibrium constant.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between the amount of reactants and products in a chemical reaction. It is based on the balanced chemical equation, which provides the mole ratios needed to calculate how much of each substance is consumed or produced. For example, in the provided reaction:
\[2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{O}_{2}(g)\]
The stoichiometry tells us that 2 moles of NO₂ will produce 2 moles of NO and 1 mole of O₂, suggesting a molar ratio of 2:2:1. Understanding and applying stoichiometry is critical in the process of setting up an ICE table and calculating the equilibrium constant, as it shows us how the changes in concentrations of one species directly affect the others.