Question

Write the equilibrium expression for each of the following heterogeneous equilibria. a. $2{\mathrm{LiHCO}}_3(s)\rightleftharpoons {\mathrm{Li}}_2{\mathrm{CO}}_3(s)+{\mathrm{H}}_2\mathrm{O}(g)+{\mathrm{CO}}_2(g)$ b. ${\mathrm{PbCO}}_3(s)\rightleftharpoons \mathrm{PbO}(s)+{\mathrm{CO}}_2(g)$ c. $4\mathrm{Al}(s)+3{\mathrm{O}}_2(g)\rightleftharpoons 2{\mathrm{Al}}_2{\mathrm{O}}_3(s)$

Short answer

The short answers for the given heterogeneous equilibria are: a. $$K_{\text{eq}}=[H_{2}O][C{O}_2]$$ b. $$K_{\text{eq}}=[C{O}_2]$$ c. $$K_{\text{eq}}=[O_2{]}^3$$

Step-by-step solution

a. Write the equilibrium expression for 2 LiHCO3(s) Li2CO3(s) + H2O(g) + CO2(g)

Since we are looking only at and including substances in their gaseous or aqueous states, the equilibrium expression for this reaction is: $$K_{\text{eq}}=[H_{2}O][C{O}_2]$$

b. Write the equilibrium expression for PbCO3(s) PbO(s) + CO2(g)

In this equilibrium, the only substance in its gaseous state is CO2, so the equilibrium expression is: $$K_{\text{eq}}=[C{O}_2]$$

c. Write the equilibrium expression for 4 Al(s) + 3 O2(g) 2 Al2O3(s)

In this case, the only substance in its gaseous state is O2, so the equilibrium expression is: $$K_{\text{eq}}=[O_2{]}^3$$

Key concepts

Equilibrium Expressions

In chemical reactions, the concept of equilibrium expressions is crucial in understanding how different substances interact at equilibrium. The equilibrium expression, also known as the equilibrium constant ($K_{\text{eq}}$), is derived from the concentrations of the gaseous or aqueous reactants and products involved in a reaction. It allows us to predict the ratio of products to reactants at equilibrium.
For heterogeneous equilibria, where the states of the reactants and products differ (such as solids, liquids, and gases), the expression includes only the concentrations of gases and aqueous solutions. Solids and pure liquids are omitted because their concentrations do not change under normal conditions, and as such, they don't affect the equilibrium constant.

• In reaction a: Only $[H_{2}O]$ and $[C{O}_2]$ appear in the expression.
• In reaction b: Only $[C{O}_2]$ is included.
• In reaction c: Only $[O_2]$ appears in the expression as $[O_2{]}^3$.

Hence, the equilibrium expression is a tool that helps in understanding and controlling chemical reactions.

Heterogeneous Equilibria

Heterogeneous equilibria occur when reactants and products are in different phases or states. For instance, a reaction may involve solids, gases, and possibly liquids interacting with each other. Understanding heterogeneous equilibria is significant for processes such as the formation of certain minerals, the manufacturing of chemicals, and various environmental processes.
In heterogeneous reactions, like those shown in this exercise, only the concentrations of gaseous components are used in the equilibrium expression. This is because the concentrations of solids remain constant and do not affect the dynamic balance of the reaction.

• Solid reactants and products, such as $ext{LiHCO}_3$, $ext{Li}_{2}ext{CO}_3$, and $extPbO$,. do not appear in the equilibrium expressions.
• Gaseous components like $ext{CO}_2$ and $ext{H}_{2}extO$ are crucial as their concentrations change with the progression of the reaction.

Thus, understanding which components to include is essential when working with heterogeneous equilibria.

Gaseous States

Gaseous state components play a critical role in the formation of equilibrium expressions through their variable concentrations. Unlike solids, gases can expand, contract, and interact significantly with their surroundings, which makes them crucial in calculating equilibrium constants. In any balanced chemical equation, it's important to point out the gases as they equalize between forward and reverse reactions, influencing the final equilibrium mixture.
In the examples given:

• Reaction a and b include gas molecules $ext{CO}_2$ contributing to the equilibrium constant.
• In reaction c, the oxygen gas $ext{O}_2$ influences the equilibrium state due to its gaseous nature and enters the expression as $[O_2{]}^3$.

This underscores the significance of identifying gaseous substances in an equilibrium reaction, as they determine the dynamic interplay of forces within the reaction.

Solid States

The contribution of solid-state substances in equilibrium reactions is unique in that they do not directly feature in the equilibrium expression. They act as either reactants or products but remain constant throughout the reaction in terms of concentration or activity. This constancy means they don't appear in the equilibrium expression calculations that include only gaseous or aqueous substances.
For the given examples:

• Solids such as $ext{LiHCO}_3,$ $ext{Li}_{2}ext{CO}_3,$ $ext{PbCO}_3,$ $extPbO,$ and $ext{Al}_{2}ext{O}_3$ are present.
• These substances are crucial for the reaction's structure but remain outside the equilibrium constant's mathematical formula.

The exclusion of solid states from the equilibrium expression simplifies calculating the equilibrium constant while ensuring the practical integrity of the reaction is maintained. This unique interplay is fundamental for accurately predicting and managing chemical reactions.