Problem 16 - Introductory Chemistry Atoms First

Question

Write the equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(s) \leftrightarrows \mathrm{CaO}(s)+\) \(\mathrm{CO}_{2}(g)\)

Step-by-Step Solution

Answer

The equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(s) \leftrightarrows \mathrm{CaO}(s)+ \mathrm{CO}_{2}(g)\) is given by: \[K = [\mathrm{CO}_{2}]\]

Step 1: Identify the gas-phase species

In this reaction, we have only one gas-phase species that is \(\mathrm{CO}_{2}(g)\). So, we will include only this in our equilibrium constant expression.

Step 2: Write the equilibrium constant expression

The equilibrium constant expression K is given by: \[K = \frac{[\mathrm{CO}_{2}]^{1}}{1}\] Here, we have used the concentration of \(\mathrm{CO}_{2}(g)\) raised to the power of its stoichiometric coefficient 1 (assuming one mole each of reactants and products) in the numerator. We don't include \(\mathrm{CaCO}_{3}(s)\) and \(\mathrm{CaO}(s)\) in the expression, as their concentrations do not change during the reaction. So, the equilibrium constant expression for the given reaction is: \[K = [\mathrm{CO}_{2}]\] The equilibrium constant K will give information about the concentration of \(\mathrm{CO}_{2}\) at equilibrium for different temperatures.

Key Concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes the state of a reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. It is a dynamic equilibrium, meaning both reactions continue to occur, but because they’re at the same rate, the overall concentrations remain constant.

Crucially, equilibrium does not mean that the reactants and products are present in equal amounts. Instead, it indicates that their ratios are stable. The position of equilibrium is represented by the equilibrium constant \( K \), which is a ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. In the exercise with the decomposition of \(\mathrm{CaCO}_{3}(s)\), the equilibrium constant expression helps us understand how much \(\mathrm{CO}_{2}(g)\) will be present at equilibrium at a given temperature.

To experimentally determine \( K \), one would typically measure the concentrations of the gaseous products and reactants at equilibrium. Then, these values are plugged into the equilibrium expression to solve for the equilibrium constant. This constant is crucial in predicting how the system will react to changes in conditions, such as changes in concentration, pressure, or temperature, often illustrated through Le Chatelier's Principle.

Gaseous Reactions and Equilibrium

Gaseous reactions such as the decomposition of \(\mathrm{CaCO}_{3}(s)\) to form \(\mathrm{CaO}(s)\) and \(\mathrm{CO}_{2}(g)\) are particularly interesting in the context of equilibrium. The behavior of gases under changing conditions is governed by the ideal gas law, which relates the pressure, volume, temperature, and moles of a gas. At equilibrium, the partial pressure of the gaseous products and reactants is used in the equilibrium expression.

For a reaction involving gases, the equilibrium constant can be expressed in terms of partial pressures (usually designated as \( K_p \)) or concentrations (designated as \( K_c \)). In the exercise, only \(\mathrm{CO}_{2}(g)\) is a gas, and its concentration is used in the equilibrium expression. In real-world applications, this information is vital because it can predict how changing the pressure (by changing the volume or the amount of gas) affects the system's position of equilibrium, and consequently, the yields of a chemical process.

Stoichiometry and the Equilibrium Constant

Stoichiometry is the branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. It is essential for calculating the equilibrium constant because the stoichiometric coefficients from the balanced chemical equation are used as exponents in the equilibrium expression. In our example with the reaction of \(\mathrm{CaCO}_{3}(s)\), the stoichiometric coefficient for \(\mathrm{CO}_{2}(g)\) is 1, reflecting the mole ratio in which \(\mathrm{CO}_{2}(g)\) is produced.

When writing the equilibrium constant expression, solid and liquid reactants and products (like \(\mathrm{CaCO}_{3}(s)\) and \(\mathrm{CaO}(s)\) in our case) are typically omitted because their concentrations are constant and do not affect the equilibrium. Only species in the gaseous or aqueous states are included. Stoichiometry is also employed when using the equilibrium constant to calculate unknown concentrations at equilibrium. If one value is known, stoichiometry allows for the calculation of the others, keeping in mind the mole ratios established by the balanced chemical equation.