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Chem. 412 – Phys. Chem. I H e a t F low O th Law In te rn a l E nergy Enthalpy H e a t & W ork E n erg y 1 st L aw R e a c tio n Spontaneity E n e rg y D istribution Entropy 2 n d Law Z e ro K 3 rd Law L a w s o f T herm odynam ics
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Chem. 412 – Phys. Chem. I

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Chem. 412 – Phys. Chem. I. Sign Convention. Δ U = Internal Energy. q = heat flow; transfer of energy between two objects. w = work; product of force applied to an object over a distance. Surroundings. System. Work – Basic Formulation. Work Done = Force x (Distance Moved) - PowerPoint PPT Presentation
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Page 1: Chem. 412 – Phys. Chem. I

Chem. 412 – Phys. Chem. IChem. 412 – Phys. Chem. I

Heat Flow

Oth Law

Internal EnergyEnthalpy

Heat & W ork

Energy

1st Law

Reaction Spontaneity

Energy Distribution

Entropy

2nd Law

Zero K

3rd Law

Law s of T herm odynam ics

Page 2: Chem. 412 – Phys. Chem. I

Sign ConventionSign Convention

System

Surroundings

ΔU = Internal Energy.

q = heat flow; transfer of energy between two objects.

w = work; product of force applied to an object over a distance.

Page 3: Chem. 412 – Phys. Chem. I

Work – Basic FormulationWork – Basic Formulation

Work Done = Force x (Distance Moved)

Differentially => dw = F • dS

Page 4: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – I (F14)Work – Gas Expansions and Compressions – I (F14)

Page 5: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – II (F14)Work – Gas Expansions and Compressions – II (F14)

Page 6: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – III (F14)Work – Gas Expansions and Compressions – III (F14)

Page 7: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – IV (F14)Work – Gas Expansions and Compressions – IV (F14)

Page 8: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – V (F14)Work – Gas Expansions and Compressions – V (F14)

Page 9: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – I (F13)Work – Gas Expansions and Compressions – I (F13)

Work is area under curve of P-V Diagram.

dw = F • dS = - P • dV

Page 10: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – II (F13)Work – Gas Expansions and Compressions – II (F13)

Work is area under curve of P-V Diagram.

dw = F • dS = - P • dV

Page 11: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – III (F13)Work – Gas Expansions and Compressions – III (F13)

Work is area under curve of P-V Diagram.

Reversible versus Irreversible Work

Page 12: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – IV (F13)Work – Gas Expansions and Compressions – IV (F13)

Work is area under curve of P-V Diagram.

Reversible versus Irreversible Work

Page 13: Chem. 412 – Phys. Chem. I

Work – Gas Expansions and Compressions – V (F13)Work – Gas Expansions and Compressions – V (F13)

Reversible versus Irreversible Work

Page 14: Chem. 412 – Phys. Chem. I

Tutorial Problem on ‘Work’Tutorial Problem on ‘Work’

Consider one mole of an ideal gas kept in a right-circular cylinder with a movable piston cap at constant temperature. The cap is attached to an external engine that is capable of moving the piston in either directions.

(a) Initially, the cylinder has a volume of 1.0 m3 at a pressure of 10. Pa. The cap was moved to give a final pressure of 1.0 Pa. In this case, the piston cap was moved at an infinitely slow rate to achieve this final pressure, thereby following the ideal gas law. Calculate the work involved in this process. This work is referred to as reversible work, wrev . [ 23 J ]

(b) The initial and final volume/pressure was kept the same as in part (a) but the piston cap was suddenly released to get to the final pressure value. Calculate the work involved in this process. This work is referred to as irreversible work, wirrev . [ 9 J ]

(c) Draw one P-V diagram for parts (a) and (b).

(d) Repeat parts (a), (b), and (c) by reversing the initial and final volume/pressure conditions; that is, compression instead of expansion. [ 23 J, 90 J ]

(e) Compare/Contrast/Discuss the above results.

