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 mol/L. Calculate the value of $K$ for this reaction.

Short answer

The equilibrium constant, K, for the given reaction is approximately 0.97.

Step-by-step solution

Write down the balanced chemical equation and K expression

The balanced chemical equation is already given: $$2{\mathrm{NO}}_2(g)\rightleftharpoons 2\mathrm{NO}(g)+{\mathrm{O}}_2(g)$$ Now let's write down the expression for the equilibrium constant K: $$K=\frac{[\mathrm{NO}{]}^2[{\mathrm{O}}_2]}{[{\mathrm{NO}}_2{]}^2}$$

Calculate the initial concentrations of NO₂, NO, and O₂

We can find the initial concentration of ${\mathrm{NO}}_2$ using the given moles and the volume of the container: Initial concentration of ${\mathrm{NO}}_2=\frac{\text{moles of }{\mathrm{NO}}_2}{\text{volume of container}}=\frac{8.1mol}{3.0L}=2.7mol/L$ Initially, there are no moles of $\mathrm{NO}$ and ${\mathrm{O}}_2$ since the reaction hasn't started yet. So the initial concentrations of $\mathrm{NO}$ and ${\mathrm{O}}_2$ are both 0 mol/L.

Calculate the change in concentrations for each species at equilibrium

We're given the equilibrium concentration of NO, which is 1.4 mol/L. Since the balanced equation shows a 1:1 ratio between NO and NO₂ reacting, we know that 1.4 mol/L of NO₂ reacted to form 1.4 mol/L of NO. The change in concentration of NO₂ is -1.4 mol/L due to reaction. The change in concentration of NO and O₂ both are 1.4 mol/L.

Calculate the equilibrium concentrations of NO₂, NO, and O₂

Now we add the changes in concentrations to the initial concentrations to obtain the equilibrium concentrations: Equilibrium concentration of ${\mathrm{NO}}_2=\text{initial concentration of }{\mathrm{NO}}_2+\text{change in concentration of }{\mathrm{NO}}_2=2.7-1.4=1.3mol/L$ Equilibrium concentration of $\mathrm{NO}=1.4mol/L\text{(already given)}$ Equilibrium concentration of ${\mathrm{O}}_{₂}=\frac{1}{2}\times \text{change in concentration of }\mathrm{NO}=\frac{1}{2}\times 1.4=0.7mol/L$ (since one mole of O₂ is formed for every two moles of NO)

Use the equilibrium concentrations to find the value of K

Now we can plug in these equilibrium concentrations into the K expression: $$K=\frac{[\mathrm{NO}{]}^2[{\mathrm{O}}_2]}{[{\mathrm{NO}}_2{]}^2}=\frac{(1.4{)}^2(0.7)}{(1.3{)}^2}=0.97$$ The equilibrium constant, K, for the reaction is approximately 0.97.

Key concepts

Equilibrium Constant (K)

Understanding the equilibrium constant, denoted as K, is central to grasping the meaning behind chemical equilibrium. It's a value that expresses the ratio of the concentration of the products to the concentration of the reactants once a chemical reaction has reached equilibrium. This value is determined at a specific temperature and remains constant, given that the temperature is unchanged.

In a balanced chemical reaction such as the decomposition of ${\mathrm{NO}}_2$, the equilibrium constant equation is expressed as $$K=\frac{[\mathrm{NO}{]}^2[{\mathrm{O}}_2]}{[{\mathrm{NO}}_2{]}^2}$$ where the square brackets represent the molarity, or concentration, of each species. To calculate K, one must know the equilibrium concentrations of all the reactants and products involved.

Concentration of Reactants and Products

In chemical processes, the concentration of reactants and products is crucial for determining how a reaction progresses. Concentration is defined by the amount of a substance within a certain volume of solution, usually given in moles per liter (mol/L). At the start of a reaction, you'll have initial concentrations, but as the reaction proceeds towards equilibrium, the concentrations of reactants decrease while those of products increase.

During an equilibrium state, the concentration of the reactants and products remains constant — not equal, unless K equals 1. As showcased in the ${\mathrm{NO}}_2$ example, these equilibrium concentrations are essential for calculating the equilibrium constant. It's important to pay attention to the stoichiometry of the balanced chemical equation since it defines the relationship between the changes in concentration of reactants and products.

Le Chatelier's Principle

Le Chatelier's principle is a valuable tool in predicting the effect of changes in concentration, temperature, or pressure on a system at equilibrium. It posits that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This helps maintain balance within the system.

For example, if more ${\mathrm{NO}}_2$ is added to the system at equilibrium, the reaction would shift to produce more $\mathrm{NO}$ and ${\mathrm{O}}_2$ according to the principle. Conversely, the removal of a product from the reaction's mixture would result in more reactants breaking down to replenish the removed product, thus shifting equilibrium towards the products side. Understanding how Le Chatelier's principle affects equilibrium is essential for controlling chemical reactions in various industrial and laboratory processes.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, ensuring the law of conservation of mass is upheld in chemical reactions. This process involves adjusting the coefficients of the reactants and products so that the number of atoms of each element is the same on both sides of the equation.

In our exercise, the balanced equation was $$2{\mathrm{NO}}_2(g)\rightleftharpoons 2\mathrm{NO}(g)+{\mathrm{O}}_2(g)$$ which indicates that two moles of ${\mathrm{NO}}_2$ react to form two moles of $\mathrm{NO}$ and one mole of ${\mathrm{O}}_2$. Proper balance ensures that when calculating concentration changes and the equilibrium constant, the stoichiometry is accurately reflected, and the resulting calculations are precise.