Transcript Thermochemistry1

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
potential
kinetic
 energy possessed by an object in virtue of
its motion.
 Ekin=1/2 mv2
Never confuse
 Ekin=3/2 RT
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!
Attraction causes the ball to fall, h smaller, Epot smaller.
Epot=mgh
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.
h=900 m
h=650 m
Final state
Irbid
Initial state
Irbid → Amman
Dh=hfinal-hinitial
Dh=hamman-hirbid
Dh=900 m-650 m=250 m
Sea level
Amman
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
DE = Q + W
The internal energy of a system can be changed
1. by gaining or losing heat, Q
2. Work, W, done on the system or by the System
DE = Q + W
surroundings
sys
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
Qsys > 0 : endothermic process
surroundings
sys
Q
for Sys: Q > 0 (pos.), DE > 0
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
Qsys < 0 : exothermic process
for Surr: Q > 0 (pos.), DE > 0
Work done by surroundings on system
surr
surr
m1
m1
m2
surr
sys
m1
sys
m2
sys
p   p
sys i

Fopposite
A
opposite
m1 g

A
popposite
m  m2  g
 1
p   p
sys i
A
opposite
Ep=mgh
h of m1 and m2 ↓
Esurr ↓
h↓, Ep↓
Ep of m1 and m2 ↓
Esys 
Vsys 
p 
psys 
sys f
 popposite
popposite

m1  m2  g

A
(Esys)f > (Esys)i
DEsys > 0
wsys > 0
Work done by system on surroundings
surr
m1
surr
m1
surr
m2
m1
sys
p   p
sys i

Fopposite
A
sys
sys
opposite

m1  m2  g

A
m g
popposite  1
A
 psys i  popposite
Ep=mgh
h of m1 
Esurr 
h, Ep
Ep of m1 
Esys ↓
Vsys 
psys 
sys f
 popposite
p 
m1 g
popposite 
A
(Esys)f < (Esys)i
DEsys < 0
wsys < 0
• 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
surr
m1
surr
m1
sys
sys
hf
hi
final
initial
w  F r
F
popp 
A
w  popp  A  r
F  popp  A
w  popp  A  h f  A  hi 
w  popp  A  h f  hi 
w  popp  V f  Vi 
w  - popp  DV
Q and w
are path functions (Depend on path).
DE1  Q1  w1
Path 1
DE1  DE2
full
Q1  w1  Q2
empty
final
initial
Path 2
DE2  Q2
• 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.
15 atm
46 L
15 atm
64 L
o Expansion against the external pressure
o External pressure opposes the expansion
o popp=15 atm = constant
w   popp  DV  15atm  V f  Vi 
w  15atm  64 L  46 L   15atm 18L  270atm.L
• Ex. 6.3
Given a balloon with a volume of
1.3x108
Vi
4.00x106
Vf
Q
L. It was heated by
J until the volume became 4.5x106L. Assuming the
popp
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.
1 atm
1 atm
4.50x106 L
4.00x106 L
DE  Q  w
DE  Q  poppDV
DE  Q  popp V f  Vi 
DE  1.3 108 J  1 atm  4.5 106 L  4.5 106 L   1.3 108 J  0.5 106 atm.L
1 atm.L  101325 Pa  10 3 m3  101.325 Pa .m3  101.325 J
DE  1.3 108 J  0.5 106 101.325 J  1.3 108 J  5.07 107 J  8.0 107 J
Chemical Energy
CH4(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 atoms
o Redistribution of Bonds (change in bonding)
o Change in attraction & repulsion forces between the atoms
o Change in the potential energy Ep of molecules
Energy is conserved!
R
Ep
Energy difference released as heat.
P
Heat of reaction (Qv, Qp).
Reaction exothermic.
DHreaction= Hf – Hi = HP – HR < 0
N2(g) + O2(g) → 2NO(g)
N≡N
P
Ep
R
O=O
N=O
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
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.
N2(g) + O2(g) → 2NO(g)
DHo=+180.5 kJ
DHo: Standard heat of reaction: Standard conditions: T=25oC, p=1atm.
Calorimetry
calorie
cal
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
DE  Q  w
DE  Q  popp  DV
V  cons tan t
DV  0
DE  QV
Heat measured at constant volume:
o Equal to DE
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
H  E  pV
dH  dE  pdV  Vdp
Hf
Ef
Vf
pf
Hi
Ei
Vi
pi
 dH   dE   pdV   Vdp
pi  p f  p
H HH
f
i
dp  Dp  p f  pi  0
 E Eif  p  V Vif  0
E
V
H f  H i  E f  Ei   p.V f  Vi 
DH  DE  p  DV
DH  Q  p  DV   p  DV
DH  Q p
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.
csp
1g
T increase
by 1ºC:
C  m  csp
m gr.
?=C
Q : heat
heat needed to raise the temperature of substance (m gram)
by a given temperature difference, DTºC.
m gr.
T increase
by 1ºC
C
T increase
by DTºC
?=Q
Q  C  DT
Q  m  csp  DT
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.
kJ
Q  QV  C  DT  11.3   2.25 C  25.4 kJ
C
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
114.2 g
QV
? = DE
DE=-(114.2 g x 25.4 kJ)/0.5269g=-5505 kJ
Mwt  QV
QV
QV
DE  


