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

Short answer

The equilibrium constant, K, for the given reaction at the specified temperature is approximately 6.91.

Step-by-step solution

Write the balanced reaction

Write the balanced chemical reaction for the given process: \[ \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \]

Write the equilibrium expression

Based on the balanced reaction, we can write the expression for the equilibrium constant K: \[ K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]} \]

Insert the equilibrium concentrations

Now we substitute the given equilibrium concentrations of each component: \[ K = \frac{(4.7 \times 10^{-4}\ \mathrm{M})^2}{(0.041\ \mathrm{M})(0.0078\ \mathrm{M})} \]

Calculate the value of K

Perform the calculations to find the value of K: \[ K = \frac{(2.209\times 10^{-7}\ \cancel{\mathrm{M^2}})}{(3.198 \times 10^{-4}\ \cancel{\mathrm{M^2}})} \] \[ K = 6.907 \] So the equilibrium constant, K, for the given reaction at the specified temperature is approximately 6.91.

Key concepts

Nitrogen Monoxide

Nitrogen monoxide, commonly referred to as nitric oxide, is a chemical compound with the formula NO. It's a colorless gas under standard conditions and plays a significant role in various chemical processes.
It is particularly important in atmospheric chemistry, biology, and industrial processes. In the atmosphere, nitrogen monoxide is involved in the formation of smog and acid rain, contributing to environmental pollution.
In this exercise, nitrogen monoxide is produced when elemental nitrogen (\(\mathrm{N}_{2}\)) and oxygen (\(\mathrm{O}_{2}\)) react, especially at high temperatures. This reaction is important as it demonstrates how nitrogen monoxide can form in significant amounts under certain conditions, affecting both natural and human-made environments.

Chemical Equilibrium

Chemical equilibrium is a state in which the forward and reverse reactions occur at the same rate, resulting in no net change in the concentrations of the reactants and products over time.
In this context, for the reaction involving nitrogen monoxide formation, equilibrium is reached when the amounts of nitrogen (\(\mathrm{N}_{2}\)), oxygen (\(\mathrm{O}_{2}\)), and nitrogen monoxide (\(\mathrm{NO}\)) become constant, although the reactions still continue.
This doesn't mean all concentrations are equal, but rather that they have reached a constant ratio as described by the equilibrium constant, K.
Factors like temperature and pressure play important roles in the chemical equilibrium state. Changing these can shift the balance, favoring either the forward or reverse reaction.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of reactants and products in a chemical reaction once it has reached equilibrium. These concentrations are essential to determine the equilibrium constant, K.
For the formation of nitrogen monoxide, we are given equilibrium concentrations: nitrogen at 0.041 M, oxygen at 0.0078 M, and nitrogen monoxide at 4.7 x 10^-4 M.
These values allow us to plug into the expression for the equilibrium constant and solve for K, reflecting the ratio of the product concentrations to the reactant concentrations at equilibrium.
Equilibrium concentrations are not fixed and can vary with system conditions such as temperature, which is why they are crucial for calculating reaction dynamics and K at specific conditions.

Reaction Quotient

The reaction quotient, Q, is a measure similar to the equilibrium constant, K, but it can be calculated at any point during a reaction. The formula for Q is the same as for K, using the current concentrations of reactants and products.
Comparing Q with K helps to determine the direction in which a reaction will proceed to reach equilibrium:

• If Q < K, the reaction will move forward, meaning more products will form.
• If Q = K, the system is at equilibrium, and no net change occurs.
• If Q > K, the reaction will shift backwards, meaning more reactants will form until equilibrium is reached.

In the context of the exercise, although Q isn't directly calculated, understanding Q is crucial. It tells you whether the given concentrations are at equilibrium or if adjustments are required, helping to analyze reaction dynamics effectively.