2.9 Joule-Thomson experiments

2.9 Joule-Thomson experiments

2 1U U U W

The of real gases:

Q=0

1 1W p V 1 1 1 1 ( =0 )pV V V V

In compressing process:

2 2W p V 2 2 2 2 ( = 0 )p V V V V

In expanding process:

The enthalpy keep constant in the whole process

1 2 1 1 2 2W W W pV p V

so 2 1 1 1 2 2U U pV p V

and 2 2 2 1 1 1U p V U pV

J-T ( )HT

p

Define

Joule-Thomson coefficient

Experimentally determined isenthalpic curve

Inversion temperature of real gases

Inversion curve

μ =0, P↓, P<0, T=0,T not change△ △μ >0, P↓, P<0, T△ △ <0,T ↓decrease

μ <0, P↓, P<0, T△ △ >0,T increase

He         40 K      N2         621 K      O2         764 K      Ne         231 K

Joule-Thomson Apparatus

Gas μJ-T/K.MPa-1

Ar 3.66

C6H14 -0.39

CH4 4.38

CO2 10.9

NH3 2.69

H2 -0.34

273K, 1atm

Work principle of a refrigerator

Application of the Joule-Thomson effect

Liquefying GASes using an isenthalpic expansion

Why μ>0, =0 or <0 ?( , )H H T p d ( ) d ( ) dp T

H HH T p

T p

( )

( )( )

T

H

p

H

T pHpT

J-T( ) ,HTp

,H U pV

( ) p pH CT

J-T( ) [ ] / pTU pV Cp

Discussions next class: Application of the first law

Group 1: Understanding about the atmosphere and climate phenomenon

(1) Marine climate/ Continental climate (2) Altitude/temperature

Group 2: Is it possible water being used as fuel?

Group 3: Food and energy reserves.

2.9 The thermochemistry The energy changes in chemical reactions the heat produced or required by chemical reactions

HreacantHproductHQ rp ）） (-(

UUUQ rV reactan

t ）product ） (-(

At the same temperature ( reactants, products)

Measurable predictable

aA bB yY zZ

* * * *( ) ( ) ( ) ( )r Y m Z m A m B mH n H Y n H Z n H A n H B

*/ ( )r m r B mB

H H H B molar enthalpy of reaction

Standard molar enthalpy of reaction( )r m B mB

H H B y y

The enthalpy change per mole for conversion of reactants in their standard states into products in their standard states, at a specified temperature and pressure Pθ

H2(g,p) + I2(g,p)=2HI(g,p)

△rHm(298.15K) = -51.8kJ·mol

For example:

Calculation of standard enthalpy of reactions (1) By standard molar enthalpy of formation f mH y

The enthalpy change when one mole of the compound is formed at 100 kPa pressure and given temperature from the elements in their stable states at that pressure and temperature.

(298.15K) B

mfB BH )(△rHm(298.15K)

( )r m B mB

H H B y y

Understanding standard molar enthalpy of formation

KJ.mol-1

-300

-200

-100

100

200

300F(g)+H(g)

F2 Cl2 H2

H(g)

+218

Cl(g)+H(g)

HF(g)

HCl(g)-564

-431

-92

-269

Standard molar enthalpy of formation of the stable forms of the elements is zero;Any form of elements other than the most stable will not be zero; such as C (diamond), C (g), H (g), and S (monoclinic).

The ∆fH depend on state of substances

KJ.mol-1

-250

-200

-150

-100

-50

0

-300

H2+O2

H2O2

H2O(g)

H2O(l)

-188-242

-285

Example: Calculate the standard enthalpy of following reaction at 25 by using standard molar enthalpy of formation℃

(g)HO(g)H2(g)HCOH(g)HC2 226452

f m1kJ mol

H

y C2H5OH(g) C4H6(g) H2O(g) H2(g)

-235.10 110.16 -241.81 0

r m f m 4 6 f m 2

f m 2 5

(C H , g)+2 (H O, g)

2 (C H OH, g)

H H H

H

y y y

y

1

1

molkJ72.96

molkJ)]10.2352(818.2412[110.16=

f m 2 f m 2 vap m 2(H O, g, ) (H O, l, ) (H O, )H T H T H T y y y

Definition: The enthalpy change when a mole of substance is completely burnt in oxygen at a given temperature and standard pressure.

