Thermodynamics

Energy: Ability to do work or produce heat.

Work=force x distance

force causes the object to move

Gravitational force causes the water to fall. can generate electricity

Energy

kinetic potential energy possessed by an object in virtue of its motion.

Ekin=1/2 mv2

Ekin=3/2 RTNever confuse

T and heat

Heat is the energy transferred from one object to another in virtue of T-difference

Potential energy:

energy possessed by an object due to its presence in a force field i.e. under the effect of external force. Object attracted/repelled by external force. stored energy!

Epot=mgh

Attraction causes the ball to fall, h smaller, Epot smaller.

Attraction causes the potential energy to decrease.

Repulsion causes the potential energy to increase.

Law of conservation of Energy (Axiome):

• Energy can neither be created nor destroyed.

• Energy of universe is constant.

• Energy can be converted from one form to another.

Ekin ↔ Epot

Heat ↔ Work

• Thermodynamics: the study of energy transformation

from one form to another.

• First Law of TD.

System

Part of universe under

investigation.

sys

sys

surroundings

surroundings

sys + surr =

universe

State Function

Change in state function depends

only on initial and final state.

Irbid

Amman

Sea level

h=650 m

h=900 m

Irbid → Amman

Dh=hfinal-hinitial

Dh=hamman-hirbid

Dh=900 m-650 m=250 m

Initial state

Final state

Change doesn’t depend on path

• Examples of state functions:– Temperature– Volume– Pressure– Altitude– Mass– Energy– Concentration

State function

xy

xy

xy

zzz

zdzdzd

dyy

zdx

x

zzd

aldifferentiexact

yxz

yxfz

342

),(

3

),( 11 yx

),( 21 yx

),( 22 yx

(DU)x

(DU) y

xy

z

yx

z

x

yz

y

xz

y

z

x

z

yxfz

xy

xy

22

),(

Schwartz theoremEquality of cross-derivatives

00

46

342

),(

22

2

3

x

yz

xy

z

y

xz

yx

z

y

zx

x

z

yxz

yxfz

yy

xy

2

2

2

2

2

),(

m

Tm

mmm

V

m

mTmmV

m

m

V

R

T

Vp

TV

p

V

R

V

Tp

VT

p

V

TR

V

p

V

R

T

p

V

TRp

TVfp

m

m

Show that p in the VdW equation is a state function!

zxy

xzy

zxy

x

y

y

z

x

z

z

y

y

x

x

z

x

y

y

z

x

z

yxfz

1

),(

Circular Rule

1

1

),(

2

2

RT

Vp

RT

V

R

p

V

R

p

V

V

T

T

p

R

p

V

T

V

TR

V

p

V

R

T

p

R

VpT

p

TRV

V

TRp

TVfp

mm

m

T

m

pmV

pmmTmmV

mm

m

m

m

m

Internal Energy E

Sum of Ekin and Epot of all particles in the system.

State function

First Law of TD

DE = Q + W

The internal energy of a system can be changed1. by gaining or losing heat, Q

2. Work, W, done on the system or by the System

Heat• energy flow from one system to another in virtue

of temperature difference.

– Appears only during a change in state

– Flows across the boundary from the a point of higher

temperature to a point of lower temp

– Not a state function

– Path function

QdQ

Work• Any quantity of energy that “flows” across the

boundary between the system and the surroundings as a result of force acting through a distance.

– Appears only during a change in state

– Completely convertible into lifting a weight in the surr

– Not a state function

– Path function

WdW

DE = Q + W

electricity (work) can be completely converted into lifting of weight.

Q and ware path functions (Depend on

path).

full

initial

empty

final

Path 1

111 wQE

Path 2 22 QE

211

21

QwQ

EE

sys

surroundings

Q

heat transferred from surr to

sys.

Surr loses heat, loses E, Esurr↓

Sys gains heat, gains E, Esys

for Surr: Q < 0 (neg.), DE < 0

for Sys: Q > 0 (pos.), DE > 0

sys

surroundings

Q

heat transferred from sys to

surr.

