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