Page 15: Chem. 412 – Phys. Chem. I
Page 16: Chem. 412 – Phys. Chem. I

First Law of ThermodynamicsFirst Law of Thermodynamics

wqdU

wqU

Page 17: Chem. 412 – Phys. Chem. I

First Law of ThermodynamicsFirst Law of Thermodynamics

U = Internal Energy = Heat Flow under Constant Volume

H = Enthalpy = Heat Flow under Constant Pressure

H = U + PV

Page 18: Chem. 412 – Phys. Chem. I

Two Heat Capacities from q Two Heat Capacities from q

dT

qC

dTCdU V

dTCdH P

Page 19: Chem. 412 – Phys. Chem. I

CalorimetryCalorimetry

TCq calrxn

• Reaction carried out under constant volume.

• Use a bomb calorimeter.• Usually study combustion.

Bomb Calorimetry (Constant Volume Calorimetry)

Uq v

Page 20: Chem. 412 – Phys. Chem. I

Constant Pressure (Solution) Calorimetry• Atmospheric pressure is constant!

CalorimetryCalorimetry

T

qq

HqP

solution of grams

solution ofheat specificsolnrxn

TmSq sorxn ln

Page 21: Chem. 412 – Phys. Chem. I

CalorimetryCalorimetry

Constant Pressure Calorimetry

TSqrxn solution) of mass total(

interest of species of moles**mol

qH rxnrxn

Page 22: Chem. 412 – Phys. Chem. I

Calorimetry Examples

1. In an experiment similar to the procedure set out for Part (A) of the Calorimetry experiment, 1.500 g of Mg(s) was combined with 125.0 mL of 1.0 M HCl. The initial temperature was 25.0oC and the final temperature was 72.3oC. Calculate: (a) the heat involved in the reaction and (b) the enthalpy of reaction in terms of the number of moles of Mg(s) used. Ans: (a) –25.0 kJ (b) –406 kJ/mol

2. 50.0 mL of 1.0 M HCl at 25.0oC were mixed with 50.0 mL of 1.0 M NaOH also at 25.0oC in a styrofoam cup calorimeter. After the mixing process, the thermometer reading was at 31.9oC. Calculate the energy involved in the reaction and the enthalpy per moles of hydrogen ions used. Ans: -2.9 kJ , -58 kJ/mol [heat of neutralization for strong acid/base reactions]

CyberChem video

Page 23: Chem. 412 – Phys. Chem. I

Calorimetry Examples: Hints50.0 mL of 1.0 M HCl at 25.0oC were mixed with 50.0 mL of 1.0 M NaOH also at 25.0oC in a styrofoam cup calorimeter. After the mixing process, the thermometer reading was at 31.9oC. Calculate the energy involved in the reaction and the enthalpy per moles of hydrogen ions used. Ans: -2.9 kJ , -58 kJ/mol [heat of neutralization for strong acid/base reactions]

Page 24: Chem. 412 – Phys. Chem. I
Page 25: Chem. 412 – Phys. Chem. I

Comparison of Heat Capacities: CP vs. CVComparison of Heat Capacities: CP vs. CV

VV

PP T

UCand

T

HC

T)f(V,UandT)f(P,H

TP

VP V

UP

T

VCC

Gas) (IdealRnCC VP Gas) (IdealRnCC VP

Page 26: Chem. 412 – Phys. Chem. I
Page 27: Chem. 412 – Phys. Chem. I
Page 28: Chem. 412 – Phys. Chem. I
Page 29: Chem. 412 – Phys. Chem. I

Example on State and Path FunctionsExample on State and Path Functions20.0 grams of Argon gas was heated from 20.0oC to 80.0oC. Considering Argon

behaving as an ideal gas, calculate q, w, U, and H for the following processes:

(a) under Constant Volume, and

(b) under Constant Pressure.

Page 30: Chem. 412 – Phys. Chem. I

Constant Volume Constant Pressure

q 374 J 623 J q

w 0 -249 J w

U 374 J 374 J U

H 623 J 623 J H

CVm 12.5 J mol-1 K-1 20.8 J mol-1 K-1 CPm

Page 31: Chem. 412 – Phys. Chem. I

Enthalpy of Phase TransitionsEnthalpy of Phase Transitions

Constant P: s g s g

sf

gv

HHfusionofheatH

HHonvaporizatiofheatH

dTCHT

HC P

PP :PConstant

Page 32: Chem. 412 – Phys. Chem. I

Enthalpy of Phase TransitionsEnthalpy of Phase Transitions

Page 33: Chem. 412 – Phys. Chem. I

Enthalpy of Phase Transitions

0

5

10

15

20

25

30

35

-50 0 50 100 150

Temperature (oC)

He

at

Flo

w a

t C

on

st.

