Problem 27 - Chemistry The Molecular Nature of Matter

Question

Write the equilibrium law for each of the following gas phase reactions in terms of molar concentrations: (a) \(2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{POCl}_{3}(g)\) (b) \(2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+6 \mathrm{H}_{2} \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{SOCl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+2 \mathrm{HCl}(g)\)

Step-by-Step Solution

Answer

The equilibrium laws for the reactions are (a) \(K = \frac{[\mathrm{POCl}_{3}]^2}{[\mathrm{PCl}_{3}]^2 [\mathrm{O}_{2}]}\), (b) \(K = \frac{[\mathrm{SO}_{2}]^2 [\mathrm{O}_{2}]}{[\mathrm{SO}_{3}]^2}\), (c) \(K = \frac{[\mathrm{NO}]^2 [\mathrm{H}_{2}\mathrm{O}]^2}{[\mathrm{N}_{2}\mathrm{H}_{4}] [\mathrm{O}_{2}]^2}\), (d) \(K = \frac{[\mathrm{NO}_{2}]^2 [\mathrm{H}_{2}\mathrm{O}]^8}{[\mathrm{N}_{2}\mathrm{H}_{4}] [\mathrm{H}_{2}\mathrm{O}_{2}]^6}\), and (e) \(K = \frac{[\mathrm{SO}_{2}] [\mathrm{HCl}]^2}{[\mathrm{SOCl}_{2}] [\mathrm{H}_{2}\mathrm{O}]}\).

Step 1: Writing Equilibrium Law for Reaction (a)

For the reaction with reactants and products in the gas phase, the equilibrium constant expression is given by the product of the molar concentrations of the products raised to the power of their stoichiometric coefficients divided by the product of the molar concentrations of the reactants raised to the power of their stoichiometric coefficients. For the reaction (a) \(2 \mathrm{PCl}_{3}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{POCl}_{3}(g)\), the equilibrium law is \(K = \frac{[\mathrm{POCl}_{3}]^2}{[\mathrm{PCl}_{3}]^2 [\mathrm{O}_{2}]}\)

Step 2: Writing Equilibrium Law for Reaction (b)

For the reaction (b) \(2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g)\), the equilibrium law is \(K = \frac{[\mathrm{SO}_{2}]^2 [\mathrm{O}_{2}]}{[\mathrm{SO}_{3}]^2}\)

Step 3: Writing Equilibrium Law for Reaction (c)

For the reaction (c) \(\mathrm{N}_{2} \mathrm{H}_{4}(g) + 2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g)\), the equilibrium law is \(K = \frac{[\mathrm{NO}]^2 [\mathrm{H}_{2}\mathrm{O}]^2}{[\mathrm{N}_{2}\mathrm{H}_{4}] [\mathrm{O}_{2}]^2}\)

Step 4: Writing Equilibrium Law for Reaction (d)

For the reaction (d) \(\mathrm{N}_{2} \mathrm{H}_{4}(g) + 6 \mathrm{H}_{2} \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) + 8 \mathrm{H}_{2} \mathrm{O}(g)\), the equilibrium law is \(K = \frac{[\mathrm{NO}_{2}]^2 [\mathrm{H}_{2}\mathrm{O}]^8}{[\mathrm{N}_{2}\mathrm{H}_{4}] [\mathrm{H}_{2}\mathrm{O}_{2}]^6}\)

Step 5: Writing Equilibrium Law for Reaction (e)

For the reaction (e) \(\mathrm{SOCl}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{SO}_{2}(g) + 2 \mathrm{HCl}(g)\), the equilibrium law is \(K = \frac{[\mathrm{SO}_{2}] [\mathrm{HCl}]^2}{[\mathrm{SOCl}_{2}] [\mathrm{H}_{2}\mathrm{O}]}\)

Key Concepts

Equilibrium Law

Understanding the equilibrium law is crucial when studying chemical reactions and how they reach a state where the rate of the forward reaction equals the rate of the reverse reaction. This state is known as chemical equilibrium. The mathematical representation of this concept is the equilibrium constant (K), which is calculated by taking the ratios of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients in the balanced equation.

For a general reversible reaction, where the uppercase letters represent the chemicals and the lowercase letters are their respective stoichiometric coefficients: \begin{center} \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\), \the equilibrium constant expression is given by: \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\). The square brackets denote the molar concentrations of the gaseous reactants and products. It's important to remember that the values of K depend on the temperature and can offer insights into the favorability of the reaction under specific conditions.

Molar Concentrations

Molar concentrations, denoted by square brackets, are a way to express how much of a substance is present in a given volume of solution. Specifically, it describes the number of moles of solute per liter of solution (mol/L). This concept is a backbone in the study of chemical equilibria, as the equilibrium law involves the molar concentrations of the reactants and products.

Calculating molar concentrations requires knowing the volume of the solution and the amount of the solute, often obtained from the mass and molar mass. For gases, the molar concentration can be determined using the ideal gas law. Equilibrium expressions utilize these concentrations to help predict the direction in which a reaction will proceed to reach equilibrium, making it an essential aspect of chemical kinetics and thermodynamics.

Gas Phase Reactions

Gas phase reactions are those that occur between reactants in the gaseous state. They are particularly interesting due to the way molecules interact in the gas phase, which is different from reactions in liquid or solid phases. When describing the equilibrium of gas phase reactions, we account for the partial pressures of the gases. However, the equilibrium expression can also be written in terms of molar concentrations, as seen in the given textbook examples, where knowledge of the Ideal Gas Law helps to convert between these measures when needed.

These reactions and their equilibria are influenced by changes in pressure and temperature due to the nature of gases to expand and compress, and this aspect must be considered when predicting the behavior of a system at equilibrium under varying conditions.

Stoichiometric Coefficients

Stoichiometric coefficients are the numbers placed in front of chemical species in a balanced chemical equation to indicate their relative amounts. In the context of equilibrium expressions, these coefficients serve as powers for the concentrations of reactants and products. For example, in a balanced reaction where a reactant or product appears with a coefficient of 2, its concentration will be squared in the equilibrium law expression.

Understanding the role of stoichiometric coefficients is essential because they directly influence the value of the equilibrium constant, K. The coefficients indicate the proportions in which reactants combine and products form, which is the basis for calculating the amount of chemicals consumed or produced in a reaction, and thus, crucial in quantitative chemical analysis.