Question

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide. $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 \mathrm{M},\) and \([\mathrm{NO}]=4.7 \times 10^{-4} \mathrm{M} .\) Calculate the value of \(K\) for the reaction.

Short answer

The value of the equilibrium constant (K) for the given reaction is as follows: \[K = \frac{[\mathrm{NO}]^{2}}{[\mathrm{N}_{2}][\mathrm{O}_{2}]} = \frac{(4.7 \times 10^{-4})^{2}}{(0.041)(0.0078)} = 0.69\]

Step-by-step solution

Write the expression for the equilibrium constant (K)

Using the balanced chemical equation provided: \(N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)\) we can write the expression for the equilibrium constant (K) as follows: \[K = \frac{[\mathrm{NO}]^{2}}{[\mathrm{N}_{2}][\mathrm{O}_{2}]}\]

Substitute given equilibrium concentrations into the formula

Now, we can substitute the given equilibrium concentrations into the expression for K: \[K = \frac{(4.7 \times 10^{-4})^{2}}{(0.041)(0.0078)}\]

Calculate K

We can now calculate K by performing the given arithmetic operation: \[K = \frac{(4.7 \times 10^{-4})^{2}}{(0.041)(0.0078)} = \frac{2.209\times10^{-7}}{3.198\times10^{-4}}\] \[K = 0.69\]

Report the value of K

The value of the equilibrium constant for this reaction is: \[K = 0.69\]

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when the rate of the forward reaction equals the rate of the reverse reaction, leading to a constant concentration of reactants and products in a closed system. In other words, it's when a reaction and its reverse are happening at the same time and at the same rate. Imagine it like a busy intersection in a city with cars constantly coming and going, but the number of cars on each street stays relatively the same.
At equilibrium, the reactants are turning into products at the same rate that the products are turning back into reactants. This doesn't mean that the quantities of reactants and products are equal, but rather that their ratios do not change over time. For example, in the textbook exercise provided, elemental nitrogen and oxygen are reacting to form nitrogen monoxide at equilibrium, which means the formation of nitrogen monoxide is balanced by the reverse reaction where nitrogen monoxide decomposes back into elemental nitrogen and oxygen.
Understand that reaching equilibrium can take different amounts of time depending on the specifics of the reaction, such as temperature, concentration, and the presence of a catalyst. The equilibrium state is dynamic and not static, even though to an observer from the outside, it may seem like nothing is changing.

Equilibrium Expressions

Understanding equilibrium expressions is essential for calculating the equilibrium constant, denoted as K. It’s a mathematical way of expressing the balance point of a reversible reaction. The equilibrium expression for a given reaction is derived from its balanced chemical equation and follows a general pattern: the concentrations of the products raised to the power of their stoichiometric coefficients are divided by the concentrations of the reactants raised to the same powers.

Creating the Expression

For instance, in our exercise, the reaction is \[N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)\],thus the equilibrium expression is \[K = \frac{[\text{NO}]^{2}}{[N_{2}][O_{2}]}\].It's important to remember that solid and liquid concentrations are not included in equilibrium expressions because their concentrations do not change appreciably in reactions.

Using Equilibrium Expressions

Once the expression is established, we can insert the equilibrium concentrations of the reactants and products to compute the numeric value of K. This is exactly what was done in the provided solution where specific concentrations were substituted into the equilibrium expression to find out the K value of 0.69 for the reaction at the given temperature. This value of K gives us vital information about the reaction’s equilibrium position—whether the products or reactants are favored.

Le Châtelier's Principle

Le Châtelier's principle is incredibly handy when predicting how a system at equilibrium will respond to external stresses, such as changes in concentration, pressure, or temperature. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. Imagine you're at a crowded party and suddenly a large group of people enters the room; the room's occupants will shift and spread out to accommodate the change. Le Châtelier's principle encompasses the same idea but in a chemical context.
For example, if we increase the concentration of one of the reactants in our nitrogen monoxide formation equation, according to Le Châtelier's principle, the equilibrium will shift to produce more nitrogen monoxide. Similarly, a temperature increase for an exothermic reaction will cause the equilibrium position to shift in favor of the reactants, as the reaction will try to absorb this added heat by carrying out the endothermic reverse reaction.
This principle helps chemists control the yield of chemical processes, optimize reaction conditions, and is fundamental to the understanding of how chemical reactions behave under different circumstances. Always remember, equilibrium will shift in the direction that will help to re-establish equilibrium, balancing out the added or subtracted component.