Problem 19 - Chemistry in 30 days

Question

For the reaction \(\mathrm{CaCO}_{3}(\mathrm{~s})=\mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\), the addition of more of \(\mathrm{CaO}(\mathrm{s})\) causes (a) decrease in the concentration of \(\mathrm{CO}_{2}(\mathrm{~g})\) (b) increase in the concentration of \(\mathrm{CO}_{2}(\mathrm{~g})\) (c) change in the concentration of \(\mathrm{CO}_{2}(\mathrm{~g})\) (d) increase in the concentration of \(\mathrm{CaCO}_{3}(\mathrm{~s})\)

Step-by-Step Solution

Answer

The concentration of \( \mathrm{CO}_2(\mathrm{~g}) \) does not change; thus, the correct answer is (c).

Step 1: Understand the reaction

The provided reaction is the decomposition of calcium carbonate \( \mathrm{CaCO}_3 \) into calcium oxide \( \mathrm{CaO} \) and carbon dioxide \( \mathrm{CO}_2 \). This is a solid to solid and gas reaction: \( \mathrm{CaCO}_3(\mathrm{s}) = \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}) \).

Step 2: Consider the impact of adding more \( \mathrm{CaO} \)

Adding more \( \mathrm{CaO} \) to the system does not affect the concentration of \( \mathrm{CO}_2 \) or \( \mathrm{CaCO}_3 \) when \( \mathrm{CaO} \) is added in solid form because solids do not appear in the equilibrium constant expression.

Step 3: Apply Le Chatelier's Principle

According to Le Chatelier's Principle, the addition of a pure solid like \( \mathrm{CaO} \) does not shift the position of the equilibrium because only the concentrations of gaseous and aqueous species appear in equilibrium expressions. Thus, the concentration of \( \mathrm{CO}_2 \) remains unchanged.

Step 4: Interpret the result

Since the addition of \( \mathrm{CaO}(\mathrm{s}) \) does not change the equilibrium position, the concentration of \( \mathrm{CO}_2 \) does not increase or decrease. Hence, option (c) is correct: there is no change in the concentration of \( \mathrm{CO}_2(\mathrm{~g}) \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant over time, although they are not necessarily equal. This state can be altered by changing certain conditions such as concentration, pressure, or temperature, which affects how the equilibrium shifts according to Le Chatelier's Principle. However, it's important to note that pure solids or liquids do not appear in the equilibrium constant expression, so altering their amounts will not shift the equilibrium.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances. This can involve breaking down into elements or simpler compounds. In the context of the given reaction, calcium carbonate ( abla aCO_3) decomposes into calcium oxide ( abla aO) and carbon dioxide ( abla rO_2). Generally, decomposition reactions require energy in the form of heat, light, or electricity to occur.

• For example, when calcium carbonate decomposes, it is often heated to drive the reaction.
• This specific type of decomposition reaction is common in the lime industry as raw material ( abla ome.g., limestone) is transformed into quicklime.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a numerical value that characterizes the ratio of product concentrations to reactant concentrations at equilibrium, emphasizing only those that appear in solution or gas phases. It is determined specifically for every chemical reaction at a given temperature. The equilibrium constant expression involves the concentrations of the products and reactants raised to the power of their respective coefficients in the balanced reaction equation. For example, in a reaction such as \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant \(K\) is expressed as: \[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\] In this equation, \(\mathrm{[X]}\) denotes the concentration of \(X\). Since solids and pure liquids do not appear in this expression, adding or removing a solid will not affect the equilibrium constant or the equilibrium position.

Effect of Solids on Equilibrium

Understanding the role of solids in a chemical equilibrium is crucial as it determines how certain changes will or will not affect a reaction. In gaseous or aqueous phase reactions, changes in concentration or pressure can shift the equilibrium. However, for reactions involving solids, like the decomposition of calcium carbonate, the amount of solid does not impact the equilibrium position. This is because solids have fixed concentrations that do not change, as their concentration is derived from their density, which remains constant. Therefore, adding more of a solid like calcium oxide ( abla aO) will not change the equilibrium state or the concentrations of other components in the mixture—specifically, the gas concentration will stay the same, due to the presence and behavior of the solid components remaining outside the equilibrium expression.