Problem 144 - 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 \(\mathrm{CuSO}_{A} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_{4}(\mathrm{~s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta \mathrm{H}^{\circ}\) 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 enthalpy change, \( \Delta H^\circ \), is approximately \( 147.41 \, \text{kJ/mol} \), matching option (c).

Step 1: Identify relevant equation

The equilibrium constant \(K_p\) varies with temperature according to the van't Hoff equation: \[ \ln \left(\frac{K_{p2}}{K_{p1}}\right) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where \(K_{p1}\) and \(K_{p2}\) are the equilibrium constants at temperatures \(T_1\) and \(T_2\) respectively.

Step 2: Convert Kp values to logarithmic form

Calculate the natural logarithm of the ratio \(\frac{K_{p2}}{K_{p1}}\):\[ \ln \left(\frac{10^{-4}}{10^{-6}}\right) = \ln(10^2) = 2 \]

Step 3: Plug values into van't Hoff equation

Substitute the known values into the van't Hoff equation:\[ 2 = -\frac{\Delta H^\circ}{R} \left( \frac{1}{323} - \frac{1}{298} \right) \]Use \( R = 8.314 \, \text{J/mol K} \).

Step 4: Calculate the temperature difference

Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \):\[ \frac{1}{323} - \frac{1}{298} = \frac{298 - 323}{323 \times 298} = -\frac{25}{96354} = -0.00025955 \, \text{K}^{-1} \]

Step 5: Solve for ΔH°

Rearrange the van't Hoff equation to solve for \( \Delta H^\circ \):\[ 2 = -\frac{\Delta H^\circ}{8.314} \times -0.00025955 \]\[ 2 = \frac{\Delta H^\circ}{8.314} \times 0.00025955 \]\[ \Delta H^\circ = \frac{2 \times 8.314}{0.00025955} = 640076.28 \, \text{J/mol} \approx 147.41 \, \text{kJ/mol} \]

Step 6: Determine the correct answer from options

The calculated \( \Delta H^\circ \approx 147.41 \, \text{kJ/mol}\). This matches option (c).

Key Concepts

Understanding Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a crucial concept in understanding chemical reactions at equilibrium. For reactions occurring in the gas phase, we use \( K_p \), which is the equilibrium constant expressed in terms of partial pressures. This constant helps us understand the extent of a reaction when it reaches a state where the rate of the forward reaction equals the rate of the reverse reaction.
The equilibrium constant value indicates the position of equilibrium.

• If \( K \) is large, the equilibrium lies towards the products, meaning the products are favored at equilibrium.
• If \( K \) is small, it indicates that reactants are favored at equilibrium.

Temperature changes affect the value of \( K \), which is why it’s crucial to specify the conditions under which the equilibrium constant is derived. This dependence of \( K_p \) on temperature is where the van't Hoff equation comes into play.

Basics of Thermodynamics in Chemical Reactions

Thermodynamics is the branch of chemistry that deals with the relations between heat and other forms of energy involved in chemical processes. In terms of chemical reactions, it helps us determine whether a reaction is spontaneous and how temperature changes affect it. Key concepts in thermodynamics include enthalpy (\( \Delta H \)), entropy (\( \Delta S \)), and Gibbs free energy (\( \Delta G \)).
For any chemical reaction:

• \( \Delta H \) (enthalpy): Represents the heat absorbed or evolved in a reaction at constant pressure. A positive \( \Delta H \) indicates an endothermic reaction, while a negative value indicates an exothermic reaction.
• \( \Delta S \) (entropy): Measures the disorder or randomness in the system. An increase in entropy favors spontaneity.
• \( \Delta G \) (Gibbs free energy): Determines the spontaneity of a reaction. A negative \( \Delta G \) indicates that a reaction is spontaneous under constant temperature and pressure.

Thermodynamic parameters often guide the prediction of reaction feasibility and help chemists understand energy dynamics in a given process.

Reaction Heat Calculation Through van't Hoff Equation

The van't Hoff equation is essential for calculating the heat of reaction or \( \Delta H^\circ \) based on changes in the equilibrium constant with temperature. This equation connects equilibrium changes to thermodynamics, providing insights into whether reactions are endothermic or exothermic.
To use the van't Hoff equation:

• Measure equilibrium constants \( K_{p1} \) and \( K_{p2} \) at two different temperatures, \( T_1 \) and \( T_2 \).
• The equation is \( \ln \left(\frac{K_{p2}}{K_{p1}}\right) = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \), where \( R \) is the gas constant.
• After determining \( \ln \left(\frac{K_{p2}}{K_{p1}}\right) \), plug the values into the van’t Hoff equation to solve for \( \Delta H^\circ \).

This approach allows the determination of whether more energy is absorbed or released when a reaction shifts with an increase or decrease in temperature, hence providing valuable insights into reaction thermodynamics.