【中南大学 无机化学精品课件】第3讲 化学热力学基础

= -4815kJ·mol-1
rHm = Qp,m = Qv,m + = - 4815+ (7-11) 8.314 = - 4825kJ·mol-1
ng RT 10-3 298.15
8
3.2.5
aA+dD= eE+fF (
)
0= eE+fF -aA-dD
0
B BB
B
B
B
stoichiometric number
F
p=
p2
T1=273K
p1=405.2kPa
V1=1.00dm3
p2=101.3kPa, V2=4.00dm3,T2=273K
W1= - p V = -101.3 103Pa
= -304J
(4-1)
10-3m3
405.2kPa
202.6kPa
p =202.6kPa, V =2.00dm3
101.3kPa
8-
C7H16(l ) + 11O2 (g) = 7CO2(g) + 8H2O(l )
298.15K
1.250g
60.09kJ
298.15K
rHm
M=100.2g·mol-1
n=
1.250g
=1.248 10-2mol
100.2g mol 1
Qv = -60.09kJ Qv,m = Q V =
n
60.09kJ 1.248 10 2mol
(Chemical thermodynamics)
+
1876-1878
J.W.Gibbs
1
open system closed system isolated system
3.1.1
(thermodynamic system)
(surrounding)
3.1.2 state
TpmVn
initial state
t2 nB/mol 1.5 5.5
3.0
2
1
n1 N 2 N2
(2.0 3.0)mol 1.0mol 1
1
n1 H2 (7.0 10.0)mol 1.0mol
H2
3
1
n1 NH3 (2.0 0)mol 1.0mol
NH3
U +p V+V p Vp
QP= H
H
Qp
pV=nRT H=U+pV=U+nRT
U nR
3.2.3
(thermal capacity) C=Q/T
J·K 1 1mol
Cm
Cp
CV
7
3.2.4
U=CV T=QV dU=CVdT= QV
H=Cp T=Qp dH=CpdT= QP
dH=dU+d(pV)
CpdT= CVdT+ d(pV)
U = Q p V = Qp p V
U2 - U1 Qp p V2 V1
U2 + p2V2
U1 + p1V1
H = U + pV ( enthalpy)
U2 + p2V2=H2 U1 + p1V1=H1 H = H2 H1= Hr = Qp
= Qp
Hr
U pV H
U U
H=U+ pV
H= U+
pV = pV
adiabatic process
cyclic process
A A
path
3.1.4 heat
work
Q W
Q W> 0
Q
Q<0
Q>0
W<0
3
We= -p
volume work
expension work
We
V We= - p dV
W
T Q
S
p p =p1
L
p = p2
p,V,T
p W=-F L= p S L= -p V
B
B
A=-a D=-d
B
E=e F=f
extent of reaction
mol
= nB( ) nB(0)
nB(0)
B
nB
=B
B =0
t
d
= (t)-Leabharlann Baidu(0)= nB
B
d = dnB
B
N2 g 3H2 g 2NH3 g
t0 nB/mol 3.0 10.0
0
0
t1 nB/mol 2.0 7.0
2.0
1
(final state)
state function
(extensive property)
Vnm
UHS
G
(intensive property) Tp
=m/v
pV=nRT
2
3.1.3 (process)
isobar process isothermal process isovolumic process
202.6kPa
W2=W WII
= -202.6 103 (2-1) 10-3 (-101.3 103 (4-2) 10-3)
= -405J (
)
dp,
W3= Wr=
p V2 dV
V1
:
nRT ln V2 V1
4
n
n= p1V1 =0.178mol RT
W3=Wr= - 0.178mol 8.314J·mol 1·K 1 273K 4.00dm3
V2 =nRT2 /p2 =(1 8.314 414.6)/101.325=34( L) W = 101.325 (34 – 20)= 1420.48( J)
U = Q + W = 0 – 1420.48 = 1420.48 (J)
U
3.2.1 3.2.2 3.2.3 3.2.4 3.2.5 3.2.6 3.2.7 3.2.8
(H.von Helmholtz) 1847
19 law of thermodynamics
the first
U1 U2
U = U2 - U1 = Q W
dU = Q W
3-1
1mol
101.325kPa
WQ
“
”
487.8K 20L 101.325kPa 414.6K U
Q=0
W = p V = p (V2 V1)
d(pV)=nRdT
Cp=CV+nR
Cm,p=Cm,V+R
Cm,p R
Cm,V
Cm,V =3R/2
Qv
Cm,V=5R/2
Cm,V 3R
Qv Qp
1-
Qp Qv
2-
H = U + (pV)
3-
H = Qp
U=Qv
4-
Qp =Qv + (PV)= Qv + n(RT)
5-
6-
n
7-
ng
Qv Qp
Qp Qv + ng RT
1.00dm3
= -560J
—
—
(reversible process)
W Q
3.1.5
— — —
(internal energy) (thermodynamic energy)
U
U U
T
EK
3 KT 2
EK
3 RT 2 NA
5
3.1.6
(
)
1842 J.R.Mayer (J.P.Joule)1840—1848
3.2.1
(thermochemistry)
(heat of reaction)
(reactant)
(product)
6
3.2.2
W’=0 W = We+ W’= We= -p V U= Q W=Q p V
V=0 U = Qv
U = Q - p V = Qv
Qv Ur QV
Qp
p = p = p1= p2