Chapter 6 Thermochemistry
I.
Energy
A.
Energy = the capacity to do work or produce heat
1. Conservation of Energy = energy can’t be created or destroyed; it can
only change forms
2. Potential Energy = due to position or composition
a. Water behind a dam has energy that can be converted to work
b. Attractive and Repulsive forces (gravity) govern this type
c. Gasoline burns to make heat: forces holding atoms together
3. Kinetic Energy = due to motion of the object
a. KE = ½mv2
b. m = mass and v = velocity of the object
4. Conversion of energy types: PE  KE  PE + frictional heating
B. Heat and Work
1. Heat = q = transfer of energy between objects at different temperatures
a. Temperature = KE of particles in random motion
b. Heat is not a substance, but our language often treats it that way
2. Work = w = force acting through a distance = F x d
3. Pathway: energy transfer might be the same with different results
a. A rough surface: mostly heat and little work
b. Smooth, steep surface: little heat and mostly work
c. State Function = property that only depends on present state
i. Value doesn’t depend on the path of how it got that way
ii. Energy is a state function: DE = q + w
iii. Work and Heat are not
C. Chemical Energy
1. Mechanical Energy = energy of the movement of objects (see above)
2. Chemical Energy = energy of the change in chemical bonds
a. CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g) + heat
b. Heat energy is liberated by rearranging the chemical bonds
3. Dividing up the Universe
a. System = the specific reactants and products we are investigating
b. Surroundings = everything else in the Universe
4. Heat flow in chemical reactions
a. Exothermic = energy flows out of the system as heat
i. Products have a lower PE than the reactants
ii. Heat can be viewed as a product
iii. Heat released results in an increase in KE of surrounding particles
b. Endothermic = energy flows into the system
i. N2(g) + O2(g) + heat -------> 2NO(g)
ii. Heat can be viewed as a reactant
iii. Heat absorbed results in less KE of the surrounding particles
iv. Products have more PE than the reactants
D. Doing thermodynamics problems
1. Thermodynamics = study of energy and its interconversions
2. 1st Law of Thermodynamics = The energy of the universe is constant
3. Internal Energy = E = sum of KE and PE;
4. Thermodynamic quantities
a. Consist of two parts: a number (magnitude) and a sign (direction)
b. The sign reflects the system’s point of view
i. Heat flowing into a system has +q (endothermic)
ii. Heat flowing out of a system = -q (exothermic)
iii. Work is done on the system = +w
iv. The system does work on the surrounding = -w
c. DE = q + w sums up the changes in energy during a process
5. Example: Find DE of an endothermic reaction if 15.6 kJ is flowing and 1.4
kJ work is done on the system. DE = q + w = 15.6 + 1.4 = 17.0kJ
6. PV Work
a. Gases do work (expand) or have work done on them (compress)
b. Expansion of gas in engine cylinder provides mechanical work
w  Fd  F (Dh)  PA(Dh)  PDV
i. Expanding gas = -w = (+P)(+DV)
ii. Compressed gas = +w = (+P)(-DV)
w  PDV
7.
Example: Find w for expansion of a gas from 46L to 64L at 15 atm.
w = -PDV = -(15atm)(18L) = -270Latm
8. Example: Find DE if 1.3 x 108 J of heat expands a balloon from
4.00 x 106 L to 4.50 x 106 L at 1.0 atm.
DE = q + w = q + (-PDV) = (1.3 x 108 J) - (1atm)(0.50 x 106 L)
= (1.3 x 108 J) - (5 x 105 Latm)(101.3 J/Latm) = 8 x 107 J
II.
Enthalpy and Calorimetry
A.
Enthalpy = H = E + PV
1. E, P, and V are all state functions, so H is also a state function
2. Consider a constant P process where only PV work is allowed
a.
DE = q + w = q – PDV → q = DE + PDV
b. DH = DE + D(PV)
c. Since P is constant: D(PV) = PDV → DH = DE + PDV
d. DH = q At constant P, DH of a system = energy flow as heat
e. Heat of reaction = change in enthalpy = DH = Hproducts – Hreactants
i. If DH = + heat will be absorbed by the system = Endothermic
ii. If DH = - heat will be released by the system = Exothermic
3. Example: a. Find DH if 1 mol CH4 burns at const. P and 890 kJ heat is
released. b. Also find DH if 5.8 g of CH4 burns.
a. q = DH = -890 kJ
 1mol   890kJ 
b.
5.8g 

