Heat of reaction from heat of formation data

Products Reactants

Elements

-n

D

H

n

D

H

Suppose we first decompose the reactants to their elements (reverse of the formation reaction), then we recompose the products from the elements,

Heat of reaction from heat of formation data

I II

 D  D   D - D 

D_rH _rH I( ) _rH II( ) _{B f}H_B _{A f} H_A D D_r ( _fH)

_rH

3C₂H₂  C₆H₆

6 6 3 2 2

D_rH  D_fH_{C H} -  D_f H_{C H}

T

(spontaneity)

– H₂+O₂→H₂O and not the reverse

– gases uniformly fill the space available (expand) – a hot object cools down to the temperature of its

environment (heat is dissipated)

Ordered Disordered

In the processes occurring spontaneously energy is dissipating.

We introduce a new state function, which can be used as the measure of the disorder. In spontaneous processes in

isolated systems its change should be positive:

Q

= T·S [ S ] = J/K

Heat input: the motion becomes more disordered Work input: makes the system more ordered

Any changes can be characterized by an entropy change.

S: entropy, the measure of disorder

State function, extensive property (depends of the amount)

S=nS_m molar entropy

Spontaneous macroscopic processes in isolated systems always increase the entropy. The system gets into equilibrium when its entropy reaches its maximum value.

(This is the 2nd law of thermodynamics.)

For pure and perfect crystals at T ^ 0 K S = 0. (This is the 3rd law of thermodynamics.)

forrás fp

Q T

hőmérséklet (K)

szilárd folyadék gáz

forrás

olvadás fázisátalakulás 0

0

abszolút entrópia, S

forrás fp

Q T

at T=0 K

- No motion: S_thermal=0

- Organisation (configuration) of the atoms might be disordered:

Sconfigurationl>0

35

. .

( .)

D  ^{phase tr}

phase tr

S phase tr Q

T

 Unlike U and H, the absolute value of entropy is known.

Temperature, K

Absoluteentropy, S

liquid

solid gas

Boiling Q_boiling

T_bp

Melting

Solid phase transition

T dependence of entropy:

S increases S decreases

expansion compression

evaporation condensation

melting freezing

heating cooling

Disorder increases Disorder decreases

Entropy at phase transitions (isothermal-isobaric processes)

D  D

melting

H( melting ) S( melting )

T

D  D

boiling

H( evap ) S( evap )

T

e.g.

chemical S(evap), JK^–1mol^–

1

bromine 88.6

benzene 87.2

carbon

tetrachloride

85.9

cyclohexane 85.1

H₂S 87.9

ammonia 97.4

water 109.1

mercury 94.2

Entropy of evaporation at the normal boiling point (p=1 atm)

Problem:

The entropy of evaporation of cyclohexane at its normal boiling point (1 atm, 197.3 °C) is 85.1 J/(molK).

Calculate its heat of evaporation at this temperature. EXERCISE 1

Solution:

Problem:

The melting point of nitrogen is -196 °C.

What will be the change of entropy if 15 liter of liquid nitrogen is evaporated at atmospheric pressure ? The density of the

liquid nitrogen is 0.81 g/cm³ ?

What will be the sign of the change and explain why. EXERCISE 2

Solution:

EXERCISE 3

Problem:

How much heat should be removed from the system if we intend to cool 5 m³ ethane gas from 140 °C to 30 °C ?

The temperature dependence of the molar heat can be neglected.

Solution:

EXERCISE 4

Problem:

The mass of a single cube of sugar (C₁₂H₂₂O₁₁) is ca. 1.5 g.

How much heat is evolved when a cube is completely burned in excess oxygen?

Solution:

If not isolated

S_system ^ 0

S_system + Ssurrounding  0

In spontaneous macroscopic processes the entropy always increases.

In isolated system

Spontaneity  rate ^{m graphite m diamond} kJ

G G

 mol

, – , – 3

In a closed system at constant T and p in spontaneous processes G decreases. When equilibrium is reached, it has a minimum (if no work occurs).

endothermic exothermic

if p and T are constant

 –

G H TS Gibbs free energy

változás iránya Gibbs energia

Összes entrópia

The entropy change of an arbitrary process:

D Q^rev  DH

S T T

system surrounding

S H

T

D  – D

syste

total m system

TDS  – DH  DT S

total system surrounding

S S S

D  D  D

system

total system

S H S

T

D  – D  D /·T

- D T S

D H T S – D  D G

S_total

G

Direction of changes

Most important properties of G^:

1. State function

 – G H TS

Energy stored by the

thermal motion of the atoms/molecules

G nG m 2. Extensive quantity

3.

Total energy

stored in the system

The spontaneity of a process depends on the sign of G during the transition:

G = n·Gm

n·G (2) - n·G (1) = n [G (2) - G (1)] < 0 ? PHASE 1 PHASE 2

e.g., in phase transition (no chemical changes)

G(T) p = const.

–

p

G S

T

  

  



 

dG - SdT G H TS –

 

H U pV

46

T

G V

p

  

   

  T = const.

dG Vdp

G(p⁾ G H TS^{ –}

 

H U pV

 _{1 }  ₂

m m

G G

 ₁  ₁  ₂  ₂

m m m m

G dG G dG

 ₁  ₂

m m

dG dG

 ₁  _{1 –}  ₁

m m m

dG V dp S dT

 ₂  _{2 –}  ₂

m m m

dG V dp S dT

 _{1 –}  ₁  _{2 –}  ₂

m m m m

V dp S dT V dp S dT

 _{2 –}  ₁  _{2 –}  ₁

m m m m

S S dT V V dp

    

   

 D D

m m

S dp

dT V

Clapeyron

m m – m

dG V dp S dT

D 

D D

 D ^m

m m

Sm

V

H dp

dT T V

The phase transition is an isothermal and isobaric process:

D _m  DH^m

S T

m m

V T

H dT

dp

D



 D

Clapeyron equation (the equation of one component phase equilibrium).

It is valid for: liquid-vapor solid-liquid solid-vapor

solid-solid equilibrium Nothing was

neglected in the

derivation.

CO₂

S/L reaction to increasing p









 D D

m m

dp S

dT V

Water

19,7 cm³/mol water: 18,0 cm³/mol ice:

Ice skating glaciers

D

 D

evap m m

dp H

dT T V

D _m  DH^m

S T

56

The liquid – gas transition: evaporation and condensation

     

 

m m m

m

V V gas V liq V gas

V gas RT

p

D  - 



evap m

p H

dp dT

 D

' 2

Clausius-Clapeyron

Let’s apply the Clapeyron equation for liquid-vapor equilibrium:

1. We neglect the molar volume of the liquid (compared to vapor).

2. We regard the vapor as ideal gas.

change of molar

volume at vaporization molar heat

of vaporization

 ln dp d p

p

pv Tv evap

pk Tk

d p H dT

RT

 D

  ₂

ln '

evap mH

p C

R T

 - D  

' 1 ln

d(1/T)/dT = -1/T²

2  -

1 dT d

T T

57

evap m

p H

dp dT

p RT

 D

' 2

The saturation pressure

of a pure liquid only depends on T.

T p

 

ln p B

 -T 

 

Pa Pa p p

1

)

 (

If the logarithm of the vapor pressure is plotted against the reciprocal of temperature, we obtain a straight line:

1/T lg{p}

a



 ^B

T p  - A  lg

A, B: constants tana = -A



 Pa

Pa p p

1

)

 (

evap m v

k k v

p H

p R T T

D  

  

 

' 1 1

ln –