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 \(1 \mathrm{~L}\) of volume) are constant.

Short answer

The concentrations of the two solids, Calcium Carbonate (CaCO3) and Calcium Oxide (CaO), in the reaction are constant. Using density information from a handbook, we found that the concentration of CaCO3 is \(27.09\frac{mol}{L}\) and the concentration of CaO is \(59.55\frac{mol}{L}\).

Step-by-step solution

Calculate mass of 1 L of each substance

To find the mass of each substance, we will first find the respective density information in a reference handbook. Given densities: Calcium Carbonate, CaCO3: \(2.71 \frac{g}{cm^3}\) or \(2710\frac{g}{L}\) (converted to the mass of 1 L) Calcium Oxide, CaO: \(3.34 \frac{g}{cm^3}\) or \(3340\frac{g}{L}\) (converted to the mass of 1 L)

Find the molar mass of each substance

To find the moles contained in 1 L, we will need the molar mass of each substance, which can be calculated using the atomic weights of the elements found in a periodic table. Molar mass of CaCO3: \(40.08\frac{g}{mol}(\mathrm{Ca}) + 12.01\frac{g}{mol}(\mathrm{C}) + 3 \times 16.00\frac{g}{mol}(\mathrm{O}) = 100.09 \frac{g}{mol}\) Molar mass of CaO: \(40.08\frac{g}{mol}(\mathrm{Ca}) + 16.00\frac{g}{mol}(\mathrm{O}) = 56.08 \frac{g}{mol}\)

Calculate the number of moles of each substance per 1 L of volume

To find the number of moles per 1 L of volume, we will divide the mass of 1 L of each substance by their respective molar mass. Moles of CaCO3 in 1 L: \(\frac{2710\frac{g}{L}}{100.09\frac{g}{mol}} = 27.09\frac{mol}{L}\) - This value is constant for CaCO3. Moles of CaO in 1 L: \(\frac{3340\frac{g}{L}}{56.08\frac{g}{mol}} = 59.55\frac{mol}{L}\) - This value is constant for CaO.

Conclusion

The concentrations of the two solids, Calcium Carbonate (CaCO3) and Calcium Oxide (CaO), are 27.09 mol/L and 59.55 mol/L respectively, which are constant values. Therefore, the concentrations of the two solids in the reaction remain constant.

Key concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant, represented as \( K \), plays a critical role. It helps us determine the position of equilibrium for a reversible reaction. This specific constant is valuable because it is derived from the concentration of products divided by the concentration of reactants, each raised to the power of their stoichiometric coefficients.
In the reaction \( \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \), the equilibrium constant \( K \) is special because it includes only the concentration of \( \mathrm{CO}_{2}(g) \), the gaseous product. Why is that so? It's because in heterogeneous equilibria, solids and liquids are pure substances and their "concentration" doesn't change—they can be thought of as being at constant concentration and thus they don’t appear in the equilibrium expression.
Understanding this concept helps in predicting how changes in concentration, pressure, or temperature affect the equilibrium state in the system.

Solid Concentration

When dealing with reactions involving solids, like \( \mathrm{CaCO}_{3} \) and \( \mathrm{CaO} \), the concept of solid concentration becomes essential. Although we don't include solids in the equilibrium expression, understanding their concentration in terms of moles in a given volume is still critical.
For solids, we often look at their density and molar mass to determine concentration. This is because solids have a fixed volume and mass, leading to a constant concentration. As calculated in the solution, the number of moles of \( \mathrm{CaCO}_{3} \) and \( \mathrm{CaO} \) per liter remains constant as 27.09 mol/L and 59.55 mol/L respectively.
This constancy is why we don't include these concentrations in the equilibrium constant expression for the reaction. So, in heterogeneous equilibria, remembering that the concentration of solids is constant helps simplify how you approach equilibrium problems.

Molar Mass Calculation

Accurate molar mass calculations are fundamental in chemistry, especially when converting between grams and moles. This task is crucial when determining the concentration of substances.
To carry out a molar mass calculation, like the one needed for \( \mathrm{CaCO}_{3} \) and \( \mathrm{CaO} \), you add up the atomic masses of each component from the periodic table. For example, the molar mass of \( \mathrm{CaCO}_{3} \) is calculated by summing the atomic masses: 40.08 g/mol for Calcium (Ca), 12.01 g/mol for Carbon (C), and 3 multiplied by 16.00 g/mol for Oxygen (O), giving a total of 100.09 g/mol.
With the molar mass, you can then determine how many moles of a substance are present in a given mass by dividing the mass by the molar mass. This allows you to convert the density of a substance to its molar concentration, as seen in the calculation steps for both \( \mathrm{CaCO}_{3} \) and \( \mathrm{CaO} \). Understanding this process helps you solve equilibrium and other stoichiometry problems efficiently.