2.9 Joule-Thomson experiments

Presentation on theme: "2.9 Joule-Thomson experiments"— Presentation transcript:

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

The enthalpy keep constant in the whole process
so and Define Joule-Thomson coefficient Experimentally determined isenthalpic curve

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

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

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

Why μ>0, =0 or <0 ?

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 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)

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:

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)

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).

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

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)

(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.

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

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

(3) By bond energies and bond enthalpies
C 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

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

(4) Solublization heat

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

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）

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

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