Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010

This course is approximately at this level

CHEMISTRY E182019 CH6

Entropy, Gibbs energy, 2nd law of thermodynamics

Entropy s CH6

ENTROPY

dq = T ds

…heat dq transferred to a system (ideally, reversibly)

Entropy s CH6

s-entropy represents probability of a macroscopic state (macrostate is something, that can be measured, for example temperatures, concentrations, length of macromolecules). This probability is proportional to the number of microstates (possible configurations, e.g. arrangements of macromolecular chains, location of molecules in space). The probability of a macrostate can be calculated, but it is not quite easy, see next.

Spontaneous processes (chemical reactions, phase changes…) proceed in the direction of increasing probability and therefore are characterised by entropy increase. This is the second law of thermodynamics.

Entropy of insulated system always increases.

Entropy – microstates spatial distributionCH6

Let us assume, that the system is a vessel containing only 2 molecules. We distinguish how many molecules are in the left and in the right part of the vessel. There exist only 3 possible macrostates: 1) two molecules are left, 2) one left and one right and 3) two molecules are right. In the first macrostate there exists only one microstate (both molecules , are left). The second macrostate can be realised by 2 microstates (molecule is left in the first, or is right in the second microstate). The macrostate with just one molecule in both parts has the highest number of possible arrangements (2), the highest probability and the highest entropy.

In this way, for example diffusion is described: the highest entropy corresponds to the uniform distribution of components in space.

Similar case is irreversible expansion of gas to vacuum or flow of gas between two vessels at different pressures.

p1 p2

Entropy – microstates energy distribution CH6

Gas contained in a vessel increases its entropy when heated. In this case the macrostate corresponds to overall energy distributed to discrete energy levels of individual molecules. Microstate is a particular distribution of molecules to different energies.

Example of microstates corresponding to one macrostate (the same total energy). Different microstates correspond for example to collision of molecules (one is slow down the other accelerated)

Energy level

Energy level

Temperature T1

Lower temperature T2

Less number of possible arrangements (microstates) corresponding to lower total energy.

Entropy decreases with decreasing temperature.

Entropy – microstates conformation CH6

Macromolecule of elastomer (rubber) can be assumed as a chain of hinge-connected rods of monomers. Macrostate is the distance of endpoints of this chain. As soon as the macromolecule is fully extended there exists the only linear arrangement of hinges and only one microstate. Unloaded macromolecule has lower end-to-end distance, and number of possible arrangements increases. Entropy of stretched polymer decreases and temperature of insulated filaments increases when stretched or compressed. Entropy decrease (dq=Tds) would correspond to the heat removal from the system. But alas! System is insulated therefore the heat cannot escape and this is manifested by the temperature increase. This is quite different situation comparing classical materials (metallic strip cools down when stretched, and metals expand with increasing temperature, while elastomers shrink).

Remark: Position of hinges in the coiled chain is restricted not only by the overall length, but also by discretized relative angles between segments due to directional bonds. When counting number of possible arrangements (microstates) the hinges should be localized in a lattice.

Entropy - Boltzmann CH6

Previous analysis can be generalised to Boltzmann equation

S- entropy, k-Boltzmann constant, W-number of microstates.

It means that as soon as there exists only one microstate (one possible arrangement of elements with respect the space and energy levels) the entropy is ZERO. This state corresponds zero thermodynamic temperature (absence of motion) and ideal crystallographic lattice (zero energy and only one possible arrangement of atoms in a lattice).

WkS ln

Remark: Entropy S is extensional and additive quantity. If entropies of subsystems are S1,S2, resulting entropy of the whole system is S=S1+S2=k(lnW1W2), because probability of 2 independent phenomena is product of probabilities.

Entropy - Clausius CH6

Thermodynamic definition of entropy was suggested by Clausius

where ds is the specific entropy change of system corresponding to the heat dq [J/kg] added in a reversible way at temperature T [K].

revT

dqds )(

δq < T ds heat transferred to the system is less than the product T.ds

Equality δq=T ds holds only for ideal reversible processes (without friction). Entropy increase ds is given by heat supplied from environment δq/T plus internal heat generated by viscous friction.

For irreversible processes the entropy is defined by INEQUALITY

in a reversible process it is possible to return back to to the initial state without any change of the system environment

Entropy - Clausius CH6

Using first law of thermodynamic it is possible to evaluate entropy change between two different macrostates (initial macrostate p1,T1,v1,h1,s1 and final macrostate p2,T2,v2,h2,s2) from the Clausius definition as

2

1

2

1

2

1

12

v

v

u

u T

pdv

T

du

T

dqsss

1

2

1

212 lnln

2

1

2

1

2

1

2

1v

v

M

R

T

Tc

Mv

Rdv

T

dTc

T

pdv

T

dTcsss v

v

v

T

T

v

v

v

T

T

v

T

hsss LG

12

Entropy change can be calculated easily for IDEAL GAS

Entropy change for isobaric processes (e.g phase changes, evaporation)

Entropy - Clausius CH6

Example: Consider insulated closed system that contains two bodies at different temperatures. Heat dQ is reversibly removed from hot body at temperature T1 and added to the colder body at temperature T2.

