Chapter 7: Thermodynamics & Thermochemistry

Chapter 7: Thermodynamics & Thermochemistry

7.1States, 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

7.2First 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

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

(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)

7.3Heat 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)

7.4First 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)

7.5Thermochemistry

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)

StandardState 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

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)

7.6Reversible 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