Question

At equilibrium, the pressure of the reacting mixture $$\mathrm{CaCO}_{3}(s) \rightleftarrows \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ is 0.105 atm at \(350^{\circ} \mathrm{C}\). Calculate \(K_{P}\) and \(K_{c}\) for this reaction.

Short answer

\(K_P = 0.105 \text{ atm}, K_c \approx 2.05 \times 10^{-3}.\)

Step-by-step solution

Identify the Reaction Components

The reaction involves solid calcium carbonate (\(\mathrm{CaCO}_{3}(s)\),) turning into solid calcium oxide (\(\mathrm{CaO}(s)\),) and gaseous carbon dioxide (\(\mathrm{CO}_{2}(g)\).) Only the gaseous component, \(\mathrm{CO}_{2}\), will affect the equilibrium constants because solids are not included in the equilibrium expression.

Determine the Equilibrium Pressure

From the problem, we know the equilibrium pressure of \(\mathrm{CO}_{2}\) is 0.105 atm. This will be used directly to calculate \(K_P\), since the equilibrium constant in terms of pressure is defined by the partial pressures of the gaseous components.

Calculate the Equilibrium Constant in Terms of Pressure

The equilibrium constant \(K_P\) is given by the partial pressure of \(\mathrm{CO}_{2}(g)\). Therefore, \[K_P = P_\mathrm{CO_2} = 0.105 \text{ atm}.\]

Relate Kp and Kc

The relationship between \(K_P\) and \(K_c\) is given by the equation: \[ K_P = K_c (RT)^{\Delta n}, \] where \(R = 0.0821 \text{ L atm/mol K},\) and \(T\ =\ 350^{\circ}C = 623 K.\) Here, \(\Delta n = \text{moles of products} - \text{moles of reactants} = 1 - 0 = 1.\). Thus, the conversion formula becomes: \[ K_c = \frac{0.105}{(0.0821 \times 623)}. \]

Calculate the Equilibrium Constant in Terms of Concentration

Solve for \(K_c\) using the derived formula: \[ K_c = \frac{0.105}{(0.0821 \times 623)} \approx 2.05 \times 10^{-3}. \]

Key concepts

Equilibrium Constant (Kp and Kc)

In chemical equilibrium, the equilibrium constant is a crucial concept. It measures the ratio of concentrations or pressures of products to reactants at equilibrium. There are two types: \(K_p\) and \(K_c\).

**Kp** is the equilibrium constant calculated using partial pressures of gases. It applies to gaseous equilibria and is particularly useful when reactions involve changes in the number of moles of gas.

To derive \(K_p\), only gaseous reactants and products are considered. In the reaction \(\text{CaCO}_3 (s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2 (g)\), \(\text{CO}_2\) is the only gas. Therefore, \(K_p\) is simply equal to the pressure of \(\text{CO}_2\), which is given as 0.105 atm in the exercise.

**Kc** is the equilibrium constant based on concentrations. Unlike \(K_p\), it uses molarity for liquids and solutions. When converting between \(K_p\) and \(K_c\), the formula is:
\[ K_p = K_c (RT)^{\Delta n} \]
where \(R\) is the gas constant, \(T\) is temperature in Kelvin, and \(\Delta n\) is the change in moles of gas. This allows us to convert from \(K_p\) at 0.105 atm to \(K_c\) using the conditions provided, resulting in \(K_c \approx 2.05 \times 10^{-3}\).

• \(K_p\) and \(K_c\) highlight the dynamic nature of equilibrium in gases versus solutions.
• The simplicity of the relationship helps in moving between gas pressures and concentrations easily.

Gas Pressure Calculations

Understanding gas pressures is key to calculating equilibrium constants. Here, the total pressure of a gas in a reaction contributes to \(K_p\).

Gases exert pressure due to their molecules hitting the container's walls. At equilibrium, this happens at a balance between opposing reactions. For \(\text{CaCO}_3 (s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2 (g)\), only \(\text{CO}_2\) contributes to \(K_p\), calculated using its partial pressure. In this case, it was directly provided as 0.105 atm.

Performing such calculations involves:

• Identifying which components are gaseous.
• Assessing partial pressures of these gases.
• Using these pressures in the equilibrium constant expression.

Gas-related reactions are dynamic and their calculations often hinge on the conditions such as temperature and the presence of gases in the mixture.

Chemical Thermodynamics

Chemical thermodynamics underpins the understanding of equilibria. This branch of chemistry deals with energy changes during chemical reactions.

At equilibrium, reactions appear static but are dynamically balanced at the molecular level. Thermodynamics introduces the concepts of enthalpy, entropy, and Gibbs free energy, which determine how and why reactions proceed.

The equation \( K_p = K_c (RT)^{\Delta n} \) links thermodynamics and equilibrium constants. Here:

• \(R\) is the universal gas constant.
• \(T\) is temperature, pivotal as it affects both reaction rate and equilibrium position.
• \(\Delta n\) (change in moles of gas) shows reactions often involve shifts in moles between products and reactants.

This relationship highlights how thermal conditions (through \(R\) and \(T\)) impact equilibrium. Understanding this is essential for predicting how changes in conditions might shift reaction equilibria. Thus, chemical thermodynamics provides a lens to predict and rationalize how reactions reach balance.