δQT1 T2

dS= - δQ/T1+ δQ/T2 = δQ (T1-T2)/(T1T2)

Entropy change of two whole system is positive dS>0 , even if there is no heat transferred from outside. Positive value is a consequence of process IRREVERSIBILITY.

Entropy Boltzmann-Clausius CH6

Relationship between thermodynamical and statistical definition can be explained on example of a cylinder filled by n-molecules of gas and closed with piston, maintaining constant pressure.

pV=nRT

V ~ T~ V

123

vv-1

Number of cells v=T

Number of microstates

W1=v for 1 molecule W2=v2 for 2 molecules W3=v3 for 3 molecules

…

W=vn for n-moleculesStatistical approach.

T

dTnkdS

TnknkTkWkS n

lnln)ln(ln

Thermodynamical approach.

T

dTcn

T

dQdS p

~

Entropy change in chemical reactionCH6

As soon as the reactants and products are at standard conditions (298 K and 100 kPa) the standard entropy change can be calculated from tabulated values of entropy in a similar way like the enthalpy of reaction

R

RRP

PPr ssS 2980

2980

2980 )~()~()(

where are stoichiometric coefficients of reactants (R) and products (P).

R P 0 0,

The entropy change during chemical reactions is caused by the fact that energy of products is different than energy of reactants (manifested by released/absorbed heat) and also by change of spatial configurations (reactions producing more molecules increase entropy due to increased number of possible configurations).

Spontaneity of chemical reactionCH6

H heat transferred from environment

0

0))

~(

)~((

0

,

2980,

2980

T

HS

T

hs

SSS

reactionreaction

PRi

ifii

gsurroundinreactor

Second law insists that spontaneous reaction proceeds in closed and insulated systems only if entropy increases – but reactor is not an insulated system, and entropy change of surroundings must be also considered

Entropy change due to reaction heat (enthalpy) supplied to or removed from environment for exothermic or endothermic reactions respectively

Entropy change of reactants and products inside reactor. Affected by reaction heat and by irreversible precesses inside reactor.

Chemical reactor

Spontaneity of chemical reactionCH6

0

T

HS reactionreaction

Entropy change is usually positive if there are more

molecules of products than reactants

This term is positive for endothermic reactions

and negative for exothermic reactions

The energy changes are usually greater than the entropy changes. Therefore exothermic reactions are usually spontaneous (but not always, sometimes negative entropy changes prevail).

Combustion of methane

CH4+2O2CO2+2H2O

exothermic (H<0)entropy change is nearly zero (S<0) (three molecules of reactants form three molecules of products)

Potential energy change prevails (forward reaction is spontaneous at any temperature)

Spontaneity of chemical reactionCH6

(Adiabatic flame temperature 1950 oC)

Spontaneity of chemical reactionCH6

Reduction of sulphur oxides (technology for SO2 removal from flue gas)

CaO+SO3CaSO4

exothermic (H<0)decreases chaos (S<0) (two molecules form only one molecule)

Potential energy change prevails (formard reaction is spontaneous up to 2400 K)

Air

Coal+CaO

Fluidised bed combustor

CaSO4

Spontaneity of chemical reactionCH6

Steam reforming

CH4+H2OCO+3H2

endothermic (H>0)increases chaos (S>0) (2 molecules form 4 molecule)

The reaction can be spontaneous only at high temperatures (>900K)

CO,H2

Reaction tubeswith catalyst Ni

CH4,H2O

Spontaneity of chemical reactionCH6

Ammonia synthesis

N2+3H22NH3

exothermic (H<0)decreases chaos (S<0) (4 molecules form 2 molecule)

Potential energy change prevails (up to 400 K) – in practice reaction temperature is much higher 800K but pressure is also higher 20-60 MPa

NH3

Nicatalyst

N2,H2

Gibbs energyCH6

Combined effect of entropy and energy upon the reaction spontaneity (or other processes, e.g. phase changes), carried out at constant pressure and temperature can be expressed by specific Gibbs energy

g = h - Ts or

[J/kg] [J/kg] [K][J/kgK]

, g h Ts G H TS

0reaction reactionreaction

H GS

T T

At constant temperature the Gibbs energy change ΔG=ΔH-TΔS is negative for spontaneous processes. See previous analysis of the chemical reaction spontaneity

Gibbs energy change at standard conditions can be calculated in the same way as reaction enthalpy or entropy changes using tabulated Gibbs energies of formation 0 0 0

298 , 298 , 298( ) ( ) ( )r P f P R f RP R

G g g

Gibbs energy N2+3H22NH3CH6

forward

back

S<0, H<0

S>0, H>0N2

H2

H2

H2

NH3

NH3

G=H-TS<0forward reaction prevails

G=H-TS>0

400 K T

G=H-TS<0backward reaction prevails

G=H-TS>0

G=0

Reaction rate of forward reaction

Reaction rate of back reaction

Principal question is: At what temperatures the forward reaction prevails and when prevails the backward reaction?

