Problem 144 - A Complete Resource Book in Chemistry for JEE Main

Question

\(K_{p}\) 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 \(\mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta_{r} H^{\circ}\) 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

We are given the reversible reaction: \( \mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \) with equilibrium constants \( K_{p} \) for different temperatures. We need to find \( \Delta_{r} H^{\circ} \) (the enthalpy change for the reaction).

Step 2: Applying Van't Hoff Equation

The Van't Hoff equation, \( \ln \left( \frac{K_{2}}{K_{1}} \right) = -\frac{\Delta_{r} H^{\circ}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right) \), relates the change in the equilibrium constant with temperature to the enthalpy change of the reaction. We will use \( K_{1} = 10^{-6} \mathrm{~atm}^{3} \), \( K_{2} = 10^{-4} \mathrm{~atm}^{3} \), \( T_{1} = 298 \mathrm{~K} \), and \( T_{2} = 323 \mathrm{~K} \).

Step 3: Solving for Enthalpy Change

Rearrange the Van't Hoff equation to solve for \( \Delta_{r} H^{\circ} \):\[\Delta_{r} H^{\circ} = -R \left(\frac{\ln(K_{2}/K_{1})}{(1/T_{2})-(1/T_{1})}\right)\]Using \( R = 8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \) and solving:\[\ln\left( \frac{10^{-4}}{10^{-6}} \right) = \ln(10^{2}) = 2\ln(10)\]\[\Delta_{r} H^{\circ} = -8.314 \left( \frac{2 \cdot 2.302}{\frac{1}{323} - \frac{1}{298}} \right)\]Calculate each step to find \( \Delta_{r} H^{\circ} \).

Step 4: Numerical Calculations

Compute \( \frac{1}{323} - \frac{1}{298} = \frac{298 - 323}{323 \times 298} \approx -2.35 \times 10^{-4} \mathrm{~K}^{-1} \).Compute \( 2 \cdot 2.302 \approx 4.604 \).Substitute into the enthalpy equation:\[\Delta_{r} H^{\circ} = -8.314 \left( \frac{4.604}{-2.35 \times 10^{-4}} \right) \approx 147410 \mathrm{~J} / \mathrm{mol} = 147.41 \mathrm{~kJ} / \mathrm{mol} \].

Step 5: Selecting the Correct Answer

Compare the calculated \( \Delta_{r} H^{\circ} = 147.41 \mathrm{~kJ} / \mathrm{mol} \) with the options. The correct choice is (c) \(147.41 \mathrm{~kJ} / \mathrm{mol}\).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_p \) in the context of reactions involving gases, provides valuable information about the ratio of products to reactants at equilibrium. In the given 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 determines the balance between hydrated copper sulfate and its gaseous components at equilibrium.

\( K_p \) values signify the pressure of the gaseous water in equilibrium with the solid reactants:

• At different temperatures, \( K_p \) varies; for instance, here \( K_p \) is \( 10^{-6} \ \mathrm{atm}^3 \) at 298 K and \( 10^{-4} \ \mathrm{atm}^3 \) at 323 K.
• The change indicates that the equilibrium shifts towards producing more gaseous water as temperature increases.

Understanding \( K_p \) helps predict how changes in conditions affect the reaction's position, crucial for controlling industrial processes.

Temperature Dependence of Reaction

Temperature is a critical factor that influences chemical equilibrium. The Van't Hoff equation specifically provides a mathematical approach to understand this dependence.

This equation shows that as temperature varies, the equilibrium constant \( K_p \) changes, reflecting a shift in the equilibrium position.

• For endothermic reactions like the dissociation of \( \mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O} \), an increase in temperature often results in a higher \( K_p \), favoring the formation of products.
• In contrast, for exothermic reactions, increasing temperature usually decreases \( K_p \), pushing the equilibrium towards reactants.

By using the Van't Hoff equation, \( \ln \left( \frac{K_{2}}{K_{1}} \right) = -\frac{\Delta_{r} H^{\circ}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right) \), we can calculate the change in \( K_p \) with temperature and explore the nature of the reaction.

Enthalpy Change of Reaction

The enthalpy change, \( \Delta_{r} H^{\circ} \), reveals whether a reaction absorbs or releases heat. For any given chemical reaction, understanding \( \Delta_{r} H^{\circ} \) offers insights into the thermal characteristics of the process:

• In the provided exercise, the positive value of \( \Delta_{r} H^{\circ} = 147.41 \ \mathrm{kJ/mol} \) indicates the reaction is endothermic, meaning it requires heat to proceed.
• This is reflected in the shift of equilibrium toward more product formation as temperature increases, consistent with Le Chatelier's Principle.
• Solving the Van't Hoff equation gives this value, reinforcing the relationship between temperature, \( K_p \), and enthalpy.
• It's also key in designing industrial processes, like optimizing conditions for desired product yields.

Overall, knowing \( \Delta_{r} H^{\circ} \) not only aids in understanding reaction spontaneity but also assists in predicting temperature effects on equilibrium.