2.9 Joule-Thomson experiments

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# 2.9 Joule-Thomson experiments - PowerPoint PPT Presentation

The of real gases:. Q =0. In compressing process:. In expanding process:. 2.9 Joule-Thomson experiments. so. Define. and. Joule-Thomson coefficient. The enthalpy keep constant in the whole process. Experimentally determined isenthalpic curve. He         40 K      N2         621 K

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The of real gases:

Q=0

In compressing process:

In expanding process:

2.9 Joule-Thomson experiments

so

Define

and

Joule-Thomson coefficient

The enthalpy keep constant in the whole process

Experimentally determined isenthalpic curve

He         40 K

N2         621 K

O2         764 K

Ne         231 K

Inversion temperature of real gases

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

Joule-Thomson Apparatus

Work principle of a refrigerator

Application of the Joule-Thomson effect

Liquefying GASes using an isenthalpic expansion

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.

=

S

S

=

D

Q

U

(

-

U

(

U

product）

reactant）

V

r

2.9 The thermochemistry
• The energy changes in chemical reactions
• the heat produced or required by chemical reactions

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:

△rHm(298.15K)

(298.15K)

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.

Cl(g)+H(g)

F(g)+H(g)

300

H(g)

200

-431

+218

100

KJ.mol-1

F2 Cl2 H2

-92

-564

-100

HCl(g)

-269

-200

-300

HF(g)

Understanding standard molar enthalpy of formation

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).

H2+O2

0

-50

-188

-242

-100

KJ.mol-1

-285

-150

H2O2

-200

H2O(g)

-250

H2O(l)

-300

The ∆fH depend on state of substances

C2H5OH(g) C4H6(g) H2O(g) H2(g)

-235.10 110.16 -241.81 0

Example: Calculate the standard enthalpy of following reaction at 25℃ by using standard molar enthalpy of formation

(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

C+O2

400

-110

CO

300

200

-394

KJ.mol-1

-284

100

CO2

0

298.15K, 100kPa：

CH3COOH(l)+2O2(g)=2CO2(g)+2H2O(l)

1/2N2(g)+3/2H2(g) NH3(g)

N(g)+3H(g)

314kJ.mol-1

NH(g)+2H(g)

390kJ.mol-1

1170kJ.mol-1

1124kJ.mol-1

NH2(g)+H(g)

466kJ.mol-1

NH3(g)

1/2N2(g)+3/2H2(g)

46 kJ.mol-1

• A sample of biphenyl (C6H5)2 weighing 0.526 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 0.825 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

For example：（1）

（2）

（1）-（2）=（3）

（3）

Hess’s law

The enthalpy change for any sequence of reactions that sum to the same overall reaction is identical.

(Based on Enthalpy being a state function)

–(a)+2(b)+(c)=the above studied reaction

Example 2:
• C(graphite) + 2H2(g) → CH4(g)
• Consider following reactions
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