m
m Mwt
n
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)?
kJ
QV CH 4   C  DT  11.3   7.3 C  83 kJ
C
1.5 g
1g
QV=83 kJ
?
=55 kJ/g
kJ
QV H 2   C  DT  11.3  14.3 C  162 kJ
C
1.15 g
1g
QV=162 kJ
?
=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 temperature was found to rise to 31.9ºC.
Calculate the heat of the neutralization 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 +
by reaction
by solution
heat gained
by calorimeter
Qrct  m  csp  DT solution  C  DT calorimeter
Assumptions:
 Ccal=0 (very small mass)
 Solution ≈ water
 (csp)solution=(csp)water=4.18 Jg-1ºC-1
 (density)solution=(density)water=1 g/mL
m  d V  1
g
100 mL  100 g
mL
J
Qrct  100 g  4.18   6.9 C  2884.2 J
g C
nHCl=MHClxVHCl = 1 mol/L x 0.050 L = 0.050 mol
nNaOH=MNaOHxVNaOH = 1 mol/L x 0.050 L = 0.050 mol
HCl(aq) +NaOH(aq) → NaCl(aq) + H2O(l)
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  
Qp
n
DH= -57.7 kJ/molH2O
Hess’s Law
initial
path 1
N2(g) + 2O2(g)
O2(g)
DH 1  DH 2
DH
for
path 2
N 2 ( g )  2O2( g )  2 NO2 ( g )

DH
 DH
for
2NO2(g)
O2(g)
2NO(g)
DH  H f  H i
final
N 2 ( g )  O2 ( g )  2 NO( g )
for 2 NO( g )  O2 ( g )  2 NO2 ( g )
N2(g) + O2(g) → 2NO(g)
DH2a
2NO(g) + O2(g) → 2NO2(g)
DH2b
N2(g) + 2O2(g) → 2NO2(g)
DH1
DH1=DH2a+DH2b
Hess’s Law:
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).
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)
1/2x (N2(g)+3H2(g) → 2NH3(g)
1/2N2(g)+3/2H2(g) → NH3(g)
DH=-184 kJ
DH=-92 kJ)
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 kJ
1 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)
CO2(g) → CD + O2(g)
Cgraphite → Cdiamond
DH=-394 kJ
DH=+396 kJ
DH=+2 kJ
Given:
2B(s)+3/2O2(g) → B2O3(s)
B2H6(g)+3O2(g) → B2O3(s) + 3H2O(g)
H2(g)+1/2O2(g) → H2O(l)
H2O(l) → H2O(g)
Calculate DH for
2B(s) + 3H2(g) → B2H6(g)
DH=-1273 kJ
DH=-2035 kJ
DH=-286 kJ
DH=+44 kJ
2B(s)+3/2O2(g) → B2O3(s)
DH=-1273 kJ
3H2(g)+3/2O2(g) → 3H2O(l)
DH=3x(-286) kJ
B2O3(s) + 3 H2O(g) → B2H6(g)+3 O2(g)
2B(s) + 3 H2O(g) + 3 H2(g) → B2H6(g)+3 H2O(l)
3H2O(l) → 3H2O(g)
2B(s) + 3 H2(g) → B2H6(g
DH=+2035 kJ
DH=-96 kJ
DH=3x(+44) kJ
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)): ½N2(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)
DFHº(Cdiamond(s)):
Cgraphite(s) → Cdiamond(s)
DFHº(O2(g)):
DFHº(Cgraphite(s)):
O2(g) → O2(g)
DHº
DHº
DHº=0
Cgraphite(s) → Cgraphite(s)
DHº=0
 stable elements 
  0
D F H 
o
 25 C , 1 atm 
o
DH
o
rct

 n  D
i
F
H

o
i products

 n  D
i
F
H

o
i reac tan ts
DH=?
CH4(g) +2O2(g) → CO2(g) + 2H2O(g)
2x
DFH(CO2)
2xDFH(H2O)
-DFH(CH4)
-DFH(O2)=0
CG(s)+ O2(g) → CO2(g)
(H2(g)+1/2O2(g) → H2O(g) )
CH4(g) → CG(s) + 2 H2(g)
O2(g) → O2(g)
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)]
o
DH rct

 n  D
i
o
H
F
i


products
 n  D
i
o
H
F
i

reac tan ts
4NH3(g) +7O2(g) → 4NO2(g) + 6H2O(l)
DH=?
DH= 4xDFH(NO2)+ 6xDFH(H2O) – 4xDFH(NH3)
2Al(s) +Fe2O3(s) → Al2O3(s) + 2Fe(s)
DH=?
DH= DFH(Al2O3)+ 2xDFH(Fe) – [DFH(Fe2O3)+ 2xDFH(Al)]
DH= DFH(Al2O3) – DFH(Fe2O3)
Calculate the heat of combustion of methanol (CH3OH(l)) in
kJ/g and compare its value with that of octane (C8H18(l)).
2CH3OH(l) +3O2(g) → 2CO2(g) + 4H2O(l)
DH=?
DH= 2xDFH(CO2)+ 4xDFH(H2O) – 2xDFH(CH3OH)
DH= 2x(-394 kJ)+ 4x(-286 kJ) – 2x(-239 kJ)=-1454 kJ
2 mol CH3OH
2x32 g
1g
-1454 kJ
-1454 kJ
?
= -22.7 kJ/g
C8H18(l) +12.5O2(g) → 8CO2(g) + 9H2O(l)
DH= 8xDFH(CO2)+ 9xDFH(H2O) – DFH(C8H18)
DH= 8x(-394 kJ)+ 9x(-286 kJ) – (-276 kJ)=-5450 kJ
1 mol C8H18
114 g
1g
-5450 kJ
-5450 kJ
?
= -47.8 kJ/g