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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
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.
Work – Basic FormulationWork – Basic Formulation
Work Done = Force x (Distance Moved)
Differentially => dw = F • dS
Work – Gas Expansions and Compressions – I (F14)Work – Gas Expansions and Compressions – I (F14)
Work – Gas Expansions and Compressions – II (F14)Work – Gas Expansions and Compressions – II (F14)
Work – Gas Expansions and Compressions – III (F14)Work – Gas Expansions and Compressions – III (F14)
Work – Gas Expansions and Compressions – IV (F14)Work – Gas Expansions and Compressions – IV (F14)
Work – Gas Expansions and Compressions – V (F14)Work – Gas Expansions and Compressions – V (F14)
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
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
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
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
Work – Gas Expansions and Compressions – V (F13)Work – Gas Expansions and Compressions – V (F13)
Reversible versus Irreversible Work
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.
First Law of ThermodynamicsFirst Law of Thermodynamics
wqdU
wqU
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
Two Heat Capacities from q Two Heat Capacities from q
dT
qC
dTCdU V
dTCdH P
CalorimetryCalorimetry
TCq calrxn
• Reaction carried out under constant volume.
• Use a bomb calorimeter.• Usually study combustion.
Bomb Calorimetry (Constant Volume Calorimetry)
Uq v
Constant Pressure (Solution) Calorimetry• Atmospheric pressure is constant!
CalorimetryCalorimetry
T
HqP
solution of grams
solution ofheat specificsolnrxn
TmSq sorxn ln
CalorimetryCalorimetry
Constant Pressure Calorimetry
TSqrxn solution) of mass total(
interest of species of moles**mol
qH rxnrxn
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
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]
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
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.
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
Enthalpy of Phase TransitionsEnthalpy of Phase Transitions
Constant P: s g s g
sf
gv
HHfusionofheatH
HHonvaporizatiofheatH
dTCHT
HC P
PP :PConstant
Enthalpy of Phase TransitionsEnthalpy of Phase Transitions
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
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
• 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
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
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
Enthalpies of Formation: ExampleEnthalpies of Formation: Example
Example: Write the balanced reaction equation for the standard enthalpy of formation of solid ammonium carbonate.
Enthalpies of FormationEnthalpies of Formation
Substance of (kJ/mol)
C(s, graphite) 0
O(g) 247.5
O2(g) 0
N2(g) 0
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
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
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
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