Question

At \(25^{\circ} \mathrm{C}\) , the value of \(K\) is \(1.7 \times 10^{-13}\) for the following reaction. \(2 \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)\) It is determined that \(\left[\mathrm{N}_{2} \mathrm{O}\right]=0.0035 \mathrm{mol} / \mathrm{L}\) and \(\left[\mathrm{O}_{2}\right]=0.0027 \mathrm{mol} / \mathrm{L}\) . Using this information, what is the concentration of \(\mathrm{NO}(g)\) at equilibrium?

Short answer

[NO] = 1.4 \times 10^{-5} \mathrm{mol/L}

Step-by-step solution

Write the expression for the equilibrium constant (K)

For the given reaction, \(2 \mathrm{N}_{2} \mathrm{O}(g) + \mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)\), the equilibrium constant expression (K) is given by:\[ K = \frac{[\mathrm{NO}]^4}{[\mathrm{N}_{2} \mathrm{O}]^2 [\mathrm{O}_{2}]} \]

Substitute the given values into the equilibrium constant expression

Given values: \([\mathrm{N}_{2} \mathrm{O}] = 0.0035 \mathrm{mol/L}\) and \([\mathrm{O}_{2}] = 0.0027 \mathrm{mol/L}\), and \(K = 1.7 \times 10^{-13}\). Substitute these values into the expression:\[ 1.7 \times 10^{-13} = \frac{[\mathrm{NO}]^4}{(0.0035)^2 \times 0.0027} \]

Solve for the concentration of NO

First, calculate the denominator:\[ (0.0035)^2 \times 0.0027 = 3.3075 \times 10^{-8} \]Next, rearrange the equilibrium expression to isolate \([\mathrm{NO}]^4\):\[ [\mathrm{NO}]^4 = 1.7 \times 10^{-13} \times 3.3075 \times 10^{-8} \]\[ [\mathrm{NO}]^4 = 5.62275 \times 10^{-21} \]Finally, take the fourth root to find \([\mathrm{NO}]\):\[ [\mathrm{NO}] = \sqrt[4]{5.62275 \times 10^{-21}} \approx 1.4 \times 10^{-5} \mathrm{mol/L} \]

Key concepts

Equilibrium Constant

In chemical reactions, an equilibrium constant, denoted as K, helps us understand the point at which a reaction is at equilibrium. For a reversible reaction, the concentration of reactants and products stabilize in a way that the ratio of their concentrations remains constant over time. This ratio is the equilibrium constant. It shows the extent of a reaction; a large value of K means that the reaction heavily favors the formation of products, whereas a small value means it favors the reactants.
For example, in the equilibrium expression for the reaction below:
\[ 2N_2O(g) + O_2(g) \rightleftharpoons 4NO(g) \]
the equilibrium constant is given as:
\[ K = \frac{[NO]^4}{[N_2O]^2[O_2]}\]
Here, the concentration terms in the expression refer to the molar concentrations of the gases involved, and the equation shows how these concentrations relate to each other at equilibrium.

Chemical Equilibrium

Chemical equilibrium refers to the state where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products become constant. This does not mean that the reactions have stopped but rather that they continue to occur at the same rate in both directions.
Consider our reaction:
\[ 2N_2O(g) + O_2(g) \rightleftharpoons 4NO(g) \]
At equilibrium, the changes in the concentrations of \(N_2O\), \(O_2\), and \(NO\) have balanced out.
Understanding equilibrium is crucial because it allows us to predict how a system behaves under different conditions, such as changes in concentration, temperature, or pressure. If any of these conditions are altered, the system will adjust to reach a new equilibrium position, described by Le Chatelier's Principle.

Reaction Quotient

The reaction quotient, represented as Q, is a measure that helps determine the direction in which a reaction will proceed to reach equilibrium. While K is calculated using the concentrations at equilibrium, Q uses the concentrations at any given moment in time. The expression for Q is similar to that for K.
For example, using the same reaction:
\[ 2N_2O(g) + O_2(g) \rightleftharpoons 4NO(g) \]
the reaction quotient is:
\[ Q = \frac{[NO]^4}{[N_2O]^2[O_2]} \]
By comparing Q to K:

• If Q < K: the reaction will proceed forward (toward products) to reach equilibrium.
• If Q > K: the reaction will proceed in the reverse direction (toward reactants) to reach equilibrium.
• If Q = K: the system is at equilibrium.

This tool is valuable for predicting how a disrupted system will behave as it works to return to equilibrium.

Partial Pressure

In a gas mixture, each gas exerts a pressure independently of the others; this is called its partial pressure. The total pressure of the gas mixture is the sum of the partial pressures of all the constituent gases.
For the equilibrium involving gases, partial pressures can be used in place of concentrations in the equilibrium expression. If we know the partial pressures, we can calculate the equilibrium constant using a similar approach.
In our exercise, suppose we were dealing with partial pressures instead of concentrations for:
\[ 2N_2O(g) + O_2(g) \rightleftharpoons 4NO(g) \]
we would write:
\[ K_p = \frac{P_{NO}^4}{P_{N_2O}^2P_{O_2}} \]
Here, \( P_{NO} \), \( P_{N_2O} \), and \( P_{O_2} \) represent the partial pressures of NO, N_2O, and O_2 respectively.
Partial pressure can be particularly useful when the reaction involves gases at different conditions of temperature and volume.