P

Page 34: Chem. 412 – Phys. Chem. I

Relating Urxn and HrxnRelating Urxn and Hrxn

)(PVUH

PVUH

t.significan quite is (PV) :gasesfor However

omit. ,negligible is )( :liquids & solidsFor

PV

)(

)()(

Tconstant at

)()(

nRT PV :Gases IdealFor

grxnrxn

g

nRTUH

nRTPV

nRTPV

Page 35: Chem. 412 – Phys. Chem. I

• Hess’s law: if a reaction is carried out in a number of steps, H for the overall reaction is the sum of H for each individual step.

• For example:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2H2O(g) 2H2O(l) H = -88 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ

Hess’s LawHess’s Law

Page 36: Chem. 412 – Phys. Chem. I

Hess’s LawHess’s LawGiven: (i) Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) ∆H = -26.7 kJ/mol

(ii) CO(g) + ½O2(g) CO2(g) ∆H = -283.0 kJ/mol

Calculate the heat of reaction for: 2Fe(s) + 3/2 O2(g) Fe2O3(s)

Ans: -822.3 kJ/molAns: -822.3 kJ/mol

Page 37: Chem. 412 – Phys. Chem. I

If 1 mol of compound is formed from its constituent elements (standard state), then the enthalpy change for the reaction is called the enthalpy of

formation, Hof .

• Standard conditions (standard state): Most stable form of the substance at 1 atm and 25.00 oC (298.15 K).

• Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state.

• Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.

• Standard enthalpy of formation of the most stable form of an element is zero.

Enthalpies of FormationEnthalpies of Formation

Page 38: Chem. 412 – Phys. Chem. I
Page 39: Chem. 412 – Phys. Chem. I

Enthalpies of Formation: ExampleEnthalpies of Formation: Example

Example: Write the balanced reaction equation for the standard enthalpy of formation of solid ammonium carbonate.

Page 40: Chem. 412 – Phys. Chem. I

Enthalpies of FormationEnthalpies of Formation

Substance of (kJ/mol)

C(s, graphite) 0

O(g) 247.5

O2(g) 0

N2(g) 0

Page 41: Chem. 412 – Phys. Chem. I
Page 42: Chem. 412 – Phys. Chem. I

Bucky Ball drawn by HyperChem

Page 43: Chem. 412 – Phys. Chem. I

Using Enthalpies of Formation to Calculate Enthalpies of Reaction

• For a reaction

• Calculate/Compare heat of reactions for the combustion of methanol gas and ethanol gas giving carbon dioxide and water.

Enthalpies of FormationEnthalpies of Formation

reactantsproductsrxn ff HmHnH

Page 44: Chem. 412 – Phys. Chem. I
Page 45: Chem. 412 – Phys. Chem. I
Page 46: Chem. 412 – Phys. Chem. I

Temperature Dependence of CP and Hrxn Temperature Dependence of CP and Hrxn

2TcTbaCP

preactsPprodsPrxn

P

reactsf

P

prodsf

P

rxn

reactsfprodsfrxn

CCCdT

Hd

T

H

T

H

T

H

HHH

)()(

,,

,,

)(

Pconstant at

)()()(

2 :where

)(2

1

2

1

TcTbaC

dTCHd

P

T

T P

T

T rxn

Page 47: Chem. 412 – Phys. Chem. I

Temperature Dependence of CP and Hrxn Temperature Dependence of CP and Hrxn

Example: Find the heat of reaction for the following reaction at 1000. K ( Ho

rxn,1000 K ? )

NaCl(s) Na(g) + ½ Cl2(g)

[ Given: Horxn,298K = 519.23 kJ ]

Species CP / J K-1 mol-1

Na(g) 20.80

NaCl(s) 49.70

Cl2(g) 31.70 + 10.14x10-3 T – 2.72x10-7 T2

Page 48: Chem. 412 – Phys. Chem. I

Heat Flow

Oth Law

Internal EnergyEnthalpy

Heat & W ork

Energy

1st Law

Reaction Spontaneity

Energy Distribution

Entropy

2nd Law

Zero K

3rd Law

Law s of Therm odynam icswqdU dVPw

Hq

Uq

P V

VV

PP T

UC

T

HC

&

R nCC VPR nCC VP

PVUH

)( grxnrxn nRTUH