All these complete products have an enthalpy of combustion of zero.

mcH(2) By standard molar enthalpy of combustion

g)(COC 2 O(l)HH 2

g)(SOS 2

g)(NN 2

BB

(298.15K) (B,298.15K)r m c mH H

Standard molar enthalpy of combustion

KJ.mol-1

0

200

300

400

100

C+O2

CO

CO2

-110

-284

-394

BB

(298.15K) (B,298.15K)r m c mH H

298.15K, 100kPa ：CH3COOH(l)+2O2(g)=2CO2(g)+2H2O(l)

-1molkJ3.870 mrH

-13(CH COOH,l,298.15K) 870.3kJ molc mH

l)(O2Hs)()(COOCHOH(l)2CHs)(COOH)( 22332 （ A ） （ B ） （ C ） (D)

)C()B(2)A( CCC mmmmr HHHH

(3) By bond energies and bond enthalpies

-12 r m

-1r m

H O(g) = H(g)+OH(g) H (1) 502.1 kJ mol

OH(g) = H(g)+O(g) H (2) =423.4 kJ mol

1

m

-1

(OH,g) (502.1 423.4) kJ mol / 2

= 462.8 kJ mol

H

H C N O F Cl Br I Si

H 436 415 390 464 569 432 370 295 395

C 345 290 350 439 330 275 240 360

N 160 200 270 200 270

O 140 185 205 185 200 370

F 160 255 160 280 540

Cl 243 220 210 359

Br 190 180 290

I 150 210

Si 230

1/2N2(g)+3/2H2(g) NH3(g)

N(g)+3H(g)

NH(g)+2H(g)

NH2(g)+H(g)

NH3(g)

466kJ.mol-1

390kJ.mol-1

314kJ.mol-1

1124kJ.mol-1

1/2N2(g)+3/2H2(g)46 kJ.mol-1

1170kJ.mol-1

(4) Solublization heat0)}aq(H{

mf H

aq)(Claq)(H)HCl(g, -OH2 p

)g,HCl()aq,Cl()aq,H()K298(sol mfmfmfm HHHH

0)aq,H(f mH

1 175.14 kJ mol ( 92.30 kJ mol )

1167.44 kJ mol

-1f molkJ30.92)gHCl,(

mH

)aq,Cl( -f

mH

The experimental determination of standard enthalpy of combustion A sample of biphenyl (C6H5)2

weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid C6H5COOH weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant pressure is known to be 3226 kJ mol–1 (that is, ΔU° = –3226 kJ mol–1.) Use this information to determine the standard enthalpy of combustion of biphenyl.

ΔH = ΔU + Δ(PV) = Qv + ΔngRT

Hess’s law

For example ：（ 1 ） （ 2 ）

(g)CO)(OC(s) 22 g m,1rH

2 212CO(g) O (g) CO (g) m,2rH

（ 1 ） - （ 2 ） =（ 3 ） （ 3 ） CO(g)g)(OC(s) 22

1 m,3rH

m,2rm,1rm,3r HHH

The enthalpy change for any sequence of reactions that sum to the same overall reaction is identical.

(Based on Enthalpy being a state function)

Example 2: C(graphite) + 2H2(g) → CH4(g) Consider following reactions

, , , 2 r m r m a r m b r m cH H H H y y y y

-1

( 890.35) 2 ( 285.84) ( 393.51)

74.84kJ mol

r mH

－－ － －

－

y

-14 2 2 2 ,

-112 2 2 ,2

2 2 ,

CH (g) + 2O (g) CO (g) + 2H O(l) 890.35kJ mol (a)

H (g) + O (g) H O(l) 285.84kJ mol (b)

C(graphite) + O (g) CO (g) 393.5

r m a

r m b

r m c

H

H

H

y

y

y -11kJ mol (c)

–(a)+2(b)+(c)=the above studied reaction

Home work Group discussion Preview: A: 2.9 Y: 1.12 A: P67: 2.17(a, b) P68: 2.38(a,b) Y: P32: 36, 37 P42: 45