 16.0 g 
mol
  320kJ

B. Calorimetry = science of measuring heat
1. Calorimeter = device for measuring heat changes in a chemical reaction
2. Each substance changes temperature at a different rate = Heat Capacity
a. Specific Heat capacity = J/g•oC = Cs
b. Molar Heat capacity = J/mol•oC
c. Water has a high heat capacity
i. Very good coolant
ii. Much higher than metals
3. Constant-Pressure Calorimeter (q = DH)
• Experiment done under atm. Pressure
• 50.0 ml 1.0M HCl + 50.0 ml 1.0M NaOH at 25 oC (H+ + OH---> H2O)
• T = 31.9 oC after the reaction
 4.18 J 
(100.0 g )(6.9o C )  2900 J
q  Cs mDT   o
 Cg 
 2900 J
J
kJ
 58,000
 58
mol
mol
 1.0mol 
(0.050 L)

 L 
4. Example: Calculate DH/mol for the following reaction
a. Ba2+ + SO42- -------> BaSO4(s)
b. 1L 1.00M Ba(NO3)2; 1.00L 1.00M Na2SO4; 25 oC start; 28.1 oC end
 4.18 J 
(2000.0 g )(3.1o C )  26,000 J
 DH  Cs mDT   o
 Cg 
 26,000 J
J
kJ
 26,000
 26
mol
mol
 1.0mol 
(1.00 L)

 L 
5. Constant Volume Calorimetry
a. If DV = 0, then w = 0
b. DE = q + w = q
c. Combustion of 0.5269g octane (C8H18) in a bomb calorimeter with a
heat capacity = 11.3 kJ/oC with a DT = 2.25 oC. Find DE.
q  Ccal DT  (11.3kJ / oC )( 2.25o C )  25.4kJ
25.4kJ
 1.0mol
(0.5269 g )
 114.2 g




 5,500kJ  5,500
Example:
kJ
mol
III.
Hess’s Law
A. Recall that Enthalpy is a State Function
1. Path doesn’t matter
2. As long as we know reactants and products, steps don’t matter for DH
3. Example:
N2(g) + O2(g) -------> 2NO(g)
DH1 = 180 kJ
2NO(g) + O2(g) -------> 2NO2(g) DH2 = -112 kJ
N2(g) + 2O2(g) -------> 2NO2(g) DHtotal = 68 kJ
4.
Hess’s Law
a. You may sum steps in order to find overall DH
b. The DH for the reverse reaction will simply change signs (+/-)
c. If you multiply a reaction, you must multiply DH the same
5. Explanation
a. Sign of DH depends on direction of heat flow. Heat flow is reversed if
the overall reaction is reversed.
Xe(g) + 2F2(g) -------> XeF4(s) + 251kJ (DH = -251kJ)
XeF4(s) + 251kJ -------> Xe(g) + 2F2(g)
(DH = +251kJ)
b. DH is an extensive property = depends on the amount of substance
(an intensive property = depends only on identity of the substance)
Xe(g) + 2F2(g) -------> XeF4(s) + 251kJ (DH = -251kJ)
2Xe(g) + 4F2(g) -------> 2XeF4(s) + 502kJ (DH = -502kJ)
B. Examples
1. Example: Calculate DH for the conversion of graphite to diamond using
the known DH’s for the combustion of graphite and diamond.
a. Cg(s) + O2(g) -------> CO2(g)
DH = -394kJ
Cd(s) + O2(g) -------> CO2(g)
DH = -396 kJ
b. Cg(s) + O2(g) -------> CO2(g)
CO2(g) -------> Cd(s) + O2(g)
Cg(s) -------> Cd(s)
DH = -394kJ
(reverse) DH = +396 kJ
DH = +2 kJ
2. Example: Find DH for synthesis of B2H6 from B and H2.
a. Known Reactions
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)
DH = -1273 kJ
DH = -2035 kJ
DH = -286 kJ
DH = +44 kJ
b. Combining Reactions
2B(s) + 3/2O2(g) -------> B2O3(s)
+
B2O3(s) + 3H2O(g) -------> B2H6(g) + 3O2(g)
= 2B(s) + 3H2O(g) -------> B2H6(g) + 3/2O2(g)
DH = -1273 kJ
DH = +2035 kJ
DH = +762 kJ
+ 3[H2(g) + 1/2O2(g) -------> H2O(l)]
+ 3[H2O(l) -------> H2O(g)]
DH = -858 kJ
DH = +132 kJ
= 2B(s) + 3H2(g) -------> B2H6(g)
DH = +36 kJ
c. Let the final needed reaction guide how you combine other reactions