Question

State and explain the law of mass action. Apply it to the following equilibria: (i) \(\mathrm{H}_{2(\mathrm{~g})}+\mathrm{F}_{2(\mathrm{~g})} \rightleftarrows 2 \mathrm{HF}_{(\mathrm{g})}\) (ii) \(\mathrm{NH}_{4} \mathrm{HS}_{(\mathrm{S})} \rightleftarrows \mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{~S}_{(\mathrm{g})}\) (iii) \(\mathrm{PCl}_{5(s)} \rightleftarrows \mathrm{PCl}_{3(t)}+\mathrm{Cl}_{2(\mathrm{~g})}\)

Short answer

The law of mass action states that the ratio of the concentration of products raised to their respective stoichiometric coefficients over the concentration of reactants raised to their respective stoichiometric coefficients is constant at a given temperature. For the 3 given equilibrium reactions, the equilibrium constant (K) expressions are: (i) H2 + F2 ↔ 2HF: \(K = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2]}\) (ii) NH4HS(s) ↔ NH3(g) + H2S(g): \(K' = [\mathrm{NH}_3] [\mathrm{H}_2\mathrm{S}]\) (iii) PCl5(s) ↔ PCl3(l) + Cl2(g): \(K' = [\mathrm{Cl}_2]\)

Step-by-step solution

(i) Law of Mass Action for H2 + F2 ↔ 2HF

First, write the balanced chemical equation for the reaction: \(\mathrm{H}_{2}(\mathrm{g}) + \mathrm{F}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{HF}_{(\mathrm{g})}\) Now, apply the law of mass action to determine the equilibrium expression, K: \(K = \frac{[\mathrm{HF}]^2}{[\mathrm{H}_2][\mathrm{F}_2]}\) where [HF], [H2], and [F2] are the equilibrium concentrations of HF, H2, and F2, respectively.

(ii) Law of Mass Action for NH4HS(s) ↔ NH3(g) + H2S(g)

First, write the balanced chemical equation for the reaction: \(\mathrm{NH}_{4} \mathrm{HS}_{(\mathrm{s})} \rightleftarrows \mathrm{NH}_{3(\mathrm{g})} + \mathrm{H}_{2} \mathrm{S}_{(\mathrm{g})}\) Now, apply the law of mass action to determine the equilibrium expression, K: \(K = \frac{[\mathrm{NH}_3][\mathrm{H}_2\mathrm{S}]}{[\mathrm{NH}_{4}\mathrm{HS}]}\) However, since the reaction has a solid reactant, its concentration remains constant and is incorporated into the equilibrium constant. Thus, the equilibrium expression becomes: \(K' = [\mathrm{NH}_3] [\mathrm{H}_2\mathrm{S}]\)

(iii) Law of Mass Action for PCl5(s) ↔ PCl3(l) + Cl2(g)

First, write the balanced chemical equation for the reaction: \(\mathrm{PCl}_{5(s)} \rightleftarrows \mathrm{PCl}_{3(l)} + \mathrm{Cl}_{2(\mathrm{g})}\) Now, apply the law of mass action to determine the equilibrium expression, K: \(K = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]}\) Since the reaction has a solid product, its concentration can be assumed to be constant, and we can redefine the equilibrium constant: \(K' = [\mathrm{Cl}_2]\)

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes the state of a system where the concentrations of reactants and products remain constant over time. This occurs in a reversible reaction when the forward and reverse reactions proceed at the same rate. Though the individual molecules are continuously reacting, there's no net change in the overall concentration of the involved substances.
To understand chemical equilibrium, think about it as a busy two-way road where cars move smoothly in both directions; though cars are constantly in motion, the number of cars around you at any given time stays the same. This is similar to how, in a chemical system at equilibrium, the rate of formation of products from reactants equals the rate of formation of reactants from products.
Key characteristics of chemical equilibrium include:

• Dynamic nature: The movement is ongoing, yet concentrations are unchanging.
• Constant ratios: The ratio of product concentrations to reactant concentrations remains constant.
• Response to disturbances: According to Le Chatelier's principle, if an external change like temperature or pressure affects the system, the equilibrium will shift to counteract the change and restore a new state of balance.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a numerical value that provides insight into the extent of a chemical reaction at equilibrium. It is derived from the concentrations of the products and reactants and depends solely on the temperature of the system. Essentially, \(K\) gives us a snapshot of where the equilibrium lies, whether it is more product-favored or reactant-favored.
For a general reversible reaction of the form \(aA + bB \rightleftarrows cC + dD\), the equilibrium expression is written as:
\[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]
Where the brackets \([ ]\) indicate molar concentrations at equilibrium and the letters represent stoichiometric coefficients. The value of \(K\) varies:

• If \(K > 1\), products are favored at equilibrium.
• If \(K < 1\), reactants are favored at equilibrium.
• If \(K \approx 1\), there is a roughly equal concentration of reactants and products at equilibrium.

This concept helps chemists predict how a system behaves under certain conditions and is crucial for industrial processes where maximizing product yield is essential.

Balanced Chemical Equation

A balanced chemical equation is a written representation of a chemical reaction that ensures the conservation of mass, meaning the same number of atoms for each element are present on both sides of the equation. This is critical because it allows for the correct interpretation of stoichiometric relationships in the reaction.
Balancing equations does not change the identity of the reactants or products but adjusts the stoichiometric coefficients (coefficients before molecule symbols) to adhere to the law of conservation of mass.
The steps to balance a chemical equation include:

• Write the unbalanced equation to show the reactants and products.
• Count atoms for each element in reactants and products.
• Use coefficients to balance each element on both sides, starting with the element that appears in the largest quantity.
• Double-check to ensure all elements and overall charges are balanced.

Consider the reaction \(\mathrm{H}_{2(\mathrm{~g})}+\mathrm{F}_{2(\mathrm{~g})}\rightleftarrows2\mathrm{HF}_{(\mathrm{g})}\). This equation is balanced because there are equal numbers of hydrogen and fluorine atoms on each side.
Balancing allows chemists to utilize the equations to calculate reactant and product quantities, vital for laboratory reactions and industrial production. It ensures that all calculations involving moles, mass, and volume in chemical reactions are accurate and reliable.