Sys loses heat, loses E, Esys↓

Surr gains heat, gains E, Esurr

for Sys: Q < 0 (neg.), DE < 0

for Surr: Q > 0 (pos.), DE > 0

DE = Q + W

Qsys > 0 : endothermic process

Qsys < 0 : exothermic process

dQsys > 0

dQsys < 0

dE = dQ + dW

A

gm

A

F

pp

opposite

oppositeisys

1

m1

sys

surr

m1

sys

surr

m2

oppositeisys

opposite

ppA

gmmp

21

m1 m2

surr

sys

A

gmmp

pp

pV

opposite

oppositefsys

syssys

21

Ep=mgh

h↓, Ep↓

h of m1 and m2 ↓

Ep of m1 and m2 ↓

Esurr ↓

Esys

Work done by surroundings on system

(Esys)f > (Esys)i

DEsys > 0

wsys > 0

A

gmm

A

F

pp

opposite

oppositeisys

21

m1

sys

surr

oppositeisys

opposite

ppA

gmp

1

m1 m2

surr

sys

A

gmp

pp

pV

opposite

oppositefsys

syssys

1

Ep=mgh

h, Ep

h of m1

Ep of m1

Esurr

Esys ↓

Work done by system on surroundings

(Esys)f < (Esys)i

DEsys < 0

wsys < 0

m1

surr

sys

• Ex. 6.1

A system undergoes an endothermic process in which 15.6 kJ

of heat flows and where 1.4 kJ work is done on the system.

Calculate the total change in the internal Energy of the

system.

Qsys > 0 Q=+15.6 kJ

wsys > 0 w=+1.4 kJ

DEsys = Qsys + wsys

DEsys = (+15.6 kJ) + (+1.4 kJ) = +17 kJ

m1

sys

surr

hi

m1

surr

sys

hf

initial final

Vpw

VVpwhAhApw

hhApwrApw

ApFA

Fp

rFw

opp

ifoppifopp

ifoppopp

oppopp

-

m1 m2

surr

sys

(Vf, p

f)

p

Vm

(Vi, p

i)

popp

=pf

W

Ideal GasOne-stage isothermal Expansion

T=constant

m1

sys

surr

hi

hf

initial final

-

m1 m2

surr

sys

m1 m2-dm

surr

sys

dVpdhApw

dxFw

oppopp

In each step: - dm is removed - Piston rises by dh - psys = popp

- dw=-poppdV - dw=-psysdV

Reversible isothermal Expansion

reversible: System in equilibrium with surroundings at each point of process oppsys pp

i

f

V

V

sysopp

opp

V

VnRTw

V

VdnRTdV

V

nRTdww

stepeachofworkofsumworktotal

dVV

nRTdw

V

nRTpp

dVpdw

f

i

ln

Ideal gas

i

f

V

VnRTw ln

Reversible Expansion

Ideal gasVf > Vi

w < 0

Reversible Compression

Ideal gasVf < Vi

w > 0

Reversible work larger than irreversible work.

But impractical.Requires infinity long time.

• Ex. 6.2

Calculate the work associated with the expansion of a gas

from 46 L to 64 L at constant external pressure of 15 atm.

46 L 64 L

15 atm 15 atm

o Expansion against the external pressureo External pressure opposes the expansion

o popp=15 atm = constant

LatmLatmLLatmw

VVatmVpw ifopp

.2701815466415

15

• Ex. 6.3

Given a balloon with a volume of 4.00x106 L. It was heated by

1.3x108 J until the volume became 4.5x106L. Assuming the

balloon is expanding against a constant external pressure of

1 atm, calculate the change in the internal energy of the

gas confined by the balloon.

4.00x106 L

1 atm1 atm

4.50x106 L

Vi

Vf

Q

popp

JJJJJE

JmPamPaLatm

LatmJLLatmJE

VVpQEVpQEwQE ifoppopp

77868

333

68668

100.81007.5103.1325.101105.0103.1

325.101.325.10110101325.1

.105.0103.1105.4105.41103.1

HeatMolar heat capacity

Electrical work w=V.I.t → system gains this energy as heat qKinetic energy of system rises → temperature rises

Translational, rotational, vibrational energy increase

cp,m: molar heat capacity at constant pressure heat needed to raise the temperature of 1 mole of substance by 1ºC at constant pressure.

cV,m: molar heat capacity at constant volume heat needed to raise the temperature of 1 mole of substance by 1ºC at constant volume.

cp: heat capacity at constant pressure heat needed to raise the temperature of n mole of substance by 1ºC at constant pressure.

cV: heat capacity at constant volume heat needed to raise the temperature of n mole of substance by 1ºC at constant volume.

mpp cnc ,

mVV cnc ,

pmp

pp dT

q

nc

dT

qc

.

