1. 2.9 Joule-Thomson experiments
The of real gases:
Q=0
In compressing process:
In expanding process:

2. The enthalpy keep constant in the whole processso
and
Define
Joule-Thomson coefficient
Experimentally determined isenthalpic curve

3. Inversion temperature of real gases
He         40 K
     N2         621 K
     O2         764 K
     Ne         231 K
Inversion curve
μ >0, P↓, △P<0, △T<0,T ↓decrease
μ =0, P↓, △P<0, △T=0,T not change
μ <0, P↓, △P<0, △T>0,T increase

4. 273K, 1atm Gas μJ-T/K.MPa-1 Ar 3.66 C6H14 -0.39 CH4 4.38 CO2 10.9 NH3
2.69 H2
-0.34
Joule-Thomson Apparatus

5. Application of the Joule-Thomson effect
Work principle of a refrigerator
Liquefying GASes using an isenthalpic expansion

6. Why μ>0, =0 or <0 ?

7. 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.

8. 2.9 The thermochemistry U Q D = S ( -
The energy changes in chemical reactions
the heat produced or required by chemical reactions U Q r V D = S
reactant）
product） ( -
Measurable
predictable
At the same temperature ( reactants, products)

9. molar enthalpy of reaction
Standard molar enthalpy of reaction
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:

10. Calculation of standard enthalpy of reactions
(1) By standard molar enthalpy of formation
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)
△rHm(298.15K)

11. Understanding standard molar enthalpy of formation
KJ.mol-1
-300
-200
-100
100
200
300
F(g)+H(g)
F Cl 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).

12. The ∆fH depend on state of substances
KJ.mol-1
-250
-200
-150
-100
-50
-300
H2+O2
H2O2
H2O(g)
H2O(l)
-188
-242
-285

13. Example: Calculate the standard enthalpy of following reaction at 25℃ by using standard molar enthalpy of formation
C2H5OH(g) C4H6(g) H2O(g) H2(g)

14. (2) By standard molar enthalpy of combustion
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.

15. Standard molar enthalpy of combustion
400
-110 CO
300
200
-394
KJ.mol-1
-284
100
CO2

16. CH3COOH(l)+2O2(g)=2CO2(g)+2H2O(l)
298.15K, 100kPa：
CH3COOH(l)+2O2(g)=2CO2(g)+2H2O(l)
（A） （B） （C） (D)

17. (3) By bond energies and bond enthalpiesC N O F Cl Br I Si
436
415
390
464
569
432
370
295
395
345
290
350
439
330
275
240
360
160
200
270
140
185
205
255
280
540
243
220
210
359
190
180
150
230

18. 1/2N2(g)+3/2H2(g) NH3(g) N(g)+3H(g) 314kJ.mol-1 NH(g)+2H(g)

19. (4) Solublization heat

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

21. Hess’s law For example：（1） （2） （1）-（2）=（3） （3）
The enthalpy change for any sequence of reactions that sum to the same overall reaction is identical.
(Based on Enthalpy being a state function)
For example：（1）
（2）
（1）-（2）=（3）
（3）

22. Example 2: C(graphite) + 2H2(g) → CH4(g) Consider following reactions
–(a)+2(b)+(c)=the above studied reaction

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