Chapter 7: Problem 29
For the reaction \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4} .\) \(3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), which one is the correct repre- sentation? (1) \(\mathrm{K}_{\mathrm{P}}=\mathrm{P}_{\mathrm{H}_{2} \mathrm{O}}^{2}\) (2) \(\mathrm{K}_{\mathrm{C}}=\left[\mathrm{II}_{2} \mathrm{O}\right]^{2}\) (3) \(\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}|R T|^{2}\) (4) \(\Lambda \mathrm{ll}\)
Short Answer
Expert verified
Option 1 is correct: \(\mathrm{K}_{\mathrm{P}} = \mathrm{P}_{\mathrm{H}_{2} \mathrm{O}}^{2}\).
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chemical Equilibrium
Chemical equilibrium is the state in which both the reactants and products of a chemical reaction are present in concentrations that have no net change over time. In other words, the forward and reverse reactions occur at the same rate. At equilibrium, the system's macroscopic properties, such as pressure, temperature, and concentration, remain constant. Consider the reaction: \[ \text{CuSO}_4 \text{ · 5H}_2\text{O}(\text{s}) \rightleftharpoons \text{CuSO}_4 \text{ · 3H}_2\text{O}(\text{s}) + 2 \text{H}_2\text{O}(\text{g}) \]Here, individual molecules continue to react in both the forward and reverse directions, but because these rates are equal, no net change in the concentration of reactants and products takes place. Achieving equilibrium does not mean the reactants and products are equal in concentration—it simply means their concentrations have become constant. Factors such as temperature and pressure can shift the equilibrium position, leading to a new balance point.
Equilibrium Expressions
An equilibrium expression is a mathematical equation representing the relationship between the concentrations or partial pressures of the reactants and products in a reaction at equilibrium. For reactions involving gases, we often use partial pressures to express this relationship. The equilibrium constant, denoted as \(K\), can be expressed in terms of concentrations (\(K_C\)) or partial pressures (\(K_P\)). For the reaction given: \[ \text{CuSO}_4 \text{ · 5H}_2\text{O} (\text{s}) \rightleftharpoons \text{CuSO}_4 \text{ · 3H}_2\text{O} (\text{s}) + 2 \text{H}_2\text{O} (\text{g}) \]we write the equilibrium expression considering only the gaseous part since solids have constant concentration that do not appear in equilibrium expressions. Therefore, the equilibrium constant in terms of partial pressures (\(K_P\)) is: \[ K_P = (P_{\text{H}_2O})^2 \]where \(P_{\text{H}_2O}\) is the partial pressure of water vapor. This expression tells us how the equilibrium position will change with a change in the partial pressure of the products.
Partial Pressure
Partial pressure represents the pressure that a single gas in a mixture would exert if it occupied the entire volume by itself. It's essential for reactions involving gases, as in our equilibrium example. The total pressure in a container can be imagined as the sum of the partial pressures of each gas present. Using the ideal gas law, partial pressure can be expressed as: \[ P = \frac{nRT}{V} \]where \(n\) is the number of moles of the gas, \(R\) is the universal gas constant, \(T\) is the temperature, and \(V\) is the volume.In chemical equilibrium, the partial pressures of reactants and products determine the position of equilibrium as per the equilibrium expression. In the given reaction, \[ \text{CuSO}_4 \text{ · 5H}_2\text{O} (\text{s}) \rightleftharpoons \text{CuSO}_4 \text{ · 3H}_2\text{O} (\text{s}) + 2 \text{H}_2\text{O} (\text{g}) \] the partial pressure of water vapor (\(P_{\text{H}_2O}\)) plays a crucial role in defining the equilibrium constant \(K_P\). Hence, knowing how to calculate and use partial pressures is essential for analyzing chemical equilibria involving gases.
