Thermochemistry

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 FunctionChange in state function depends only on initial

and final state.

Irbid

Amman

Sea level

h=650 m

h=900 m

Irbid → Amman

h=hfinal-hinitial

h=hamman-hirbid

h=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

Internal Energy E

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

State function

First Law of TD

E = 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

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.), E < 0

for Sys: Q > 0 (pos.), E > 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.), E < 0

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

E = Q + W

Qsys > 0 : endothermic process

Qsys < 0 : exothermic process

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

Esys > 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

Esys < 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

Esys = Qsys + wsys

Esys = (+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

-

Q and w are path functions (Depend on path).

full

initial

empty

final

Path 1

111 wQE

Path 2 22 QE

211

21

QwQ

EE

• 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

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.

Hreaction= 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=E,

H).

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

Ereaction= 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) Ho=+180.5 kJ

CH4(g) +2O2(g) → CO2(g) + 2H2O(g) Ho=-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.

Ho: 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 E

o 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 H

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, TºC.

m gr.

CT increase by 1ºC

T increase by TºC

? = 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

E of the combustion reaction of octane.

kJCC

kJTCQQ V 4.2525.23.11

E defined for the reaction as written!!!!!!!!!!!

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

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

0.5269 g QV

114.2 g ? = E

E=-(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

H= -57,684 J/molH2O

H= -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) H2a

2NO(g) + O2(g) → 2NO2(g) H2b

N2(g) + 2O2(g) → 2NO2(g) H1

H1=H2a+H2b

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 H by this factor.

N2(g)+3H2(g) → 2NH3(g) H=-92 kJ

2x (N2(g)+3H2(g) → 2NH3(g) H=-92 kJ)

2N2(g)+6H2(g) → 4NH3(g) H=-184 kJ

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

1/2N2(g)+3/2H2(g) → NH3(g) H=-46 kJ

- If equation is reversed, change the sign of H

2NH3(g) → N2(g) + 3H2(g) H=+92 kJ

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

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

Calculate H for the reaction:

Cgraphite → Cdiamond

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

CG + O2(g) → CO2(g) H=-394 kJ

CD + O2(g) → CO2(g) H=-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) H=-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) H=+396 kJ

Sum the resulting equations and their H values:

CG + O2(g) → CO2(g) H=-394 kJ

CO2(g) → CD + O2(g) H=+396 kJ

Cgraphite → Cdiamond H=+2 kJ

Given:

2 B(s)+3/2 O2(g) → B2O3(s) H=-1273 kJ

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

H2(g)+1/2 O2(g) → H2O(l) H=-286 kJ

H2O(l) → H2O(g) H=+44 kJ

Calculate H for

2 B(s) + 3 H2(g) → B2H6(g)

2 B(s)+3/2 O2(g) → B2O3(s) H=-1273 kJ

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

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

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

3 H2O(l) → 3 H2O(g) H=3x(+44) kJ

2 B(s) + 3 H2(g) → B2H6(g H=+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 = H of formation reaction = FH

Standard heat of formation = Hº of formation reaction = FHº

FHº(NO(g)): ½ N2(g)+½ O2(g) → NO(g) Hº

FHº(CO(g)): Cgraphite(s)+½ O2(g) → CO(g) Hº

FHº(O(g)): ½ O2(g) → O(g) Hº

FHº(Cdiamond(s)): Cgraphite(s) → Cdiamond(s) Hº

FHº(O2(g)): O2(g) → O2(g) Hº=0

FHº(Cgraphite(s)): Cgraphite(s) → Cgraphite(s) Hº=0

01,25

atmC

elementsstableH o

oF

tsreac

oiFiproducts

oiFi

orct HnHnH

tan

CH4(g) +2O2(g) → CO2(g) + 2H2O(g)H=?

CG(s)+ O2(g) → CO2(g) FH(CO2)

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

CH4(g) → CG(s) + 2 H2(g) -FH(CH4)

O2(g) → O2(g) -FH(O2)=0

CH4(g) +2O2(g) → CO2(g) + 2H2O(g)

H=FH(CO2)+ 2xFH(H2O) - FH(CH4) - FH(O2)

H=FH(CO2)+ 2xFH(H2O) – [FH(CH4) +FH(O2)]

tsreac

oiFiproducts

oiFi

orct HnHnH

tan

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

H= 4xFH(NO2)+ 6xFH(H2O) – 4xFH(NH3)

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

H= FH(Al2O3)+ 2xFH(Fe) – [FH(Fe2O3)+ 2xFH(Al)]

H= FH(Al2O3) – FH(Fe2O3)

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

H= 2xFH(CO2)+ 4xFH(H2O) – 2xFH(CH3OH)

H= 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)

H= 8xFH(CO2)+ 9xFH(H2O) – FH(C8H18)

H= 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