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Chapter 16: Problem 9

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {surr }}\) ? \(\Delta S ?\) \(\Delta S_{\text {univ }}\) ? Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

Short Answer

Expert verified
In this process, ΔS of the system is greater than zero (option i) because the phase change from liquid water to water vapor increases disorder. The ΔS_surr is less than zero (option ii) because the surroundings lose heat, leading to decreased disorder. The ΔS_univ cannot be determined without more information (option iv), as it depends on the magnitudes of ΔS and ΔS_surr.

Step by step solution

01

Understand the given situation and definitions

As liquid water is heated above 100°C, it will undergo a phase change and become water vapor (steam). This is important to consider when determining the changes in entropy. Entropy (S) is a measure of the disorder or randomness in a system. During a process, there are three possible changes in entropy: (i) ΔS > 0 (entropy increases), (ii) ΔS < 0 (entropy decreases), and (iii) ΔS = 0 (entropy remains constant). Now we will analyze the changes in entropy for the system, surroundings, and universe.
02

Entropy change for the system (ΔS)

For the system which is heating the liquid water, the entropy change is a result of a phase transition: liquid water turning into water vapor. This change of phase increases the disorder of the system because water molecules in the vapor phase are more disordered than they are in the liquid phase. This means that the entropy change for the system (ΔS) is greater than zero (positive).
03

Entropy change for the surroundings (ΔS_surr)

The surroundings of the system are losing heat to the system, as heat flows from the surroundings into the system to heat the liquid water. As a result, the entropy of the surroundings decreases, because the energy (heat) that has left the surroundings causes less disorder. Therefore, the entropy change for the surroundings (ΔS_surr) is less than zero (negative).
04

Determine the entropy change for the universe (ΔS_univ)

The entropy change for the universe (ΔS_univ) is the sum of the entropy changes for the system and the surroundings: ΔS_univ = ΔS + ΔS_surr. Since ΔS is positive (greater than zero) and ΔS_surr is negative (less than zero), the overall entropy change for the universe depends on whether the magnitude of the positive entropy change (ΔS) is greater than, equal to, or less than the magnitude of the negative entropy change (ΔS_surr). If the magnitudes are equal, then ΔS_univ = 0. If the magnitude of ΔS is greater than the magnitude of ΔS_surr, then ΔS_univ > 0. And if the magnitude of ΔS is less than the magnitude of ΔS_surr, then ΔS_univ < 0.
05

Find the correct option for the given situation

Now we can determine which of the given options (i-iv) is correct for the changes in entropy: ΔS_surr: less than zero (option ii) ΔS: greater than zero (option i) ΔS_univ: cannot be determined without more information (option iv) In conclusion, the correct option for ΔS_surr is less than zero, for ΔS is greater than zero, and for ΔS_univ is that it cannot be determined with the given information.

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Most popular questions from this chapter

Consider the following reaction at \(800 . \mathrm{K}:\) $$\mathrm{N}_{2}(g)+3 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{atm}, P_{\mathrm{NF}_{3}}=0.48\) atm. Calculate \(\Delta G^{\circ}\) for the reaction at \(800 .\) K.

What information can be determined from \(\Delta G\) for a reaction? Does one get the same information from \(\Delta G^{\circ},\) the standard free energy change? \(\Delta G^{\circ}\) allows determination of the equilibrium constant \(K\) for a reaction. How? How can one estimate the value of \(K\) at temperatures other than \(25^{\circ} \mathrm{C}\) for a reaction? How can one estimate the temperature where \(K=1\) for a reaction? Do all reactions have a specific temperature where \(K=1 ?\)

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{kJ} / \mathrm{mol} .\) In a particular experiment, equal moles of \(\mathrm{H}_{2}(g)\) at 1.00 atm and \(\mathrm{Br}_{2}(g)\) at 1.00 atm were mixed in a 1.00-L flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ},\) and \(\Delta S^{\circ} .\)

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. \(\operatorname{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)\) b. \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

Which of the following involve an increase in the entropy of the system? a. melting of a solid b. sublimation c. freezing d. mixing e. separation f. boiling
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