材料科学基础08-2-计算吉布斯自由能 (2)

dQ dH cP ≡ = dT P dT
∆H T2 − ∆H T1 = ∫ CP dT
T1
T2
dH = c P dT
Cp：热容
∆H T2 = ∆H T1 + ∫ CP dT
T1
T2
C P = a + bT + cT −2
a、b、c in data manuals
Calculate Enthalpy of Formation ∆HT
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Standard State of Liquid and Solid
• Standard pressure: 1 atm (101.325kPa) • Standard state : 1 atm and T (any temperature) • ∆H0, ∆S0, ∆G0 –––0 represents the values of materials under standard state • ∆H, ∆S, ∆G also represents the values of standard state of liquid and solid materials because pressure does not influence much the values of ∆H, ∆S, ∆G
(2)
(1)
∆H T2 = ∆H T1 + ∫ ∆C P dT
T1
T2
(3) (4) (5) (6)
∆C P = Σ(C P ) products − Σ(C P )反应物 = ( mC P,M + nC P,N ) − ( aC P,A + b∆C P,B )
C P = a + bT + cT −2 ∆C P = ∆a + ∆bT + ∆cT −2
C P = a + bT + cT −2
T2
ST2 = ST1 + ∫
T2
cP dT T T1
ST2 = ST1 + ∫ (
T1
a + b + cT −3 )dT T
1 ST2 = ST1 + a ln(T2 / T1 ) + b(T2 − T1 ) − c (T2− 2 − T1− 2 ) 2 Is integration right?
1 ∆H S = a (T − 298 ) + b T 2 − 298 2 2
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Compute Enthalpy of Formation ∆HT
For the liquid phase:
L ∆H L (T ) = ∆H ref + T L P
Solid transforms into liquid at 1234K (between 298 and 1500K)
Standard Molar Enthalpy of Formation
• ∆G = ∆H – T∆S • ∆H: Standard molar enthalpy/heat of formation 标准摩尔生成焓/热 •Standard molar enthalpy of formation for elements （元素）
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Compute Entropy of Formation at 1500K
Last lecture
∆S =
QR T
dS =
dQ T
dQ = TdS
Entropy can be computed in a similar fashion:
Why do We Need to Study thermodynamics?
Could the following reaction happen at 1000°C? 2Al + 3CO = Al2O3 + C How do you know?
Calculate standard free Gibbs energy (吉布斯自由能）: ∆G0 = ∆H0 – T∆S0 Chemical thermodynamics（化学热力学）
Reaction: aA + bB = mM + nN
∆C P = ∆a + ∆bT + ∆cT −2
T2
∆H T2 = ∆H T1 + ∫ ∆C P dT = ∆H T1 + ∆H T1 →T2
Standard State（标准状态）
• Standard pressure: Ø 1 atm (101.325kPa) • Standard state （标准状态）: Ø 1 atm and T (any temperature) 0 • ∆H , ∆S0, ∆G0 –––0 represents standard state • Stable state （稳定状态）：at 1 atm and any temperature Example: 1 atm at 1500K • Gaseous state of O2(g), N2(g), H2(g) is standard state as their Tb is below 1500k • Solid state of Al2O3(s) is standard state because Tm is higher than 1500K • Liquid state of Al(l) is standard state as Tm = 933K and Tb = 2767K
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Compute Entropy of Formation at 1500K
A phase transition occurs at 1234K (between 298 and 1500K)
S S →L ∆H L (T ) = ∆H T + ∆H T + trans trans
(
) ∫ c (T )dT
L P Ttrans T Ttrans
T
∆H L (T ) =
Ttrans
Hale Waihona Puke Baidu
298
S S→L ∫ cP dT + ∆H Ttrans +
∫c
L P
dT
Compute Enthalpy of Formation ∆HT
298
S S→L ∫ cP (T ) dT + ∆H1234 +
∫c
L P
dT '
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Compute Enthalpy of Formation ∆HT
∆H L
→L ∆H TStrans
∆H S
298 K
1234 K
For the liquid phase:
L ∆H L (T ) = ∆H ref + T L ∫ cP dT
Tref
NOTE: HL is a straight line in this case because cP is a constant (i.e. cP=a).
Tref
∫ c (T ')dT '
S S →L ∆H L (T ) = ∆H T + ∆H T + trans trans
(
) ∫ c (T ')dT '
L P Ttrans T 1234
T
∆H L (T ) =
1234
We need a reference relative to the solid phase, for which we made an assumption.
1 C ( s) + O2 ( g ) = CO( g ) 2
0 (kJ/mole) ∆H 298
- 110.54 - 393. 51 - 825.50
3 Fe (s ) + O2 ( g ) = Fe2O3 ( s) 2
0 ∆H 298 is available for many materials in data manuals
a, b in (5) are different from those in (1) ~ (4)
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Computer Enthalpy of Reaction at T
0 ∆H 298 = 0 for graphite C(s), O2(g), Fe(s)
•Standard (298 k and 1 atm) molar enthalpy of formation for compounds
Reaction
C ( s ) + O2 ( g ) = CO2 ( g )
dQ cP ≡ dT P c dS = P dT T
T2
dQ = c P dT = TdS ∆S =
T2
T1
∫T
cP
dT
C P = a + bT + cT −2
T2
c ST2 − ST1 = ∫ P dT T T1
ST2 = ST1 + ∫
cP dT T T1
Compute Entropy of Formation at 1500K
∆H (T ) = 0 +
S
298
∫c
S P
dT
T
∆H (T ) = 0 +
S
298
∫ (a + bT )dT
T
1 ∆H S (T ) = aT + bT 2 2
298
1 ∆H S (T ) = a (T − 298 ) + b T 2 − 298 2 2
Thus, we obtain a general function for the enthalpy of solid silver as a function of temperature.
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)
R.E. Napolitano – Iowa State University
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Calculate Enthalpy of Formation ∆HT
298k
∆H S T
ST2 = ST1 +
Ttrans
T1
∫
S 2 cL cP S →L dT + S trans + ∫ P dT T T Ttrans
T
Computer Enthalpy of Reaction at T
• Reaction: aA + bB = mM + nN
• Materials A, B, M, N with number of molecule a, b, m, n respectively ∆H T1 = Σ(∆H T1 ) products − Σ(∆H T1 )反应物 = ( m∆H M + n∆H N ) − ( a∆H A + b∆H B ) Most reactions do not happen at 298K
∆H L
→L ∆H TStrans
∆H S
1234 K
1500 K
So, at 1500 K:
∆H L (1500) =
1234
298
S S →L ∫ c P dT + ∆H1234 +
1500
1234
∫c
L P
dT
1 = a S (1234 − 298 ) + b S 1234 2 − 298 2 2 + 2855 + a L (1500 − 1234 )
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Calculate Enthalpy of Formation ∆HT
物质的生成焓：通常只有298°C的数据；那么不同温度的 生成焓怎么得到？
Qp = △H
∆H = ∫ C p dT
Compute standard enthalpy of liquid Ag at 1500K and atmospheric pressure. Let’s start with the solid phase:
S ∆H S (T ) = ∆H ref + T T
Tref
∫c
S P
dT
By convention, we assume that the enthalpy of the lowest energy phase is zero at 298K.