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Chapter 18: Problem 54

Crystallization of sodium acetate from a supersaturated solution occurs spontaneously (see p. 426). What can you deduce about the signs of \(\Delta S\) and \(\Delta H ?\)

Short Answer

Expert verified
The sign of the change in entropy (\(\Delta S\)) is negative because the process is going from a state with higher entropy (disorder) to a state with lower entropy (more order). The sign of the change in enthalpy (\(\Delta H\)) is also negative because releasing heat is an exothermic reaction, which decreases the system’s total energy.

Step by step solution

01

Understanding the terms

Start by understanding what is meant by the changes in entropy (\(\Delta S\)) and enthalpy (\(\Delta H\)). The change in entropy is a measure of how much energy is spread out or how spread out energy becomes, while change in enthalpy refers to the total energy change of a system.
02

Evaluating change in entropy

Crystallization is a process where a substance in gaseous or liquid state changes to a solid state. It’s a process where disorder (or high entropy state) changes to order (or low entropy state). Since we are moving from high entropy to low entropy, the change in entropy (\(\Delta S\)) would be negative.
03

Evaluating change in enthalpy

Crystallization is an exothermic process – heat is released when a substance crystallizes. So the system loses heat to the surroundings and thus the change in enthalpy (\(\Delta H\)) would be negative.

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Most popular questions from this chapter

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Explain the following nursery rhyme in terms of the second law of thermodynamics. Humpty Dumpty sat on a wall; Humpty Dumpty had a great fall. All the King's horses and all the King's men Couldn't put Humpty together again.

For reactions carried out under standard-state conditions, Equation (18.10) takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-\) \(T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation $$\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C} K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\)

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The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.40 at \(2000 \mathrm{~K}\). (a) Calculate \(\Delta G^{\circ}\) for the reaction. (b) Calculate \(\Delta G\) for the reaction when the partial pressures are \(P_{\mathrm{H}_{2}}=0.25 \mathrm{~atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{~atm}\) \(P_{\mathrm{H}_{2} \mathrm{O}}=0.66 \mathrm{~atm},\) and \(P_{\mathrm{CO}}=1.20 \mathrm{~atm}\)
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