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AP CHEMISTRY NOTES
Ch 6 Thermochemistry
Ch 8.8 Covalent bond energies
Ch. 16 Spontaneity, Entropy, and Free Energy
6.1 The Nature of Energy
• Energy?
– The capacity to do work to produce heat
• Law conservation of energy
– Energy can be converted from one form to
another but can be either created nor
destroyed.
6.1 The Nature of Energy
• Kinetic energy
– 1/2mv2
• Temperature
– Property that reflects the random motions of
the particles in a particular substance.
• Heat
– Transfer of energy b/w two objects due to a
temperature difference.
6.1 The Nature of Energy
• Work
– Force acting over a distance
• Transfer of energy
1. Through work
2. Through heat
In other words, the energy transferred into
two ways.
6.1 The Nature of Energy
- Energy change is independent of the
pathway; however, work and heat are
dependent on the pathway.
6.1 The Nature of Energy
*Chemical energy
Released or Absorbed through
chemical reaction
*System
*Surroundings (anything other than
reactants and products)
Sign
- Or +
Exothermic Process
• Energy flows out of the system
• Delta PE is the change in PE
• In any exothermic reaction, some of the
potential energy stored in the chemical
bonds converted to thermal energy via
heat
Endothermic Process
• Energy flow is into a system
Thermodynamics
• First law of thermodynamics
– The law of conservation of energy
– (The energy of the universe is constant)
Internal energy
• The internal energy E of a system
• = the sum of the kinetic and potential
energies in the system
• Delta E = q + w
• (q = heat, w = work)
• Consist of a number (magnitude) and sign
(direction of the flow)
q = -x (out of the system)
q = +x (into the system)
Work from the system’s point of view
Figure 6.4 The Piston, Moving a Distance Against
a Pressure P, Does Work On the Surroundings
P=F/A
Work = F x ∆ h
= PA x ∆ h
=Px ∆V
Sign?
Work and ∆V must
have opposite sign
W= - P x ∆ V
6.2 Enthalpy and Calorimetry
• Enthalpy (H)
• H=E + PV
E is internal energy (KE+PE)
PV is pressure-volume work @constant P
Enthalpy is heat at constant pressure
• ∆H=∆E + ∆(PV)
• ∆H=qP (heat or energy flow)
Ex 6.4
Calorimetry
• Calorimeter
• Calorimetry
• Heat capacity C
– (heat absorbed /increase in temp)
• Specific heat capacity s
Coffee cup
calorimeter
– (per gram or per mol)
Cwater=4.18 J/oC∙g
4.18J is required to increase 1oC of 1g water
Extensive or intensive
properties?
• Heat=an extensive prop., depends on
amount of substance
• Temp=an intensive prop., not related to
the amount of substance
Ex 6.5 Constant pressure
calorimetry
• q=mc∆t
• Assume solution specific heat capacity is
same as water
• ∆t= final t – initial t
Example
• Consider a calorimeter at constant
pressure. 50.0 ml of a 4.0 M HCl solution
is mixed with 50.0 ml of a 2.0 M NaOH
solution at 25˚C. After mixing, the
temperature of the solution is 31.9˚C.
• Calculate the energy released.
• Consider a calorimeter at constant pressure. 50.0 ml of
a 4.0 M HCl solution is mixed with 50.0 ml of a 2.0 M
NaOH solution at 25˚C. After mixing, the temperature of
the solution is 31.9˚C.
• Calculate the energy released per mole of
H+ neutralized
Constant volume calorimetry
Coffee cup and Bomb
calorimeter 1
Constant volume calorimetry
• ∆E=q+w (w=0, constant V)
• ∆E=q
• Ex 6.6
– E released / mass
Constant volume
• A calorimeter has a heat capacity of 195
J/K. It contains 100. grams of water at
25˚C. 50.0 grams of aluminum (heat
capacity of 0.900 J/gK) at 100.0 ˚C was
added to it. What is the final temperature
of the solution?
• 30.1oC
Constant volume
• A calorimeter has a heat capacity of 50.
cal/K. 100 grams of cold water at 15˚C is
added to it. Then some hot water at 80˚C
is added to it. The final temperature of the
water is 30. ˚C. What is the mass of hot
water added?
• 45g
Constant pressure
• A calorimeter contains water at 25˚C.
Describe the heat flow when a metal at
50.0˚C is placed into the water.
Constant pressure
• A 15.0g sample of nickel metal is heated
to 100.0˚C and dropped into 55.0g of
water, initially at 23.0˚C. Assuming that all
the heat lost by the nickel is absorbed by
the water, calculate the final temperature
of the nickel and water. Nickel has a
specific heat of 0.444 J/˚Cg.
• In a coffee cup calorimeter, 1.60g of
ammonium nitrate is mixed with 75.0g of
water at an initial temperature of 25.00˚C.
After dissolution of the salt, the final
temperature of the calorimeter contents is
23.34˚C. Assuming the solution has a heat
capacity of 4.18 J/˚Cg and assuming no
heat loss to the calorimeter, calculate the
enthalpy change for the dissolution of
ammonium nitrate in units of kJ/mol.
6.3 Hess’s law
• ∆H is not dependent on the reaction
pathway.
