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

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Presentation on theme: "2.9 Joule-Thomson experiments The of real gases: Q=0 In compressing process: In expanding process:"— Presentation transcript:

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 process so and Define Joule-Thomson coefficient Experimentally determined isenthalpic curve

3 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,T increase He 40 K N2 621 K O2 764 K Ne 231 K

4 Joule-Thomson Apparatus Gasμ J-T /K.MPa -1 Ar3.66 C6H CH44.38 CO210.9 NH32.69 H K, 1atm

5 Work principle of a refrigerator Application of the Joule-Thomson effect 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 The energy changes in chemical reactions the heat produced or required by chemical reactions UUUQ rV  reactant ) product ) (-( At the same temperature ( reactants, products) Measurable predictable

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 H 2 (g,p  ) + I 2 (g,p  )=2HI(g,p  ) △ r H m  (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) △ r H m  (298.15K)

11 Understanding standard molar enthalpy of formation KJ.mol F(g)+H(g) F2 Cl2 H2 H(g) +218Cl(g)+H(g) HF(g) HCl(g) 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 ∆ f H depend on state of substances KJ.mol H2+O2 H2O2H2O2 H 2 O(g) H 2 O(l)

13 Example: Calculate the standard enthalpy of following reaction at 25 ℃ by using standard molar enthalpy of formation C 2 H 5 OH(g) C 4 H 6 (g) H 2 O(g) H 2 (g)

14 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. (2) By standard molar enthalpy of combustion

15 Standard molar enthalpy of combustion KJ.mol C+O2 CO CO

16 298.15K, 100kPa : CH 3 COOH(l)+2O 2 (g)=2CO 2 (g)+2H 2 O(l) ( A ) ( B ) ( C ) (D)

17 (3) By bond energies and bond enthalpies HCNOFClBrISi H C N O F Cl Br I Si230

18 1/2N 2 (g)+3/2H 2 (g) NH 3 (g) N(g)+3H(g) NH(g)+2H(g) NH2(g)+H(g) NH 3 (g) 466kJ.mol kJ.mol kJ.mol kJ.mol -1 1/2N2(g)+3/2H2(g) 46 kJ.mol kJ.mol -1

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) = Q v + Δn g RT

21 Hess ’ s law For example :( 1 ) ( 2 ) (1)-(2)=(3) (3) (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)

22 Example 2: C(graphite) + 2H 2 (g) → CH 4 (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) P68: 2.38(a,b) Y: P32: 36, 37 P42: 45


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