Question

At a particular temperature a 2.00-L flask at equilibrium contains \(2.80 \times 10^{-4} \mathrm{~mol} \mathrm{~N}_{2}, 2.50 \times 10^{-5} \mathrm{~mol} \mathrm{O}_{2}\), and \(2.00 \times 10^{-2}\) mol \(\mathrm{N}_{2} \mathrm{O} .\) Calculate \(K\) at this temperature for the reaction $$ 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2} \mathrm{O}(g) $$ If \(\left[\mathrm{N}_{2}\right]=2.00 \times 10^{-4} M,\left[\mathrm{~N}_{2} \mathrm{O}\right]=0.200 M\), and \(\left[\mathrm{O}_{2}\right]=0.00245\) \(M\), does this represent a system at equilibrium?

Short answer

The equilibrium constant, K, for the given reaction at this temperature is approximately 64.20. When plugging in the provided concentrations, the reaction quotient, Q, is calculated to be approximately 19854.12. Since Q is not equal to K, the system is not at equilibrium.

Step-by-step solution

Write the equilibrium expression for K

First, write the equilibrium expression for the reaction: \(K = \frac{[\mathrm{N}_{2}\mathrm{O}]^2}{[\mathrm{N}_{2}]^2 \times [\mathrm{O}_{2}]}\)

Convert moles to molarity

Since we're given the amounts of substances in moles and the volume of the flask, we can find the molarity of each substance: Molarity of \(\mathrm{N}_{2} = \frac{2.80 \times 10^{-4}\,\mathrm{mol}}{2.00\,\mathrm{L}} = 1.40 \times 10^{-4}\,\mathrm{M}\) Molarity of \(\mathrm{O}_{2} = \frac{2.50 \times 10^{-5}\,\mathrm{mol}}{2.00\,\mathrm{L}} = 1.25 \times 10^{-5}\,\mathrm{M}\) Molarity of \(\mathrm{N}_{2}\mathrm{O} = \frac{2.00 \times 10^{-2}\,\mathrm{mol}}{2.00\,\mathrm{L}} = 1.00 \times 10^{-2}\,\mathrm{M}\)

Calculate K using the equilibrium concentrations

Plug the equilibrium concentrations into the expression for K: \(K = \frac{(1.00 \times 10^{-2})^2}{(1.40 \times 10^{-4})^2 \times (1.25 \times 10^{-5})} = 64.20\)

Plug the given concentrations into the reaction quotient, Q

For the given concentrations, \([\mathrm{N}_{2}] = 2.00 \times 10^{-4}\,\mathrm{M}\), \([\mathrm{N}_{2}\mathrm{O}] = 0.200 \,\mathrm{M}\), and \([\mathrm{O}_{2}] = 0.00245\,\mathrm{M}\): Reaction quotient, Q, is calculated the same way as K: \(Q = \frac{[\mathrm{N}_{2}\mathrm{O}]^{2}}{[\mathrm{N}_{2}]^{2}\times[\mathrm{O}_{2}]}\) Now, plug the given concentrations into the Q expression: \(Q = \frac{(0.200)^2}{(2.00 \times 10^{-4})^2 \times (0.00245)} = 19854.12\)

Compare Q and K to determine if the system is at equilibrium

The system is at equilibrium when Q = K. In this case, Q and K are not equal (Q > K), so the system is not at equilibrium.

Key concepts

Reaction Quotient (Q)

The reaction quotient, denoted as \(Q\), is a vital tool in chemical equilibrium. It helps us analyze the current condition of a reaction compared to its equilibrium state. By calculating \(Q\), we can determine whether a reaction is at equilibrium or if any shifts are required. Much like the equilibrium constant \(K\), \(Q\) is expressed in terms of concentrations or partial pressures of reactants and products. For the reaction \(2 \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{N}_2\text{O}(g)\), the reaction quotient is:- \(Q = \frac{[\text{N}_2\text{O}]^2}{[\text{N}_2]^2 \times [\text{O}_2]}\)Once calculated, \(Q\) can be compared with \(K\) to make predictions:

• If \(Q < K\), the forward reaction is favored.
• If \(Q > K\), the reverse reaction is favored.
• If \(Q = K\), the system is at equilibrium.

Chemical Equilibrium

Chemical equilibrium occurs when a reversible chemical reaction proceeds at a rate such that the concentrations of reactants and products remain constant over time. At this point, the forward and reverse reactions continue to occur but at equal rates.In simple terms, the "balance" between the forward and reverse reactions means there is no net change in the concentrations of substances involved. For this reaction: \(2 \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{N}_2\text{O}(g)\), equilibrium is described by the constant \(K\):

• When a system is at equilibrium, \(Q = K\), meaning the ratio of product to reactant concentrations is constant.
• Equilibrium does not imply equal molar concentrations but rather that their ratio is stable.

Thus, equilibrium ensures the reaction does not proceed one way more than the other, maintaining a consistent state.

Molarity Conversion

Converting moles to molarity is essential for solving equilibrium problems. Molarity, represented as \(\text{M}\), measures the concentration of a solution, defined as moles of solute per liter of solution. Here's how:Given the moles of a substance and the volume of the solution, you can find molarity using:- \(\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)Consider \(\text{N}_2\): If we have \(2.80 \times 10^{-4}\) moles in a 2.00-L flask:- \(\text{Molarity of } \text{N}_2 = \frac{2.80 \times 10^{-4} \text{ moles}}{2.00 \text{ L}} = 1.40 \times 10^{-4} \text{ M}\)This conversion is necessary when plugging into equilibrium expressions or calculating the reaction quotient.

Equilibrium Expression

The equilibrium expression provides a mathematical way to represent the concentrations of reactants and products at equilibrium for a given reaction. It is key to solving equilibrium problems and finding the equilibrium constant \(K\).For the reaction \(2 \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{N}_2\text{O}(g)\), the equilibrium expression is:- \(K = \frac{[\text{N}_2\text{O}]^2}{[\text{N}_2]^2 \times [\text{O}_2]}\)To compute \(K\), calculate the concentrations of the compounds involved and substitute them into the expression. The power to which each concentration is raised corresponds to its coefficient in the balanced chemical equation. This expression serves as the foundation to compare with \(Q\) and analyze whether the reaction is at equilibrium. A precise calculation ensures that we know how the reaction behaves under various conditions.