The temperature when direction changes is characterized by the same rate of forward and backward reactions (equlibrium), therefore…

Gibbs energy at equilibriumCH6

Principal question is: At what temperatures the forward reaction prevails and when prevails the backward reaction?

The temperature when direction changes is characterized by the same rate of forward and backward reactions (equlibrium), therefore…

reaction

reaction

HT

S

0reactionG …and assuming that the reaction enthalpy as well as the entropy change depend only slightly upon temperature ….

Helmholtz energyCH6

Gibbs energy G

G = H - TSwas historically called „free enthalpy“

and the Helmoltz energy F

F = U – TS„free energy“.

The both are suitable criteria for spontaneity of processes. While criterion G<0 is relevant for processes at constant pressure (e.g. phase changes), the criterion F<0 holds for isochoric processes (F is a suitable measure of deformation energy and therefore is useful for solid mechanics).

d(H-TS)=dH-TdS-SdT=dU+pdV+Vdp-TdS-SdT=Vdp-SdT

d(U-TS)=dU-TdS-SdT=-pdV-SdT

First law TdS=dU+pdV

First law TdS=dU+pdV

Consequence: dG=0 at constant pressure and temperature (e.g. at phase or chemical changes, gibbs energy of reactants is the same as products)

Consequence: dF=0 at constant volume and temperature (e.g. at isothermal deformation of incompressible solid)

Relax – Gibbs energyCH6

Ami C

F Ami

G

Dmi E

Your glow is heating,

my temptation sweet,

not negative being,

I cannot proceed.

Enthalpy Tutorial Syringe CH6

Calculate temperature, pressure, enthalpy, internal energy, entropy changes

0v

dT R dvds c

T M v

0.4022 300 2 396 T K

2 2

1 1

ln ln 0v

T vRc

T M v

8.3140.402

28.96 0.714v

R

Mc

2 1

1 2

( ) v

R

McT v

T v

V

xD

Thermocouple

Example: V2/V1=0.5 cv=0.714 kJ/kg.K cp=1 kJ/kg.K M=28.96 kg/kmol T1=300 K

First law (ideal gas, adiabatic compression)

2 1( ) 68.8 kJ/kgvu c T T

2 1( ) 96 kJ/kgph c T T

Enthalpy Tutorial Syringe CH6

Gas temperature measurement

V

xD

Thermocouple

This is a problem, because the mass of air inside the syringe is small (20 ml ~ 0.02 g) and also corresponding energy is small

H ~ 2 J

Thermal capacity of thermocouple is higher (mass of tiny thermocouple of diameter 1mm and length 10mm is ~ 0.06 g) and therefore the recorded temperature will be significantly less than the predicted 396 K.

Entropy TutorialCH6

Given volume V0 containing the mixture of nA moles of gas A and nB moles of gas B. Calculate entropy changes necessary for separation of A and B into separate volumes V0/2.

The separation can be technically realised in two steps

1.Step using piston with semipermeable membrane (permeable only to gas B with smaller molecules)

2.Step. Permeable mebrane is replaced by an impermeable piston (or pores are closed)

Entropy TutorialCH6

Both steps are assumed ISOTHERMAL and ideal gas state equation can be applied for both gases.

1 A v A

RTdQ n c dT pdV n dV

V

11

0

lnA

VQ n RT

V

02( )

0

/ 2lnB B

VQ n RT

V

02( )

1

/ 2lnA A

VQ n RT

V

Resulting entropy change is the sum of heats

1 2 3 01

0 1

/ 2 1 1(ln ln ) ln ( ) ln

2 2A B A B

Q Q Q VVS n R n R n n R

T V V

2.Step. Heat must be added to gas A, and removed from B

1.Step. Heat must be removed by isothermal compression of gas A

Gibbs energy Tutorial Fischer TropfCH6

Ammonia synthesis N2+3H22NH3 is realized in industrial scale by Fischer Tropf synthesis.

Ammonia is used for production of fertilizers, plasts and explosives

ammonium sulfate(NH4)2SO4

ammonium phosphate(NH4)3PO4

NH4NO3 ammonium nitrate

HNO3, nitric acid

Gibbs energy Tutorial Fischer TropfCH6

The lower is temperature, the greater is yield of the chemical reaction.

Gibbs energy Tutorial Fischer TropfCH6

Gibbs energy of ammonia synthesis N2+3H22NH3 can be calculated from entropies, enthalpies anf gibbs energies of formation (notice the fact that these values are zero for nitrogen and hydrogen – they are elements in the most stable state)

3

3

3

2

2

0, 298

0, 298

0298

0298

0298

( ) 16.5 kJ/mol

( ) 46.1 kJ/mol

( ) 0.192 kJ/(mol K)

( ) 0.191 kJ/(mol K)

( ) 0.131 kJ/(mol K)

f NH

f NH

NH

N

H

g

h

s

s

s

0.191 3 0.131 2 0.193 0.2reactionS

92 298 0.2 33 kJ/molG H T S

3

3

0, 298

0, 298

2( ) 92 kJ/mol

2( ) 33 kJ/mol

reaction f NH

reaction f NH

H h

G g

92470

0.2

HT K

S

Check result using entropy

Temperature at equilibrium of forward and backward reaction