Question

For the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ the equilibrium constant \(K\) has the form \(K=\left[\mathrm{CO}_{2}\right]\) Using a handbook to find density information about \(\mathrm{CaCO}_{3}(s)\) and \(\mathrm{CaO}(s),\) show that the concentrations of the two solids (the number of moles contained in 1 L of volume) are constant.

Short answer

Using a handbook, we find the molar mass and densities of CaCO3 and CaO: CaCO3: Molar mass: 100.09 g/mol Density: \(2.71\,g/cm^3\) CaO: Molar mass: 56.08 g/mol Density: \(3.34\,g/cm^3\) Next, we calculate the volume per mole and determine the concentrations: \(Concentration_{CaCO3} = 0.027\,mol/L\) \(Concentration_{CaO} = 0.0595\,mol/L\) Thus, the concentrations of both solids are constant, as they do not change with external conditions.

Step-by-step solution

Find density information from the handbook

Look up the molar mass and density of CaCO3 and CaO in a handbook. The molar mass tells us the grams of substance present per mole, and the density represents the mass of the substance per unit volume.

Calculate Volume

For each compound, calculate the volume of the solid by dividing the molar mass by its density. Here, we found the molar mass and densities of CaCO3 and CaO: CaCO3: Molar mass: 100.09 g/mol Density: \(2.71\, g/cm^3\) CaO: Molar mass: 56.08 g/mol Density: \(3.34\, g/cm^3\) Now, calculate the volume: \(Volume_{CaCO3} = \frac{100.09\,g/mol}{2.71\,g/cm^3} = 36.9\,cm^3/mol\) \(Volume_{CaO} = \frac{56.08\,g/mol}{3.34\,g/cm^3} = 16.8\,cm^3/mol\) Notice that the calculated volumes are per mole of each substance.

Determine the concentration

The concentration is defined as the number of moles of a substance per unit volume. Since we have already calculated the volume occupied by one mole of both substances, we can now determine the concentration: \(Concentration_{CaCO3} = \frac{1\,mol}{36.9\,cm^3} \cdot \frac{1\,L}{1000\,cm^3} = 0.027\,mol/L\) \(Concentration_{CaO} = \frac{1\,mol}{16.8\,cm^3} \cdot \frac{1\,L}{1000\,cm^3} = 0.0595\,mol/L\) These concentrations are constants as they do not change with changes in temperature, pressure, or other external conditions. They are specific to the substances CaCO3 and CaO.

Conclusion

Based on the calculations from the density information of CaCO3 and CaO obtained from a handbook, we have shown that the concentrations of both solids (the number of moles contained in 1 L of volume) are constant.

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a crucial concept in chemical reactions where reversible changes take place. For the reaction \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \), the equilibrium constant is represented by the concentration of \( \text{CO}_2 \) in the system. This is because the equilibrium constant only includes the concentrations of gases and aqueous solutions.
Solids do not appear in the equilibrium constant expression because their concentrations are unchanging due to the constant density and molar volume under given conditions. Understanding this helps explain why only \( \text{CO}_2 \) concentration is used in the equilibrium expression for this system. In essence, the equilibrium constant gives insight into the ratio of products to reactants at equilibrium for reversible reactions.

Concentration of Solids

When dealing with solids like \( \text{CaCO}_3 \) and \( \text{CaO} \), their concentrations remain constant in a chemical equilibrium equation. This constancy is rooted in the idea that solids' densities are stable over a range of temperatures and pressures. Since their densities do not fluctuate significantly, neither does the volume they occupy per mole.
This stability allows chemists to omit solids from the equilibrium constant calculations. The main point is that for solids, the number of moles per unit volume—and thus their concentration—is a fixed value, unaffected by the conditions that might change the concentrations of gases or solutions.

The calculated concentrations of \( \text{CaCO}_3 \) and \( \text{CaO} \) through handbook data illustrate this constancy. By dividing their molar mass by their density, we obtain a volume that stays the same over time, solidifying their role as constants in equilibrium equations.

Molar Mass and Density

Molar mass and density are fundamental to understanding how substances are measured and quantified in chemistry. Molar mass is the weight of one mole of a substance expressed in grams per mole (g/mol), representing how much one mole of that substance weighs. For instance, the molar mass for \( \text{CaCO}_3 \) and \( \text{CaO} \) are 100.09 g/mol and 56.08 g/mol, respectively.

Density, on the other hand, is defined as the mass per unit volume, typically in grams per cubic centimeter (g/cm³). In the given reaction, \( \text{CaCO}_3 \) and \( \text{CaO} \) have densities of 2.71 g/cm³ and 3.34 g/cm³ respectively. Knowing both the molar mass and density allows you to determine how much space a mole of a given substance will occupy. This is crucial for calculating the concentration of solids in chemical reactions.

Calculation of Concentration

Calculating the concentration of a substance involves determining how many moles are present in a given volume. For solids, this calculation incorporates the use of molar mass and density. To find concentration, you first determine the volume occupied by one mole by dividing molar mass by density.
Converting that value to liters (since concentration is typically expressed in moles per liter) and inverting it will yield the number of moles per liter: the concentration. For example, \( \text{CaCO}_3 \) has a computed volume of 36.9 cm³/mol and a concentration of 0.027 mol/L.

• Start by dividing the molar mass by density: \( \frac{100.09 \text{ g/mol}}{2.71 \text{ g/cm}^3} \)
• Convert the volume to liters: \( 1 \text{ L} = 1000 \text{ cm}^3 \)
• Final concentration: \( \frac{1 \text{ mol}}{36.9 \text{ cm}^3} \cdot \frac{1 \text{ L}}{1000 \text{ cm}^3} = 0.027 \text{ mol/L} \)
  With these steps, you can determine the constant concentration of solids in the reaction, confirming their unchanging nature under standard conditions.