## 16.2: Entropy Exercises

### Q16.2.1

In the listed below Figure all feasible distributions and microsays are presented for four different pposts common between two boxes. Determine the entropy change, Δ*S*, if the pwrite-ups are initially evenly spread between the 2 boxes, however upon redistribution all finish up in Box (b).

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### Q16.2.4

Consider a mechanism equivalent to the one below, except that it includes six pshort articles rather of 4. What is the probcapacity of having actually all the pshort articles in only among the 2 boxes in the case? Compare this with the similar probcapacity for the system of 4 pposts that we have actually acquired to be equal to (dfrac18). What does this comparichild tell us about also larger systems?

### S16.2.4

The probability for all the particles to be on one side is (dfrac132). This probcapacity is noticeably lower than the (dfrac18) result for the four-pwrite-up system. The conclusion we have the right to make is that the probcapacity for all the pposts to stay in only one part of the mechanism will decrease swiftly as the number of particles boosts, and, for circumstances, the probcapability for all molecules of gregarding gather in just one side of a room at room temperature and pressure is negligible considering that the variety of gas molecules in the room is exceptionally big.

### Q16.2.5

Consider the mechanism shown in Figure. What is the readjust in entropy for the process wbelow the energy is initially associated only via pshort article A, yet in the final state the energy is dispersed between two different particles?

api/deki/files/126420/CNX_jiyuushikan.org_16_04_aceticdimr_img.jpg?revision=1" />At 25 °C, the equilibrium continuous for the dimerization is 1.3 × 103 (push in atm). What is Δ*S*° for the reaction?

### S16.4.25B

−0.16 kJ

### Q16.4.26

Nitric acid, HNO3, deserve to be ready by the complying with sequence of reactions:

How a lot warmth is advanced when 1 mol of NH3(*g*) is converted to HNO3(*l*)? Assume typical says at 25 °C.

### Q16.4.27A

Determine Δ*G* for the adhering to reactions.

(a) Antimony pentachloride decomposes at 448 °C. The reaction is:

An equilibrium mixture in a 5.00 L flask at 448 °C includes 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.

Chlorine molecules dissociate according to this reaction:

1.00% of Cl2 molecules dissociate at 975 K and a pressure of 1.00 atm.

### S16.4.27A

(a) −22.1 kJ; 61.6 kJ/mol### Q16.4.27

Given that the (ΔG^circ_cef) for Pb2+(*aq*) and also Cl−(*aq*) is −24.3 kJ/mole and −131.2 kJ/mole respectively, recognize the solubility product, *K*sp, for PbCl2(*s*).

### Q16.4.28

Determine the typical cost-free power adjust, (ΔG^circ_cef), for the development of S2−(*aq*) given that the (ΔG^circ_cef) for Ag+(*aq*) and also Ag2S(*s*) are 77.1 k/mole and also −39.5 kJ/mole respectively, and the solubility product for Ag2S(*s*) is 8 × 10−51.

### S16.4.28

90 kJ/mol

### Q16.4.29

Determine the typical enthalpy change, entropy adjust, and complimentary energy readjust for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy alters. Exordinary why diamond spontaneously changing into graphite is not oboffered.

### Q16.4.30

The evaporation of one mole of water at 298 K has actually a standard totally free power change of 8.58 kJ.

*KP*, for this physical process. By calculating ∆

*G*, recognize if the evaporation of water at 298 K is spontaneous once the partial push of water, (P_ceH2O), is 0.011 atm. If the evaporation of water were always nonspontaneous at room temperature, wet laundry would certainly never before dry as soon as placed exterior. In order for laundry to dry, what need to be the worth of (P_ceH2O) in the air?

### S16.4.30

(a) Under conventional thermodynamic problems, the evaporation is nonspontaneous; *Kp* = 0.031; The evaporation of water is spontaneous; (P_ceH2O) must always be much less than *Kp* or much less than 0.031 atm. 0.031 atm represents air saturated via water vapor at 25 °C, or 100% humidity.

### Q16.4.31

In glycolysis, the reactivity of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the complying with equation:

In this procedure, ATP becomes ADP summarized by the following equation:

Determine the conventional cost-free power change for the complying with reaction, and also define why ATP is vital to drive this process:

### Q16.4.32

One of the crucial reactions in the biojiyuushikan.orgical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to develop fructose-6-phosphate (F6P):

*M*, yet, in a typical cell, they are not even close to these values. Calculate Δ

*G*when the concentrations of G6P and F6P are 120 μ

*M*and 28 μ

*M*respectively, and discuss the spontaneity of the forward reactivity under these problems. Assume the temperature is 37 °C.

### S16.4.32

(a) Nonspontaneous as (ΔG^circ_298>0); (ΔG^circ_298=−RTln K,) (ΔG = 1.7×10^3 + left(8.314 × 335 × lndfrac28128 ight) = mathrm−2.5: kJ). The forward reaction to produce F6P is spontaneous under these problems.

### Q16.4.33

Without doing a numerical calculation, recognize which of the adhering to will minimize the cost-free power change for the reactivity, that is, make it less positive or more negative, once the temperature is boosted. Exordinary.

(a) (ceN2(g)+ce3H2(g)⟶ce2NH3(g)) (ceHCl(g)+ceNH3(g)⟶ceNH4Cl(s)) (ce(NH4)2Cr2O7(s)⟶ceCr2O3(s)+ce4H2O(g)+ceN2(g)) (ce2Fe(s)+ce3O2(g)⟶ceFe2O3(s))When ammonium chloride is included to water and also stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any type of calculations, deduce the signs of Δ*G*, Δ*H*, and also Δ*S* for this process, and justify your selections.

### S16.4.33

Δ*G* is negative as the process is spontaneous. Δ*H* is positive as through the solution ending up being cold, the dissolving must be endothermic. Δ*S* must be positive as this drives the process, and also it is expected for the dissolution of any type of soluble ionic compound.

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### Q16.4.34

An necessary source of copper is from the copper ore, chalcopoint out, a kind of copper(I) sulfide. When heated, the Cu2S decomposes to form copper and sulfur described by the adhering to equation:

*s*). The reaction of sulfur with oxygen yields sulhair dioxide as the just product. Write an equation that explains this reaction, and also recognize (ΔG^circ_298) for the process. The manufacturing of copper from chalcomention is percreated by roasting the Cu2S in air to develop the Cu. By combining the equations from Parts (a) and also (b), compose the equation that describes the roasting of the chalcocite, and define why coupling these reactions together makes for an extra reliable process for the production of the copper.

### Q16.4.35

What happens to (ΔG^circ_298) (becomes more negative or more positive) for the complying with jiyuushikan.orgical reactions when the partial press of oxygen is increased?

(a) (ceS(s)+ceO2(g)⟶ceSO2(g)) (ce2SO2(g)+ceO2(g)⟶ceSO3(g)) (ceHgO(s)⟶ceHg(l)+ceO2(g))Recommfinished articles