Chapter 7: Problem 144
\(K_{\text {p }}\) 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
Short Answer
Step by step solution
Identify Known Variables
Apply the Van't Hoff Equation
Calculate the Change in \( K_{p} \)
Calculate the Reciprocal Temperature Difference
Substitute and Solve for \( \Delta_{r} H^{\circ} \)
Identify the Correct Option
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Constant
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.
Enthalpy Change
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.
Temperature Dependence
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 \( K \).
Chemical Thermodynamics
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.