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\(K_{p}\) 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

Short Answer

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

Step by step solution

01

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).
02

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 \( T_{2} = 323 \mathrm{~K} \).
03

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 \( \Delta_{r} H^{\circ} \).
04

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:\[\Delta_{r} H^{\circ} = -8.314 \left( \frac{4.604}{-2.35 \times 10^{-4}} \right) \approx 147410 \mathrm{~J} / \mathrm{mol} = 147.41 \mathrm{~kJ} / \mathrm{mol} \].
05

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) \(147.41 \mathrm{~kJ} / \mathrm{mol}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Assertion: Adding an inert gas todissociation equilibrium of gaseous \(\mathrm{N}_{2} \mathrm{O}_{4}\) at constant pressure and temperature increases the dissociation. Reason: Molar concentration of the reactants and products decreases on the addition of inert gas.

In a reversible reaction, the catalyst (a) decreases activation energy of forward reaction (b) increases activation energy of forward reaction (c) decreases activation energy of both forward and backward reactions (d) increases activation energy of backward reaction

The equilibrium constant value for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) is \(1.48 \times 10^{-5}\), the value for the reaction \(1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) is \(\mathrm{n} \times 10^{-3}\) where \(\mathrm{n}\) is (a) \(1.85\) (b) \(3.85\) (c) \(4.85\) (d) 10

The equilibrium constants for the reactions \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) and \(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3}\) are \(K_{1}\) and \(K_{2}\) respectively. The correct relationship between \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) is (a) \(\mathrm{K}_{\mathrm{t}}=\frac{\mathrm{K}_{2}}{2}\) (b) \(\mathrm{K}_{2}=\sqrt{\mathrm{K}_{1}}\) (c) \(\mathrm{K}_{2}=\mathrm{K}_{1}{ }^{2}\) (d) \(\mathrm{K}_{1}=\mathrm{K}_{2}\)

For the following three reactions \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), equilibrium constants are given: [2008] (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{1}\)(b) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{2}\) (c) \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{3}\) Which of the following relation is correct? (a) \(\mathrm{K}_{1} \sqrt{\mathrm{K}}_{2}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{2} \mathrm{~K}_{3}=\mathrm{K}_{1}\) (c) \(\mathrm{K}_{3}=\mathrm{K}_{1} \mathrm{~K}_{2}\) (d) \(\mathrm{K}_{3} \cdot \mathrm{K}_{2}^{3}=\mathrm{K}_{1}^{2}\)

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