Question

Heating limestone (CaCO \(_{3}(s) )\) forms quicklime \((\mathrm{CaO}(\mathrm{s}))\) and carbon dioxide gas. Write the equilibrium constant expression for this reversible reaction.

Short answer

The equilibrium constant expression for the reversible reaction of heating limestone (CaCO$_{3}$(s)) to form quicklime (CaO(s)) and carbon dioxide gas (CO$_{2}$(g)) is: \( K_{c} = [\mathrm{CO_{2}(g)}] \)

Step-by-step solution

Write the balanced chemical equation for the given reaction.

The balanced chemical equation for the given reversible reaction is: \[ \mathrm{CaCO_{3}(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_{2}(g)} \]

Write the equilibrium constant expression for the reaction.

Based on the balanced chemical equation, we can write the equilibrium constant expression using the concentrations of the species involved in the reaction. Recall that only gases are included in the expression. Therefore, we will have: \[ K_{c} = \frac{[\mathrm{CO_{2}(g)}]}{[\mathrm{CaCO_{3}(s)}]} \] However, since the concentration of solids do not change during the reaction, they are not included in the equilibrium constant expression. Thus, the final equilibrium constant expression is given by: \[ K_{c} = [\mathrm{CO_{2}(g)}] \] This is the equilibrium constant expression for the reversible reaction of heating limestone to form quicklime and carbon dioxide gas.

Key concepts

Reversible Reactions

Understanding reversible reactions is crucial in chemistry. A reversible reaction is a chemical reaction that can progress in both forward and backward directions. This means that the products can reform the original reactants under certain conditions. Reversible reactions eventually reach a point of equilibrium, where the rate of the forward reaction equals the rate of the reverse reaction.

In the context of the original exercise, the decomposition of calcium carbonate to form calcium oxide and carbon dioxide is reversible. This means that under certain conditions, calcium oxide and carbon dioxide can react to reform calcium carbonate. At this balance point, known as dynamic equilibrium, the concentrations of reactants and products remain constant over time. It's this principle that allows chemists to predict and calculate the position of equilibrium using an equilibrium constant.

• Equilibrium can shift with changes in conditions like temperature and pressure.
• The reverse reaction must be feasible under the conditions being evaluated.
• Understanding equilibrium involves recognizing that it doesn't mean equal concentrations of reactants and products, but rather constant concentrations.

Balanced Chemical Equation

A balanced chemical equation is the starting point for understanding any chemical reaction, including reversible ones. It depicts the reactants and products in their stoichiometric ratios, which ensures the law of conservation of mass is followed.

In the exercise scenario, the decomposition of calcium carbonate is represented by the balanced chemical equation: \[ \mathrm{CaCO_{3}(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_{2}(g)} \]

This equation shows one molecule of calcium carbonate yielding one molecule of calcium oxide and one molecule of carbon dioxide. It’s essential to have a balanced equation to correctly write the equilibrium constant expression.

• All reactants and products in their correct proportions.
• Indicating phases (s for solid, g for gas) is also important as it influences how we write equilibrium expressions.
• The arrow symbolizes the reversible nature of the reaction, indicating equilibrium can be established.

Concentration Terms

Concentration terms are key to constructing an equilibrium constant expression. Concentration is typically represented using molarity, denoted by brackets around a chemical species (e.g., \([\mathrm{CO_2(g)}]\)).

For the decomposition of calcium carbonate, the equilibrium constant expression, known as \( K_c \), is formulated based on these concentration terms. Importantly, only the concentrations of gases and aqueous solutions are included in \( K_c \). Therefore, despite being part of the reaction, the concentrations of solids like \( \mathrm{CaCO_3(s)} \) and \( \mathrm{CaO(s)} \) are excluded from the expression. This simplifies to:

\[ K_{c} = [\mathrm{CO_{2}(g)}] \]

This reflects that the equilibrium position for this reaction is dependent solely on the concentration of the carbon dioxide gas, a product of the reaction.

• Concentration of solid materials remains constant and doesn't affect \( K_c \).
• The expression reflects the ratio of concentrations of products and reactants at equilibrium.
• Understanding these concepts is crucial for solving equilibrium problems.