1,

VmV

VV dT

q

nc

dT

qc

.

1,

Extensive and intensive properties

properties

extensive intensive

depends on amount of substance

cp, V, m, n

doesn’t depend on amount of substance

cp,m, p, T

Ratio of two extensive propertiescp,m, Mwt,

n

mMwt

n

cc p

mp ,

dTcndTcq

dT

q

nc

dT

qc

mppp

pmp

pp

,

, .1

dTcndTcq

dT

q

nc

dT

qc

mVVV

VmV

VV

,

, .1

dVpw

dVpqdE

wqdE

opp

dVpqdE opp

00

0.

ww

Vdvolumeconst

tindependenTcifTcnE

qEEE

qdTcnqdE

qdE

mVVm

Vif

V

T

T

VmV

E

E

V

f

i

f

i

,,

,

Vq

0w

Measure qV

→ change in

internal energy

of system DE

determined

VmV

V

mVV

V

TV

ccnT

E

dTcnqEd

dTT

EEddVvolumeconstat

dVV

EdT

T

EEd

VTfE

,

,

0:

),(

dVV

EdTcnEd

T

mV

,

dTcnEd

dTT

EdV

V

EdT

T

EEd

V

EgasidealFor

mV

VTV

T

,

0:

0

0

:/exp

dVV

EdT

T

EEd

dT

gasidealofncompressioansionIsothermal

TV

HqHH

VpEH

qVpEVpE

VpVpqEE

pVpVqVVpqEE

dVpqdVpqdE

dVpqdVpqdVpqdE

pif

piiifff

iiffpif

ifpifpif

V

V

p

V

V

p

E

E

psyspopp

f

i

f

i

f

i

Enthalpy, H

First law under the conditions

- Reversible popp=psys

- constant pressure psys=const

pqHd

p

p

qdH

VdpqdH

VdppdVpdVqdH

VdppdVdEdH

pVddEdH

pVEH

qHd

)(

pmp

p

mpp

p

TV

ccnT

H

dTcnqHd

dTT

HHddppressureconstat

dpp

HdT

T

HHd

pTfH

,

,

0:

),(

dpp

HdTcnHd

T

mp

,

dTcnHd

dTT

Hdp

p

HdT

T

HHd

p

HgasidealFor

mp

pTp

T

,

0:

0

0

:/exp

dpp

HdT

T

HHd

dT

gasidealofncompressioansionIsothermal

Tp

fi zzfunctionstatezdz

dHdE

0

00

dVpdTcn

dVpqdVV

EdT

T

EEd

ppdq

gasidealofncompressioansion

adiabaticreversible

mV

TV

sysopp

,.

0

/exp

V

dV

c

R

T

dT

dVV

RTndTcn

mV

mV

,

,.

i

f

i

f

mV

mp

i

f

mV

mVmp

i

f

mVmp

i

f

mVi

f

V

VmV

T

T

V

V

T

T

c

c

V

V

c

cc

T

T

Rccgasidealfor

V

V

c

R

T

T

V

dV

c

R

T

dT f

i

f

i

ln1ln

lnln

:

lnln

,

,

,

,,

,,

,

,

ii

ff

i

f

i

f

i

f

i

f

i

f

i

f

i

f

i

f

Vp

Vp

RnpV

RnpV

T

T

Rn

pVTTRnpV

V

V

T

T

V

V

T

T

V

V

T

T

1

1

lnln

ln1ln

iiff

iiifff

i

f

ii

ff

VpVp

VVpVVp

V

V

Vp

Vp

11

1

dVpdTcn mV ,.

Adiabatic compression: work done on system → internal energy increase → T↑

00 dTdV

dVpdTcn mV ,.

