Chapter 5:
Thermochemistry
Thermochemistry:
 Energy
• Kinetic & Potential
 First Law of Thermo
• internal energy, heat & work
• endothermic & exothermic processes
• state functions
 Enthalpy
 Enthalpies of Reaction
 Calorimetry
• heat capacity and specific heat
• constant-pressure calorimetry
• bomb calorimetry (constant-volume calorimetry)
 Hess’s Law
 Enthalpies of Formation
• for calculation of enthalpies of reaction
 Foods and Fuels
Energy
• work is a form of energy w = F x d
• energy is the capacity to do work or transfer heat
• Kinetic Energy
• energy of motion E = ½ mv2
• potential energy
• energy of position
• applies to electrostatic energy
• applies to chemical energy (energy of bonds)
• energy units
•one joule = energy of a 2 kg mass moving at 1
m/s
• E = ½ mv2 (½)(2 kg) (m/s)2 = kg m2/s2 = 1 J
•1 cal = 4.184 J
1 kcal = 1 food calorie (Cal)
Systems & Surroundings
• system -- chemicals in the reaction
•closed system can exchange energy (but not matter)
with its surroundings
• surroundings -- container & all outside environment
Closed System
energy
(as heat
or work)
2H2(g) + O2(g)

no exchg
of matter
with
surroundings
2H2O(l)+ energy
(system)
First Law of Thermo.
• Energy is always conserved
•any energy lost by system, must be gained by
surroundings
• Internal Energy -- total energy of system
•combination of all potential and kinetic energy of
system
•incl. motions & interactions of of all components
•we measure the changes in energy
E = Efinal - Einitial
• +  E = Efinal > Einitial system has gained E from
surroundings
• -  E = Efinal < Einitial system has lost E to
surroundings
• Relating  E to heat and work
E=q+w
q is positive if heat goes from
surroundings to system
w is positive if work is done on
system by surroundings
+q +w
-q -w
surroundings
surroundings
system
system
Endothermic
•system absorbs heat or heat flows into the
system
Exothermic
•system gives off heat or heat flows out of the
system
State Function
•a property of a system that is determined by
specifying its condition or state (T, P, etc.)
•internal energy is a state function, \ E depends
only on Efinal & Einitial
Enthalpy
• for most reactions, most of the energy exchanged
is in the form of heat, that heat transfer is called
enthalpy, H
• enthalpy is a state function
• like internal energy, we can only measure the
change in enthalpy, H
H = qp when the process occurs under constant
pressure
H = Hfinal - Hinitial = qp
• - H
• +H


