Upload
colleen-tucker
View
350
Download
5
Embed Size (px)
DESCRIPTION
The of real gases:. Q =0. In compressing process:. In expanding process:. 2.9 Joule-Thomson experiments. so. Define. and. Joule-Thomson coefficient. The enthalpy keep constant in the whole process. Experimentally determined isenthalpic curve. He 40 K N2 621 K - PowerPoint PPT Presentation
Citation preview
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