Adiabatic expansion: work done by system → internal energy decrease → T↓

00 dTdV

reversible adiabatic

work for ideal gas:

TcnnRTnRTR

cnRTnRT

cc

cw

nRTnRTcc

cnRTnRT

c

cnRTnRTw

VpVpVVpVVpw

VconstVconstVVconstV

constw

dVVconstwdVVconstw

VconstpconstpV

dVpdVpw

mVifmV

ifmVmp

mV

ifmVmp

mVif

mV

mpif

iffiifff

ifif

V

V

V

V

opp

ii

f

i

f

i

,,

,,

,

,,

,

,

,

1

1

1

1

1

1

1

1

1

1

1

11

11

1111

dVpdTcn mV ,.much easier

Dependence of internal energy on volume changes

pT

pT

V

E

dVV

EdT

T

EEd

VT

TV

For ideal

gas: V

nR

T

p

V

nRTp

V

0

pppV

nRT

V

E

T

For VdW

gas: 2

2

V

an

bnV

RTnp

bnV

nR

T

p

V

2

2

2

2

V

an

V

an

bnV

RTn

bnV

nRT

V

E

T

Two moles of a VdW gas were compressed at constant temperature of

25oC from 10 L to 1 L. Calculate the change in the internal energy E.

fiif

V

V

V

V

V

V

TTV

VVan

VVan

VanE

V

dVandV

V

anEd

dVV

andV

V

EdV

V

EdT

T

EEd

f

i

f

i

f

i

11111 222

22

2

2

2

2

Vp

Vp

V

E

p

H

VdpdVpV

Edp

p

H

dTisothermal

VdppdVdVV

EdTcndp

p

HdTcn

VdppdVdVV

EdT

T

Edp

p

HdT

T

H

VdppdVdEdH

pVEH

dpp

HdT

T

HHd

TTT

TT

T

mV

T

mp

TVTV

TV

0:

,,

Dependence of enthalpy on pressure

changes

pT

pTV

TpV

TVTVT

VT

TTT

T

VTV

p

H

T

V

p

V

T

p

p

V

V

T

T

p

Vp

V

T

pTV

p

Vpp

T

pT

p

H

pT

pT

V

E

Vp

Vp

V

E

p

H

1

For ideal

gas: p

nR

T

V

p

nRTV

p

For

liquids

and

solids:

0

p

nRT

p

nRT

p

H

T

VTV

p

H

T

pT

VdpdTcndHVp

Hmp

T

.. ,

dTcndTT

EdV

V

EdT

T

EEd

ilitycompressiblowdV

mV

VTV

..

0

,

Relation

between

cp and cv

pT

mVmp

TT

mVmp

T

mVmp

mpp

opp

T

mV

opp

T

mV

TV

T

Vp

V

Ecncn

dVpV

EdVpdV

V

EdTcncn

reversibledVpdVV

EdTcndTcn

dTcnqqpressureconstatprocessaFor

dVpdVV

EdTcnq

dVpqdVV

EdTcn

wqdVV

EdT

T

EEd

,,

,,

,,

,

,

,

..

..

)(..

.

.

.

pVpV

mVmp

VT

pT

mVmp

T

V

T

pT

T

Vpp

T

pTcncn

pT

pT

V

E

T

Vp

V

Ecncn

,,

,,

..

..

p

nR

T

V

p

nRTV

p

For ideal

gas:

Rcc

p

nRp

T

Vp

V

Ecncn

mVmp

pT

mVmp

,,

,, ..

For

liquids

and

solids:

mVmp cc ,, because is rather

smallp

T

V

Joule-Thompson Effect

p1=constant p2=constant Insulated tube →

q=0

2211

0

0

21

0

0

21

21

1

2

1

2

VpVpdVpdVpdVpdVpw

dVpdVpdwdwdw

V

V

V

V

rightleft

00

21

111222

221112

dHH

HH

VpEVpE

VpVpEE

WWQE

TmpH

TJ

TH

mp

T

mp

TV

p

H

cnp

T

p

H

p

Tcn

dpp

HdTcn

dpp

HdT

T

HHd

,

,

,

1

0

isoenthalpic

Joule-Thompson

coefficient

For ideal

gas:

0

Tp

H

01

,

TmpH

TJ p

H

cnp

T

Pressure drop: p1 > p2 dp

< 0dT = 0 no change in Temp

Tinincreasep

T

Tindropp

T

Tinchangenop

T

H

TJ

H

TJ

H

TJ

00

00

00

For VdW

gas:

Can be shown at zero-

pressure limit:

bRT

a

cnp

T

mpH

TJ

21

,

J-T inversion temperature: T at which mJ-T=0

Expansion at this Temp: no change in T

bR

aT

bRT

ab

RT

a

bRT

a

cn

TJ

TJTJ

TJmpTJ

2

20

2

021

,

TinincreasebRT

aTTif

TindropbRT

aTTif

TinchangenobRT

aTTif

TJTJ

TJTJ

TJTJ

02

02

02

If a nitrogen gas cylinder is opened at room temp:

