Problem 145 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{K}_{\mathrm{p}}\) has the value of \(10^{-6} \mathrm{~atm}^{3}\) and \(10^{-4} \mathrm{~atm}^{3}\) at \(298 \mathrm{~K}\) and \(323 \mathrm{~K}\) respectively for the reaction \(\mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta_{r} \mathrm{H}^{\circ}\) for the reaction is: (a) \(7.7 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-147.41 \mathrm{~kJ} / \mathrm{mol}\) (c) \(147.41 \mathrm{~kJ} / \mathrm{mol}\) (d) None of these

Step-by-Step Solution

Answer

The correct answer is (c) 147.41 kJ/mol.

Step 1: Understanding the Reaction

The given reaction is \( \text{CuSO}_4 \cdot 3 \text{H}_2 \text{O}(s) \rightleftharpoons \text{CuSO}_4(s) + 3 \text{H}_2 \text{O}(g) \). It involves solid hydrated copper sulfate losing water molecules to form solid copper sulfate and water vapor.

Step 2: Determine Given Values

The equilibrium constants \( \mathrm{K}_\mathrm{p} \) are provided as \( 10^{-6} \text{ atm}^3 \) at \( 298 \text{ K} \) and \( 10^{-4} \text{ atm}^3 \) at \( 323 \text{ K} \). We need to find \( \Delta_r H^{\circ} \).

Step 3: Use van't Hoff Equation

The van't Hoff equation is used to relate the change in temperature and equilibrium constant to the change in enthalpy: \[\ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta_r H^{\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]where \( K_1 = 10^{-6} \), \( K_2 = 10^{-4} \), \( T_1 = 298 \text{ K} \), \( T_2 = 323 \text{ K} \), and \( R = 8.314 \text{ J/mol K} \).

Step 4: Calculate \( \ln \left( \frac{K_2}{K_1} \right) \)

Calculate the ratio of \( K_2 \) and \( K_1 \) and find the natural logarithm:\[\ln \left( \frac{10^{-4}}{10^{-6}} \right) = \ln(10^2) = 2 \ln(10) = 2 \times 2.303 = 4.606\]

Step 5: Substitute and Solve for \( \Delta_r H^{\circ} \)

Substitute the values into the van't Hoff equation and solve for \( \Delta_r H^{\circ} \):\[4.606 = -\frac{\Delta_r H^{\circ}}{8.314} \left( \frac{1}{323} - \frac{1}{298} \right) \]Calculate the temperature difference:\[\frac{1}{323} - \frac{1}{298} = -0.000276\]Then solve for \( \Delta_r H^{\circ} \):\[\Delta_r H^{\circ} = - \frac{4.606 \times 8.314}{-0.000276} = 147416 \text{ J/mol} \approx 147.41 \text{ kJ/mol}\]

Step 6: Conclusion about \( \Delta_r H^{\circ} \)

The calculated \( \Delta_r H^{\circ} \) is approximately \( 147.41 \text{ kJ/mol} \). Comparing this value with the options provided, option (c) is correct.

Key Concepts

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \( K_p \), is a critical concept in chemical thermodynamics. It provides information about the relative concentrations of products and reactants in a chemical reaction at equilibrium, specifically in terms of partial pressures for reactions involving gases. The value of \( K_p \) can change depending on the temperature, as different temperatures can shift the balance between the forward and reverse reactions.

For the reaction \( \mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \), the equilibrium constant \( K_p \) was given as \( 10^{-6} \text{ atm}^3 \) at \( 298 \text{ K} \) and \( 10^{-4} \text{ atm}^3 \) at \( 323 \text{ K} \).

This increase in \( K_p \) with temperature indicates that the reaction favors the formation of products as the temperature rises. This behavior suggests an endothermic reaction, where heat is absorbed, shifting the equilibrium towards the side with more gas molecules.

van't Hoff Equation

The van't Hoff equation is a fundamental tool in chemical thermodynamics, used to determine how the equilibrium constant \( K \) changes with temperature. This equation relates the change in temperature to the change in \( K \) and provides insight into the enthalpy change \( \Delta_r H^{\circ} \).

The equation is expressed as: \[\ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta_r H^{\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\]
Where:

• \( K_1 \) and \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \) respectively.
• \( R \) is the ideal gas constant, \( 8.314 \text{ J/mol K} \).

This equation allows us to calculate the reaction enthalpy change when \( K \) values at different temperatures are known. In our context, the calculation uses shifts in \( K_p \) from \( 10^{-6} \text{ to } 10^{-4} \text{ atm}^3 \) over the temperature increase from \( 298 \text{ K} \text{ to } 323 \text{ K} \). This equation, hence, can help predict how reactions will behave under different thermal conditions, crucial for industrial processes and scientific research.

Reaction Enthalpy (ΔrH°)

The reaction enthalpy, denoted as \( \Delta_r H^{\circ} \), represents the change in enthalpy when a reaction occurs at standard conditions. It is a measure of the total energy change and indicates whether the reaction is endothermic or exothermic.

An endothermic reaction, as suggested by a positive \( \Delta_r H^{\circ} \), absorbs heat from the surroundings, whereas an exothermic reaction releases heat. The calculation we performed led to \( \Delta_r H^{\circ} = 147.41 \text{ kJ/mol} \), aligning with the concept of an endothermic process.

Understanding \( \Delta_r H^{\circ} \) is pivotal because it influences the temperature dependence of the reaction equilibrium. With the help of the van't Hoff equation and the provided \( K_p \) values, we calculated this enthalpy, confirming the reaction favors the production of gaseous water as temperature increases, demonstrating the entropic gains in producing more gas molecules.