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Chapter 7: Thermodynamics & Thermochemistry
7.1 States, Systems & Processes

Back to the macroscopic:
 System of interest, closed or open
 The rest = universe

Properties
 Extensive – sum of parts
 Intensive – same in each part, eg. T, P

Thermodynamic state is at equilibrium when none of its properties change with time
 Described by an equation of state, eg. PV = nRT (for ideal gas as model system)

Thermodynamic process – a change in state
 Reversible process occurs via a series of thermodynamic states
State functions: V, T, P, internal energy E (change depends only on initial and final
points, independent of pathway)
Two questions:
 Will a process occur spontaneously?
 How far will the process go?
 These addressed by the first two laws of thermodynamics, First Law here, Second Law in
chapter 8


7.2 First Law of Thermodynamics: Energy, Work & Heat
Work
 Mechanical work, w = F(rf – ri) = Ma(rf – ri), a displacement
But, distance = velocity x time, and a = (vf – vi)/t
 v - v i  v i  v f 
w  M f

t
 t  2 
M
v f - v i v i  v f 

2
M 2 M 2

vf v i   (kinetic energy)
2
2

PV-work: eg. piston in cylinder
Fi = PiA
if Pext < Pi , expansion
w = - Fext (hf – hi)
= - Pext  V
here,  V > 0, hence w < 0, i.e.- the system does work
if  V < 0 (compression), w > 0, i.e.- work is done on the system
Chem 59-110 (’02)
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(see Fig. 7.1)
Heat
 Above concerned with kinetic and potential energy
 Also internal thermal or heat energy (recall gas molecules)
 Characterized as specific heat capacity, heat required to raise temp of 1 g, 1 oC
q = McsT
= heat transferred to a body of mass M, specific heat capacity cs to cause temp change of
T
 Often given in calories (1 cal defined for 1 g of water raised from 14.5 to 15.5 oC; 1 cal =
4.184 J)
 See Example 7.2
First Law of Thermodynamics
 Energy change in a system is due to the mechanical work done and heat transferred
E = q + w
 Note, Esystem = - Esurroundings , so that Euniverse = 0
 And, E a state function, although q and w are not (pathway-dependent)
7.3 Heat Capacity, Enthalpy & Calorimetry

Heat capacity of system, C, energy required to raise temp 1 K
 Specify whether constant P, define Cp, or constant V, define Cv
 Eg. Table 7.1 for Cp’s (specific, since per g)
 Or, as molar quantities, cp, cv

Heat transfer
qv = n cv (T2 – T1) = n cv T
qp = n cp (T2 – T1) = n cv T
These determined in calorimeters
 Constant volume, “bomb” calorimeter
 Easier to do at constant pressure, determine enthalpy
E = qp + w
= qp - PextV
hence, qp = E + PextV
=  (E + PV)
= H
H = enthalpy, a state function at constant P (a correction on internal energy when
some used for expansion work rather than raising temp)

7.4 First Law & Ideal Gases


Read for interest, no testing
Eg. relates cp and cv: cp = cv + R (this for gases; for solids and liquids, cp  cv)
Chem 59-110 (’02), ch. 7, Thermodynamics & Thermochem
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7.5 Thermochemistry
Enthalpies of Reactions
 Energy changes accompanying reactions
Eg. CO(g) +  O2(g)  CO2(g) + heat
(283 kJ)
i.e.- “heat” is a product, released out of system to surroundings
therefore, H has negative sign, exothermic
(opposite: endothermic, positive sign, for reverse reaction)

Additivity of reaction enthalpies: Hess’s Law, see Fig. 7.14

Similarly, treat phase changes like reactions, eg. Hfus , Hvap , etc., Table 7.2
 Eg. could calculate Hsubl for H2O from Table)
Standard State Enthalpies
 No absolute
 First define a reference point standard state
 Solids & liquids: stable state at 1 atm & specified temp
 Gases: gas phase at 1 atm, specified temp, ideal behaviour
 Solutions: 1 M at 1 atm, specified temp, ideal behaviour
 Temp usually 25oC (298.15 K)

Then define a zero point
 Chemical elements in their standard states at 298.15 K have enthalpies of zero (use
superscript o, “naught”)

Standard enthalpy change for a reaction, Ho
 All reactants and products in their standard states

Standard enthalpy of formation, Hfo
 For 1 mole of a compound from the elements in their standard states at 1 atm and 25oC
Eg. H2(g) +  O2(g)  H2O(l); Ho = -285.83 kJ
Therefore, Hfo(H2O(l)) = - 285.83 kJ mol-1
 With a table of Hfo’s, can calculate Ho for any reaction
 Example 7.7 for: 2 NO(g) + O2(g)  2 NO2(g)

In general: for aA + bB  cC + dD
Ho = cHfo(C) + dHfo(D) - aHfo(A) - bHfo(B)

Eg. calculate heat of combustion of octane, C8H18
C8H18(l) + 12.5 O2(g)  8 CO2(g) + 9 H2O(l)
Given:
compound
Hfo (kJ mol-1)
C8H18(l)
-250.0
O2(g)
0
CO2(g)
-393.51
H2O(l)
-285.83
Chem 59-110 (’02), ch. 7, Thermodynamics & Thermochem
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Horxn = 8 Hfo(CO2) + 9 Hfo(H2O) - Hfo(C8H18) - 12.5 Hfo(O2)
= 8 (-393.51) + 9 (-285.83) - (-250.0) - 0
= -5470.6 kJ

Extension to bond enthalpies, Table 7.3 (read for interest)
7.6 Reversible Processes in Ideal Gases



Not just initial and final states in equilibrium, but every point along the path
Two situations:
 Isothermal – constant temp,
E = 0 (ideal gas), therefore, w = -q
 Adiabatic – no heat transfer into or out of system
q = 0, therefore, E = w
Read for interest
Suggested Problems

1 – 15, 23 – 35, odd
Chem 59-110 (’02), ch. 7, Thermodynamics & Thermochem