Question

Suppose that for the following reaction $$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ it is determined that, at a particular temperature, the equilibrium concentrations are as follows: \([\mathrm{NO}(g)]\) \(2.1 \times 10^{-3} \mathrm{M},\left[\mathrm{O}_{3}(g)\right]=4.7 \times 10^{-5} \mathrm{M},\left[\mathrm{NO}_{2}(g)\right]=\) \(3.5 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{O}_{2}(g)\right]=7.8 \times 10^{-3} \mathrm{M} .\) Calculate the value of \(K\) for the reaction at this temperature.

Short answer

The value of the equilibrium constant (K) for the given reaction at the specified temperature is approximately 2767.16, when calculated using the equilibrium concentrations of the species involved in the reaction.

Step-by-step solution

Write the equilibrium constant expression for the reaction

Using the stoichiometric coefficients from the balanced chemical equation, we can write the expression for the equilibrium constant (K) as: $$ K = \frac{[\mathrm{NO}_{2}][\mathrm{O}_{2}]}{[\mathrm{NO}][\mathrm{O}_{3}]} $$

Plug in the given concentrations

Now we will substitute the given equilibrium concentrations into the K expression: $$K = \frac{(3.5 \times 10^{-2})(7.8 \times 10^{-3})}{(2.1 \times 10^{-3})(4.7 \times 10^{-5})} $$

Calculate the value of K

Now we will calculate the value of K using the expression: $$K = \frac{(3.5 \times 10^{-2})(7.8 \times 10^{-3})}{(2.1 \times 10^{-3})(4.7 \times 10^{-5})} = \frac{(0.035)(0.0078)}{(0.0021)(0.000047)} = \frac{0.000273}{9.87 \times 10^{-8}} $$ $$ K = 2767.16 $$ Thus, the equilibrium constant for the reaction at the given temperature is approximately 2767.16.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a dynamic state in a reversible chemical process where the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant over time.
Understanding chemical equilibrium involves recognizing that although the concentrations do not change, molecular interactions are ongoing. In a closed system, when equilibrium is achieved, it indicates a balance in chemical activity - reactants are converting to products and products to reactants at an equal and constant rate.

Equilibrium Concentrations

The equilibrium concentrations are the amounts of reactants and products present when a reaction has reached equilibrium.
These concentrations can be measured in moles per liter (Molarity, M) and are crucial for calculating the equilibrium constant, which is a quantitative measure of a reaction's tendency to proceed in the forward or reverse direction.
For students to grasp this concept effectively, it's essential to practice determining these concentrations within various chemical systems. This helps in understanding the direct influence they have on the numerical value of the equilibrium constant.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass where the total mass of reactants equals the total mass of products in a chemical reaction.

• The coefficients in a balanced chemical equation represent the relative amounts of moles of each substance involved.

Stoichiometry is not only about balancing elements, but also about understanding the ratios and how these influence the products formed. For example, the ratio of molecules of NO and O3 to molecules of NO2 and O2 is crucial in our equilibrium constant calculation.

Balance Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that requires equalizing the number of atoms for each element on both sides of the equation. It reflects the conservation of mass and is essential for correct stoichiometric calculations.

Why Balance Equations?

• To adhere to the Law of Conservation of Mass.
• To provide the correct stoichiometric ratios, which are necessary to calculate reactant and product amounts.
• To accurately describe the qualitative and quantitative aspects of the chemical reaction.

Learning to balance equations properly paves the way for understanding more complex topics like reaction yields, limiting reagents, and indeed, equilibrium constant calculations.