– Is a state of function
Initial
Final
6.3 Hess’s law
• The change in enthalpy is the same
whether the reaction takes place in one
step or in a series of steps.
6.3 Hess’s law
• The change in enthalpy is the same
whether the reaction takes place in one
step or in a series of steps.
Rules of Hess’s law
1. If a sub-reaction is multiplied by a
coefficient, ∆Ho value is also multiplied by
the same coefficient.
2. If a sub-reaction is reversed (flipped), ∆Ho
value is switched the sign.
Hint!
1. Work backward
2. Reverse any reactions
3. Multiply reactions
Example of Hess’s law
H2 (g) + Cl2 (g)  2HCl (g)
ΔH˚ = ?
Given:
NH3 (g) + HCl (g)  NH4Cl (s)
ΔH˚ = - 176.0 kJ
N2 (g) + 3H2 (g)  2NH3 (g)
ΔH˚ = - 92.22 kJ
N2 (g) + 4H2 (g) + Cl2 (g)  2NH4Cl (s)
ΔH˚ = -628.86 kJ
• ΔH˚ = -184.64 kJ
Example of Hess’s law
C2H2(g) + 2H2(g)  C2H6(g)
ΔH˚ = ?
Given:
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l)
ΔH˚ = -2600 kJ
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l)
ΔH˚ = -3120 kJ
H2(g) + 1/2 O2(g)  H20(l)
ΔH˚ = - 286 kJ
• ΔH˚ = -312 kJ
Example of Hess’s law
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Given:
2H2(g) + C(s)  CH4(g)
ΔH˚ = ?
ΔH˚ = 74.81 kJ
2H2(g) + O2(g)  2H2O(l)
ΔH˚ = -571.66 kJ
C(s) + O2(g)  CO2(g)
ΔH˚ = -393.52 kJ
• ΔH˚ = - 1039.99
kJ
Example of Hess’s law
2NH3(g) + 3N2O(g)  4N2(g) + 3H2O(l) ΔH˚ = ?
Given:
2NH3(g) + 3/2 O2(g)  N2(g) + 3H2O(l)
ΔH˚ = - 765.5 kJ/ mole
3N2O(g) + 3H2(g)  3N2(g) + 3H2O(l)
ΔH˚ = -1102.2 kJ/ mole
3H2O(l)  3H2(g) + 3/2O2(g)
ΔH˚ = 857.7 kJ/ mole
• ΔH˚ = - 1010.0 kJ/mol
6.4 Standard Enthalpies of
Formation
• ∆Hof (under standard condition)
• The change is enthalpy that accompanies
the formation of one mole of a compound
from its elements with all substances in
their states.
• We can’t determine absolute values of
enthalpy
Reference
Element @1atm, 25oC
P gas @1atm
Conc.=1M
o
∆H
f
= per mole of product
(kJ/mol)
• See appendix 4
• Tabulated ∆Hof value – using this for
calculation
Ex. CH4(g)+2O2(g)CO2(g)+2H2O(l)
*Add all up
A: CH4 CH4 C + 2H2
∆H=-75kJ/mol
Flip!
C + 2H2  CH4
∆H=75kJ/mol
B: 2O2
∆H=0kJ (element)
C: CO2 C + O2  CO2
∆H=-394 kJ
D: 2H2O H2 + ½ O2 H2O ∆H=2x(-286) kJ
(75kJ)+(-394kJ)+(-572kJ)= -891kJ
Ex. CH4(g)+2O2(g)CO2(g)+2H2O(l)
*Sum of produced – sum of required
A: CH4 CH4 C + 2H2
∆H=-75kJ/mol
You don’t have to flip!
B: 2O2
∆H=0kJ (element)
C: CO2 C + O2  CO2
∆H=-394 kJ
D: 2H2O H2 + ½ O2 H2O ∆H=2x(-286) kJ
(-394kJ)+(-572kJ)-[0+(-75kJ)]= -891kJ
0
Hreaction
  np H 0f ( products)   nr H 0f (reactants)
Ex. 6.9
4NH3(g)+7O2(g)4NO2(g)+6H2O(l)
Table 6.2 Standard Enthalpies of Formation
for Several Compounds at 25°C
Copyright © Houghton
Mifflin Company. All
6–51
Ex. 6.9
4NH3(g)+7O2(g)4NO2(g)+6H2O(l)
• -1396kJ
Ex. 6.10
2Al(s)+Fe2O3(s)Al2O3(s)+2Fe(s)
Ex. 6.10
2Al(s)+Fe2O3(s)Al2O3(s)+2Fe(s)
• -850.kJ
6.11
CH3OH or C8H18?
•
•
•
•
Need two equations
_CH3OH (l) + _O2(g)_CO2(g)+_H2O(l)
_C8H18(l) + _O2(g)_CO2(g)+_H2O(l)
Finally divide by molar mass for both
6.11
CH3OH or C8H18?
•
•
•
•
2CH3OH (l) + 3O2(g)2CO2(g)+4H2O(l)
-1454kJ/2mol*32.0g = -22.7 kJ/g
2C8H18(l) + 25O2(g)16CO2(g)+18H2O(l)
-10914kJ/2mol*114.2g = -47.8 kJ/g
6.5 Present sources of energy
6.6 New energy sources