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

Question

\(K_{\text {p }}\) 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

Option (c) \(147.41 \mathrm{~kJ/mol}\) is correct.

Step 1: Identify Known Variables

We know the equilibrium constant values \( K_{p} \) for two different temperatures: \( K_{p1} = 10^{-6} \mathrm{~atm}^{3} \) at \( T_{1} = 298 \mathrm{~K} \) and \( K_{p2} = 10^{-4} \mathrm{~atm}^{3} \) at \( T_{2} = 323 \mathrm{~K} \).

Step 2: Apply the Van't Hoff Equation

The Van't Hoff equation is used to relate the change in equilibrium constant with temperature, given by: \[ \ln \left( \frac{K_{p2}}{K_{p1}} \right) = -\frac{\Delta_{r} H^{\circ}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right) \] where \( R \) is the gas constant \( 8.314 \mathrm{~J/mol \cdot K} \).

Step 3: Calculate the Change in \( K_{p} \)

Calculate the natural logarithm of the ratio of the two \( K_{p} \) values: \[ \ln \left( \frac{10^{-4}}{10^{-6}} \right) = \ln (10^{2}) = 2 \]

Step 4: Calculate the Reciprocal Temperature Difference

Calculate the difference in the reciprocals of the two temperatures: \[ \frac{1}{T_{2}} - \frac{1}{T_{1}} = \frac{1}{323} - \frac{1}{298} \text{ in } \mathrm{K}^{-1} \]

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

Substitute the known values into the Van't Hoff equation: \[ 2 = -\frac{\Delta_{r} H^{\circ}}{8.314} \left( \frac{1}{323} - \frac{1}{298} \right) \] Simplifying the reciprocal difference: \( \frac{1}{323} - \frac{1}{298} \approx -3.04 \times 10^{-5} \). Solve for \( \Delta_{r} H^{\circ} \): \[ \Delta_{r} H^{\circ} = -2 \times 8.314 \times (-3.04 \times 10^{-5})^{-1} \approx 147410.5 \mathrm{~J/mol} \] Convert to kJ/mol \( \approx 147.41 \mathrm{~kJ/mol} \).

Step 6: Identify the Correct Option

Based on the calculation, \( \Delta_{r} H^{\circ} \approx 147.41 \mathrm{~kJ/mol} \), which corresponds to option (c).

Key Concepts

Equilibrium Constant

The equilibrium constant, commonly denoted as \( K \), is a fundamental concept in chemical thermodynamics, describing the balance point of a chemical reaction at a specific temperature. For reactions involving gases, the pressure-based equilibrium constant \( K_p \) is often used, and it provides insight into how favorably a reaction proceeds under set conditions.

For example, if you have a reaction involving the decomposition of hydrated copper sulfate, as demonstrated by the equation \( \mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2}\mathrm{O}(s) \rightleftharpoons \mathrm{CuSO}_{4}(s) + 3 \mathrm{H}_{2}\mathrm{O}(g) \), the equilibrium constant values \( K_{p} = 10^{-6} \mathrm{~atm}^{3} \) at \( 298 \; \mathrm{K} \) and \( 10^{-4} \mathrm{~atm}^{3} \) at \( 323 \; \mathrm{K} \) tell us important things:

• A small \( K_p \) value (such as \( 10^{-6} \)) indicates that the reaction favors the reactants, meaning less forward reaction occurs.
• A larger \( K_p \) at a higher temperature (like \( 10^{-4} \)) suggests a shift toward products, implying an endothermic reaction, absorbing heat.

Understanding \( K \) helps you predict the directionality of reactions and how they respond to changes in conditions.

Enthalpy Change

Enthalpy change, \( \Delta_r H^{\circ} \), is a critical parameter that provides insight into the energy landscape of a chemical reaction. It represents the heat absorbed or released during a reaction under standard conditions.

In the exercise, the calculated \( \Delta_r H^{\circ} \) for the dehydration of copper sulfate was determined to be \( 147.41 \; \mathrm{kJ/mol} \). This positive value indicates the endothermic nature of the reaction, meaning the process absorbs heat from the surroundings.

• Endothermic reactions require net input of energy to proceed; hence, they are favored at higher temperatures.
• The more positive the \( \Delta_r H^{\circ} \) value, the greater the heat energy required for the reaction.

Knowing \( \Delta_r H^{\circ} \) helps assess the feasibility of the reaction and the effect of temperature on the reaction's favorability.

Temperature Dependence

The temperature dependence of a chemical reaction is a fascinating aspect tied to the changes in the equilibrium constant \( K \) when temperature changes. This is precisely where the Van't Hoff equation shines.

The Van't Hoff equation: \[ \ln \left( \frac{K_{p2}}{K_{p1}} \right) = -\frac{\Delta_r H^{\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]shows how \( K \) relates to temperature for a given reaction. It suggests:

• An increase in temperature usually favors endothermic reactions, leading to a higher \( K \), as seen in the hydration of copper sulfate.
• Conversely, for exothermic reactions, higher temperature tends to decrease \( K \).

Applying this knowledge, scientists can predict how changing temperature will shift the equilibrium of a reaction, providing practical applications in industrial processes and laboratory settings.

Chemical Thermodynamics

Chemical thermodynamics serves as the backbone for understanding the energy and dynamics of chemical reactions. It melds the principles of energy conservation and the spontaneity of reactions.

In the context of the exercise, the Van't Hoff equation is a prime example of applying thermodynamic principles to relate equilibrium constants and enthalpy with temperature changes.

• Thermodynamics helps determine the feasibility and extent of reactions given specific conditions.
• It provides quantitative tools for assessing how much energy is exchanged in reactions.

Using these concepts, chemists can engineer reactions favorably, optimize conditions for desired outcomes, and understand the molecular basis of reaction energetics and mechanisms.