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 \(\left[\mathrm{N}_{2}\right]=0.041 M\), \(\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calculate the value of \(K\) for the reaction.

Short answer

The equilibrium constant (K) for the given reaction can be calculated using the expression \[ K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]} \]Substituting the given equilibrium concentrations, we get \[ K = \frac{(4.7 \times 10^{-4})^2}{(0.041)(0.0078)} \approx 7.07 \times 10^{-3} \]Thus, the equilibrium constant K for this reaction is approximately \(7.07 \times 10^{-3}\).

Step-by-step solution

Write the equilibrium constant expression

We need to first write the expression for the equilibrium constant (K). For the given reaction: \[ \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{NO}(g) \] The equilibrium constant expression will be: \[ K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]} \]

Substitute the values of the equilibrium concentrations

Now we substitute the equilibrium concentration values given in the problem into the equilibrium constant expression: \[ K = \frac{(4.7 \times 10^{-4})^2}{(0.041)(0.0078)} \]

Solve for K

By solving the expression we will get the value of K: \[ K = \frac{(4.7 \times 10^{-4})^2}{(0.041)(0.0078)} \approx 7.07 \times 10^{-3} \] So the equilibrium constant K for this reaction is approximately \(7.07 \times 10^{-3}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs in a chemical reaction when the concentrations of reactants and products remain constant over time in a closed system. This state is achieved when the forward reaction, which forms products from reactants, happens at the same rate as the reverse reaction, which forms reactants from products. At this point, there is no net change in the amounts of reactants and products, even though the reactions continue to occur.
To understand chemical equilibrium, consider the reaction of nitrogen and oxygen forming nitrogen monoxide:

• The balanced equation is: \( \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \)
• At equilibrium, the rate of formation of \(\mathrm{NO}\) is equal to the rate of its decomposition back into \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\).

Understanding this concept is crucial because it provides insight into why reactions do not always proceed to completion and why some reactions are reversible.

Reaction Kinetics

Reaction kinetics involves studying the rates of chemical reactions and how different conditions affect these rates. It differs from chemical equilibrium, as kinetics focuses on how fast a reaction proceeds rather than the position of equilibrium.
A few key factors affect the kinetics, or rate, of the reaction between nitrogen and oxygen to form nitrogen monoxide:

• Temperature: Increasing temperature increases the kinetic energy of molecules, making collisions more frequent and more energetic, which can increase the reaction rate.
• Concentration: Higher concentrations of reactants lead to more frequent collisions, thus, speeding up the reaction.
• Presence of a Catalyst: Catalysts lower the activation energy needed for a reaction to occur, increasing the reaction rate without being consumed in the process.

While equilibrium deals with the final state of the reaction, kinetics helps us understand how the reaction reaches this state.

Equilibrium Concentrations

Equilibrium concentrations are the concentrations of reactants and products in a chemical reaction at the point of equilibrium. These values are crucial because they determine the position of equilibrium and are used to calculate the equilibrium constant, \(K\). The equilibrium constant is a measure of the extent to which a reaction proceeds to form products.
In the given problem, we have equilibrium concentrations for \(\mathrm{N}_{2}\), \(\mathrm{O}_{2}\), and \(\mathrm{NO}\). The importance of knowing these values is that they allow us to construct the expression for \(K\) and subsequently determine its numerical value:

• The equilibrium constant expression is: \( K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]} \)
• By substituting the equilibrium concentrations in, we calculate \(K\), which gives insight into the reaction's favorability towards products or reactants.

In summary, understanding equilibrium concentrations helps us predict how changes in conditions will shift the position of equilibrium, according to Le Chatelier’s principle.