generalchem_ls_21.pptx

8/10/2019 GeneralChem_LS_21.pptx

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Standard Reaction Entropies

rS= iSim(products) iSim(reactants)

For 2 H2(g) + O2(g) 2 H2O(l) at 298 K

rS= 2 Sm(H2O(l)) ( 2 Sm(H2(g)) + Sm(O2(g)) )

= 327 J/K. ( Why negative?)

However, the reaction is explosively spontaneous.


2/18

Need to consider the total entropy change ie., of the system

and the surroundings. For this reaction, rH298 = 572 kJ, sothat

(S)surr= 572x103/ 298 = 1.92x103J/K

and (S)tot= 1.59x103J/K

indicating that the reaction is spontaneous.

Note: This reaction is also spontaneous in an isolatedcontainer (Explain).


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Statistical Entropy

Description of entropy from microscopic point of view. Thearrangements of molecules over the available energy levels

determines the value of the statistical entropy.

Boltzmann Eqn.: S = k lnW

k = Boltzmanns constant = 1.38110-23JK-1

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4/18

Statistical Entropy


Example: Suppose a tiny system of four molecules A, B, C, and D thatcan occupy three equally spaced levels of energies 0, and 2and say

total energy is 4. The 19 arrangements are possible, so W = 19.


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Statistical Entropy


At T = 0, all molecules are in lowest possible energy level. So W = 1.

Therefore, S = 0. Boltzmanns formula agrees with Third Law.

When T > 0, the molecules can occupy energy levels above the lowest

one, and many different arrangements of the molecules will correspond

to the same total energy. That is, when T > 0, W > 1, So S > 0.Boltzmanns formula is consistent with the entropy of a substance

increasing with temperature.

When container expands, the levels occupied by the molecules getcloser together, and the number of ways of arranging the molecules for

a given total energy increases. So W increases and S increases.

Boltzmanns formula is consistent with the entropy of a gas increasing as

the volume it occupies is increased.


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Residual Entropy


Boltzmanns formula provides an explanation for the entropy of somesubstances is greater than zero at T = 0. This contribution is called the

Residual entropy of a solid.

Example: We can estimate the value of residual entropy of CO molecules

at T =0 using Boltzmanns formula. Each CO molecule can lie in either of

two orientation. Then the total no. of ways of arranging N molecules is

(222)N times= 2N

Then, S = k ln 2N or S = N k ln 2. Sm = NAVk ln 2 = R ln2 = 5.8 JK-1mol-1

Sm(0) = 5.8 JK-1mol-1


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Spontaneity and Equilibrium

The criterion for the spontaneity of a process in anisolated system is

dS > 0

Isolated system spontaneously moves to states ofever greater entropy.

The equilibrium state of an isolated system is

therefore characterized by maximum entropy forthe given internal energy U and volume V.

What about systems other than isolated?



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In the chemistry laboratory, one is most commonly

concerned with systems at constant T and p. At constant

pressure, q = = qsurrso that

(S)tot= S ()/T 0

(> for irreversible and = for reversible process).

Or TS 0

If T is constant, this may be written as

(H TS) 0

(


9/18

Gibbs Energy

Define the Gibbs (free) energy G as

G S

Extensive state function with units of energy. As just seen, atconstant T and p, G = T(S)tot and so for spontaneous

process at constant T and p, G < 0.

For infinitesimal change at given T and p, and no work otherthan that of expansion/compression, dG 0

For a closed system at given T and p, with only pdV work,G < 0 for a spontaneous process, and G is thereforeminimum at equilibrium for such a system

dG = SdT + Vdp



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11/18

Chemical Potential

In a homogeneous solution of say k components, one maywrite any extensive property E as E = njjwhere jcalleda partial molar property represents the contribution per molethat species j makes to the property E in that solution at the

given T and p.

The partial molar Gibbs energy is of special significance, isdenoted by j, and called the chemical potential of the

species j in the given solution at the given T and pFor a pure substance, the chemical potential is just themolar Gibbs energy, = G/n.



12/18

Equilibrium and Spontaneity Criteria

In terms of intensive properties, the criteria of equilibriumare:

Thermal equilibrium: Uniformity of temperature T

Heat transfer occurs spontaneously in the direction ofdecreasing temperature

Mechanical equilibrium: Uniformity of pressure

Displacement of wall occurs spontaneously in the directionof decreasing pressure

Material equilibrium (transfer of substance j): Uniformity ofthe chemical potential of j

Net transfer of substance j occurs in the direction ofdecreasing chemical potential j



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Thermodynamics of Phase Equilibrium

At given T and p, the stable phase of a pure substance isthat of minimum molar Gibbs energy Gm.

At 1 atm, ice is the stable phase of water below 0C, while

above 0C but below 100C, liquid water is the stablephase, and above 100C, steam is the stable phase.

At this pressure, ice and liquid water have the same valueof Gm at 0C, and coexist in equilibrium, while at 100C,liquid water and steam have the same Gm, and coexist inequilibrium.



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Variation of G with p and T

dG = dU + d(pV) d(TS)and with dU = TdS pdV one hasdG = SdT + VdpIt follows that

(Gm/

p)T= Vmand

(Gm/T)p= Sm



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Variation of G with p at constant T


(Gm/p)T= Vm

Vapour stable at lowpressures, liquid at

intermediate pressures,and solid at highpressures. Transitionwhen molar Gibbsenergies are equal


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Variation of G with T at constant p


(Gm/T)p= Sm

Solid stable at lowtemperatures, liquid at

intermediate temperaturesand vapour at hightemperatures. Transitionwhen molar Gibbs energies

are equal


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Phase Diagram


Map showing ranges of Tand p over which variousphases are stable.

Phase Boundaries-Melting, Vaporization andBoiling Curves

Triple Point, Critical Point

Phase Rule F = CP + 2


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Enthalpy and Entropy changes

Heat transfer accompanying a phase change at constant p isH. States must be completely specified.

Energy supplied as heat at constant pressure for the fusionof 1 mol of substance under standard condition i.e. pure solidsubstance at 1 bar changing to pure liquid at 1 bar, is thestandard enthalpy of fusion fus

Similarly , vap, the standard enthalpy of vaporization

sub= fus

+ vap(Hess Law).

If the word standard is removed, then it refers to the value atthe pressure (and T) at which there is equilibrium.

Since the transformations are at equilibrium, and hencereversible, transS = trans/T where T refers to the equilibriumtemperature at the given p.

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generalchem_ls_21.pptx

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