Frost formation on valve

If a hydrogen gas cylinder is opened at room temp:

increase of valve temp

eventual explosion

Liquefying Air

Chemical EnergyCH4(g) +2O2(g) → CO2(g) + 2H2O(g)

C-H O=O C=O O-H

Chemical reaction:o No change in the number/nature of atomso Redistribution of Bonds (change in bonding)o Change in attraction & repulsion forces between the atoms

o Change in the potential energy Ep of molecules

Ep

R

P

Energy is conserved!

Energy difference released as heat.

Heat of reaction (Qv, Qp).

Reaction exothermic.

DHreaction= Hf – Hi = HP – HR < 0

N2(g) + O2(g) → 2NO(g)

N≡N O=O N=O

Ep

R

P Energy is conserved!

Energy difference obtained

from

surroundings as heat.

Heat of reaction (Qv, Qp=DE,

DH).

Reaction endothermic.DHreaction= Hf – Hi = HP – HR > 0

DEreaction= Ef – Ei = EP – ER > 0

o Ep(R) > Ep(P), reaction exothermic

o Ep(R) < Ep(P), reaction endothermic

Thermochemical equation

N2(g) + O2(g) → 2NO(g) DHo=+180.5 kJ

CH4(g) +2O2(g) → CO2(g) + 2H2O(g) DHo=-802.3 kJ

moles

1 mole of gaseous methane (CH4) reacts with two moles of

gaseous molecular oxygen producing 1 mole of gaseous carbon

dioxide, 2 moles of water vapor and 802.3 kJ of heat.

DHo: Standard heat of reaction:Standard conditions: T=25oC, p=1atm.

Calorimetry

caloriecal

measurement

heat unit

1 cal = 4.185 J

Calorimetry = heat measurement experiments

1 Cal =1000 cal

Problem: - heat (Q) is a path function!!!

- Q differs from one way of performing

the experiment to another.

- details of the experiment must be described!!!!

heat measurement experiments

VpQE

wQE

opp

0

tan

V

tconsV

VQE

Heat measured at constant volume:

o Equal to DEo Equal to a change in a state function!!o Details of the experiment no more important.

However, heat measurement experiments are usually performed at constant pressure pVEH

VdppdVdEdH

VpVpQH

VpEH

VVpEEHH

VpEH

pppdpppp

VdppdVdEdH

ififif

VV

EE

HH

iffi

p

p

V

V

E

E

H

H

f

i

f

i

f

i

f

i

f

i

f

i

f

i

.

0

0

pQH Heat measured at constant pressure:

o Equal to DH

o Equal to a change in a state function!!o Details of the exp. no more important.

csp: specific heat (specific heat capacity) heat needed to raise the temperature of 1 gram of substance by 1ºC.

C : heat capacity heat needed to raise the temperature of substance (m gram) by 1ºC.

T increase by 1ºC:

1 g csp

m gr. ? = CspcmC

Q : heat heat needed to raise the temperature of substance (m gram) by a given temperature difference, DTºC.

m gr.

CT increase by 1ºC

T increase by DTºC

? = QTcmQ

TCQ

sp

Bomb Calorimeter

0.5269 g of octane (C8H18) were placed in a bomb calorimeter

with a heat capacity of 11.3 kJ/ºC. The octane sample was

ignited in presence of excess oxygen. The temperature of

the calorimeter was found to increase by 2.25ºC. Calculate

DE of the combustion reaction of octane.

kJCC

kJTCQQ V 4.2525.23.11

DE defined for the reaction as written!!!!!!!!!!!

C8H18(g) +12.5O2(g) → 8CO2(g) + 9H2O(g)

DE defined for the combustion of 1 mole octane (114.2 g)!!

0.5269 g QV

114.2 g ? = DE

DE=-(114.2 g x 25.4 kJ)/0.5269g=-5505 kJ

n

Q

Mwtm

Q

m

QMwtE VVV

When 1.5 g of methane (CH4) was ignited in a bomb

calorimeter with 11.3 kJ/ºC heat capacity, the temperature

rised by 7.3ºC. When 1.15 g hydrogen (H2) was ignited in

the same calorimeter, the temperature rised by 14.3ºC.