exothermic process
endothermic process
H > 0
surroundings
system
system
surroundings
H < 0
Enthalpies of Reaction
 Hrxn = Hprod - Hreact
• enthalpy is an extensive property
•magnitude of  H depends directly on the
amount of reactant
C(s) + 2H2(g)  CH4(g)
 H = -74.8 kJ/mol
2C(s) + 4H2(g)  2CH4(g)
 H = -149.6 kJ/2mol
• enthalpy change for forward rxn is equal in
magnitude but opposite in sign for the reverse rxn
CH4(g)  C(s) + 2H2(g)
 H = +74.8 kJ/mol
C(s) + 2H2(g)  CH4(g)
 H = - 74.8 kJ/mol
• enthalpy change for a reaction depends on the
state of the reactants and products
C(g) + 2H2(g)  CH4(g)
 H = -793.2 kJ/mol
2H2(g) + O2(g)  2H2O(g)
 H = -486.6 kJ/mol
2H2(g)+ O2(g)  2H2O(l)
 H = -571.7 kJ/mol
H = Hfinal - Hinitial
Enthalpy
H2O(g)
-241.8 kJ
44 kJ
+
H2O(l)
-285.8 kJ
Practice Ex. 5.2:
• Hydrogen peroxide can decompose to water and
oxygen . Calculate the value of q when 5.00 g of
H2O2(l) decomposes at constant pressure.
2H2O2(l)  2H2O(l) + O2(g)
5.00 g H2O2(l) x
1 mol
 H = -196 kJ
= 0.147 mol H2O2(l)
34.0 g H2O2(l)
0.147 mol H2O2(l) x
-196 kJ H2O2(l) =
2 mol
-14.4 kJ
Calorimetry
•experimental determination of  H using heat
flow
heat capacity
•measures the energy absorbed using
temperature change
•the amount of heat required to raise its temp. by
1K
•molar heat capacity -- heat capacity of 1 mol of
substance
specific heat
•heat energy required to raise some mass of a
substance to some different temp.
specific heat =
quantity of heat trans.
(g substance) (temp. change)
=
q
.
m T
S.H.
=
joule
g K
remember: this is
•q
=
change in temp.
(S.H.) (g substance) ( T)
Practice Ex. 5.3:
• Calculate the quantity of heat absorbed by 50.0 kg
of rocks if their temp. increases by 12.0 C if the
specific heat of the rocks is 0.82 J/gK.
S.H. x g x T = joules
What unit should be in the solution?
joules -- quantity of heat
0.82 J
gK
x
50.0 x 103 g x
12.0 K = 4.9 x 105 J
Constant-Pressure Calorimetry
•  H = qp at constant pressure as in coffee cup
calorimeter
•heat gained by solution = qsoln
\ qsoln = (S.H.soln)(gsoln)(T)
•heat gained by solution must that which is given
off by reaction
\ qrxn = - qsoln
= - (S.H.soln)(gsoln)(T)
must be opposite in sign
if T is positive then qrxn is exothermic
Practice Ex. 5.4:
• When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of
0.100 M HCl are mixed in a c.p. calorimeter, the
temp. of the mixture increases from 22.30C to
23.11C. Calculate  H for this reaction, assuming
that the combined solution has a mass of 100.0 g
and a S.H. = 4.18 J/g C.
AgNO3(aq) + HCl(aq)  AgCl(s)
+ HNO3(aq)
qsoln = 4.18 J x 100.0 g soln x 0.81C = 3.39 x 102 J
g C
qrxn = - qsoln = - 3.39 x 102 J = - 68,000 J or
0.00500 mol
- 68 kJ/mol
insulating cup
q
soln
rxn
Bomb Calorimetry (Constant-Volume)
• bomb calorimeter has a pre-determined heat
capacity
• sample is combusted in the calorimeter and  T is
used to determine the heat change of the reaction
• qrxn = - Ccalorimeter x  T
heat capacity of calorimeter
because rxn is exothermic
thermometer
insulation
water
rxn
Practice Ex. 5.5:
• A 0.5865 g sample of lactic acid, HC3H5O3, is
burned in a calorimeter with C = 4.812 kJ/C.
Temp. increases from 23.10C to 24.95C.
Calculate heat of combustion per gram and per
mole.  T = +1.85C
qrxn = - (4.812 kJ/C) (1.85C) = - 8.90 kJ per
0.5865 g lactic acid
-8.90 kJ = - 15.2 kJ/g
0.5865 g
- 15.2 kJ x
1g
90 .1 g =
1 mol
- 1370 kJ/mol
Hess’s Law
• rxns in one step or multiple steps are additive
because they are state functions
eg.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
2H2O(g) 
2H2O(l)
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
 H = - 802 kJ
 H = - 88 kJ
 H = - 890 kJ
Practice Ex. 5.6:
• Calculate  H for the conversion of graphite to
diamond:
Cgraphite 
Cdiamond
Cgraphite + O2(g)  CO2(g)
Cdiamond + O2(g)  CO2(g)
 H = -393.5 kJ
 H = -395.4 kJ
Cgraphite + O2(g)  CO2(g)  H = -393.5 kJ
CO2(g)  Cdiamond + O2(g)  H = 395.4 kJ
Cgraphite

Cdiamond
 H = + 1.9 kJ
Enthalpies of Formation
• enthalpies are tabulated for many processes
– vaporization, fusion, formation, etc.
• enthalpy of formation describes the change in heat
when a compound is formed from its constituent
elements, Hf
• standard enthalpy of formation, Hfo, are values for
a rxn that forms 1 mol of the compound from its
elements under standard conditions, 298 K, 1 atm
• For elemental forms:
eg. C(s) graphite, Ag(s) , H2(g) , O2(g) , etc.
Hfo, for any element is = 0
• used for calculation of enthalpies of reaction, Hrxn
• Hrxn = S Hfo prod - S Hfo react
Practice Ex. 5.9:
• Given this standard enthalpy of reaction, use the
standard enthalpies of formation to calculate the
standard enthalpy of formation of CuO(s)(
CuO(a) + H2(g)  Cu(s) + H2O(l) Ho = -130.6 kJ
Hrxn = S H f o prod - S H f o react
-130.6 kJ = [(0) + (-285.8)] - [(CuO) + (0)]
Hfo CuO = -155.2 kJ/mol