Work principle of a refrigerator Application of the Joule-Thomson effect 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 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
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:
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)
Understanding standard molar enthalpy of formation KJ.mol -1 -300 -200 -100 100 200 300 F(g)+H(g) F2 Cl2 H2 H(g) +218Cl(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 ∆ f H depend on state of substances KJ.mol -1 -250 -200 -150 -100 -50 0 -300 H2+O2 H2O2H2O2 H 2 O(g) H 2 O(l) -188 -242 -285
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) -235.10 110.16 -241.81 0
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
Standard molar enthalpy of combustion KJ.mol -1 0 200 300 400 100 C+O2 CO CO2 -110 -284 -394
298.15K, 100kPa ： CH 3 COOH(l)+2O 2 (g)=2CO 2 (g)+2H 2 O(l) （ A ） （ B ） （ C ） (D)
(3) By bond energies and bond enthalpies HCNOFClBrISi H436415390464569432370295395 C345290350439330275240360 N160200270200270 O140185205185200370 F160255160280540 Cl243220210359 Br190180290 I150210 Si230
The experimental determination of standard enthalpy of combustion 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) = Q v + Δn g RT
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)
Example 2: C(graphite) + 2H 2 (g) → CH 4 (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) P68: 2.38(a,b) Y: P32: 36, 37 P42: 45