Which one of the two substances has a higher specific heat

of combustion (i.e. heat evolved upon the combustion of 1

g of substance)? kJCC

kJTCCHQV 833.73.114

kJCC

kJTCHQV 1623.143.112

1.5 g QV=83 kJ

1 g ? =55 kJ/g

1.15 g QV=162 kJ

1 g ? =141 kJ/g

Coffee-Cup Calorimeter

50 mL of 1.0 M HCl at 25ºC were

added to 50 mL of 1.0 M NaOH at

25ºC in a coffee-cup calorimeter.

The tempe-rature was found to rise

to 31.9ºC. Calculate the heat of the

neutraliza-tion reaction!

Was caused the temperature to increase?

Exothermic Reaction

HCl(aq) +NaOH(aq) → NaCl(aq) + H2O(l)

H+(aq) +OH-

(aq) → H2O(l)

heat evolved = heat gained + heat

gained by reaction by solution by

calorimeter rcalorimetesolutionsprct TCTcmQ

Assumptions:

Ccal=0 (very small mass)

Solution ≈ water (csp)solution=(csp)water=4.18 Jg-1ºC-1

(density)solution=(density)water=1 g/mL

gmLmL

gVdm 1001001

JCCg

JgQrct 2.28849.618.4100

HCl(aq) +NaOH(aq) → NaCl(aq) + H2O(l)

nHCl=MHClxVHCl = 1 mol/L x 0.050 L = 0.050 mol

nNaOH=MNaOHxVNaOH = 1 mol/L x 0.050 L = 0.050 mol

0.050 mol 0.050 mol

0.050 mol 0.050 mol

0.050 mol H2O 2884.2 J mol

1 mol H2O ? Qp=57,684 J/molH2O

DH= -57,684 J/molH2O

DH= -57.7 kJ/molH2O

n

QH p

Hess’s Law

N2(g) + 2O2(g)

initial

2NO2(g)

final

2NO(g)

O2(g) O2(g)

path 1

path 2

)(2)(2)(

)()(2)(2

)(2)(2)(2

21

22

2

22

ggg

ggg

ggg

if

NOONOforH

NOONforH

NOONforH

HH

HHH

N2(g) + O2(g) → 2NO(g) DH2a

2NO(g) + O2(g) → 2NO2(g) DH2b

N2(g) + 2O2(g) → 2NO2(g) DH1

DH1=DH2a+DH2b

The enthalpy of a given chemical reaction is constant,

regardless of the reaction happening in one step or many

steps.

If a chemical equation can be written as the sum of

several other chemical equations (steps), the enthalpy

change of the first chemical equation equals the sum of

the enthalpy changes of the other chemical equations

(steps).

Hess’s Law:

Rules for manipulating thermochemical equations

- If equation is multiplied by a factor, multiply DH by this factor.

N2(g)+3H2(g) → 2NH3(g) DH=-92 kJ

2x (N2(g)+3H2(g) → 2NH3(g) DH=-92 kJ)

2N2(g)+6H2(g) → 4NH3(g) DH=-184 kJ

1/2x (N2(g)+3H2(g) → 2NH3(g) DH=-92 kJ)

1/2N2(g)+3/2H2(g) → NH3(g) DH=-46 kJ

- If equation is reversed, change the sign of DH

2NH3(g) → N2(g) + 3H2(g) DH=+92 kJ

The enthalpy of combustion of graphite is -394 kJ/mol.

The enthalpy of combustion of diamond is -396 kJ/mol.

Calculate DH for the reaction:

Cgraphite → Cdiamond

Solving Strategy• Write the given data in form of thermochemical equations:

CG + O2(g) → CO2(g) DH=-394 kJ

CD + O2(g) → CO2(g) DH=-396 kJ• Construct the equation of interest from the given data:

1 mole cgraphite is needed as reactant. Take the equation in the given data that contains cgraphite. Check the number of moles and whether it is on the reactant side. Manipulate if necessary.

CG + O2(g) → CO2(g) DH=-394 kJ1 mole cdiamond is needed as product. Take the equation in the given data

that contains cdiamond. Check the number of moles and whether it is on the product side. Manipulate if necessary.

CO2(g) → CD + O2(g) DH=+396 kJ

Sum the resulting equations and their DH values:

CG + O2(g) → CO2(g) DH=-394 kJ

CO2(g) → CD + O2(g) DH=+396 kJ

Cgraphite → Cdiamond DH=+2 kJ

Given:

2 B(s)+3/2 O2(g) → B2O3(s) DH=-1273 kJ

B2H6(g)+3 O2(g) → B2O3(s) + 3 H2O(g) DH=-2035 kJ

H2(g)+1/2 O2(g) → H2O(l) DH=-286 kJ

H2O(l) → H2O(g) DH=+44 kJ

Calculate DH for

2 B(s) + 3 H2(g) → B2H6(g)

2 B(s)+3/2 O2(g) → B2O3(s) DH=-1273 kJ

B2O3(s) + 3 H2O(g) → B2H6(g)+3 O2(g) DH=+2035 kJ

3H2(g)+3/2 O2(g) → 3 H2O(l) DH=3x(-286) kJ

2 B(s) + 3 H2O(g) + 3 H2(g) → B2H6(g)+3 H2O(l) DH=-96 kJ

3 H2O(l) → 3 H2O(g) DH=3x(+44) kJ

2 B(s) + 3 H2(g) → B2H6(g DH=+36 kJ

Heat of Formation

Formation reaction:

reaction of forming 1 mole of product from the

elements in their stable form at 25ºC and 1

atm.Heat of formation = DH of formation reaction = DFH

Standard heat of formation = DHº of formation reaction = DFHº

DFHº(NO(g)): ½ N2(g)+½ O2(g) → NO(g) DHº

DFHº(CO(g)): Cgraphite(s)+½ O2(g) → CO(g) DHº

DFHº(O(g)): ½ O2(g) → O(g) DHº

DFHº(Cdiamond(s)): Cgraphite(s) → Cdiamond(s) DHº

DFHº(O2(g)): O2(g) → O2(g) DHº=0

DFHº(Cgraphite(s)): Cgraphite(s) → Cgraphite(s) DHº=0

01,25

atmC

elementsstableH o

oF

tsreac

oiFiproducts

oiFi

orct HnHnH

tan

CH4(g) +2O2(g) → CO2(g) + 2H2O(g)DH=?

CG(s)+ O2(g) → CO2(g) DFH(CO2)

2x (H2(g)+1/2 O2(g) → H2O(g) ) 2xDFH(H2O)

CH4(g) → CG(s) + 2 H2(g) -DFH(CH4)

O2(g) → O2(g) -DFH(O2)=0

CH4(g) +2O2(g) → CO2(g) + 2H2O(g)

DH=DFH(CO2)+ 2xDFH(H2O) - DFH(CH4) - DFH(O2)

DH=DFH(CO2)+ 2xDFH(H2O) – [DFH(CH4) +DFH(O2)]

tsreac

oiFiproducts

oiFi

orct HnHnH

tan

4 NH3(g) +7 O2(g) → 4 NO2(g) + 6 H2O(l)DH=?

DH= 4xDFH(NO2)+ 6xDFH(H2O) – 4xDFH(NH3)

2 Al(s) +Fe2O3(s) → Al2O3(s) + 2 Fe(s)DH=?

DH= DFH(Al2O3)+ 2xDFH(Fe) – [DFH(Fe2O3)+ 2xDFH(Al)]

DH= DFH(Al2O3) – DFH(Fe2O3)

2 CH3OH(l) +3 O2(g) → 2 CO2(g) + 4 H2O(l)DH=?

DH= 2xDFH(CO2)+ 4xDFH(H2O) – 2xDFH(CH3OH)

DH= 2x(-394 kJ)+ 4x(-286 kJ) – 2x(-239 kJ)=-1454 kJ

2 mol CH3OH -1454 kJ

2x32 g

?

-1454 kJ

1 g = -22.7 kJ/g

Calculate the heat of combustion of methanol

(CH3OH(l)) in kJ/g and compare its value with that of

octane (C8H18(l)).

C8H18(l) +12.5 O2(g) → 8 CO2(g) + 9 H2O(l)

DH= 8xDFH(CO2)+ 9xDFH(H2O) – DFH(C8H18)

DH= 8x(-394 kJ)+ 9x(-286 kJ) – (-276 kJ)=-5450 kJ

1 mol C8H18 -5450 kJ

114 g

?

-5450 kJ

1 g = -47.8 kJ/g