Chapter 4

Thermochemistry

4.1 Energy Storage and Refer-

ence

Thermochemistry is the study of the energy released

when the atomic-level “associations” are changed. A

chemical reaction (rxn) is one where these associations

change.

We are free to choose our energy reference, or zero

of energy, as we please. We could choose the reference to

be separated atoms. However, as you will see, a smarter

choice for our standard state is the most stable elemental

form at a convenient T and P. Once the zero is cho-

sen, our job is: to follow the energy. That means, if the

atom-atom associations change to a more energetically

favorable arrangement, there must be an exactly com-

pensating release of energy which we can measure with

the aid of calorimeters.

4.2 Chemical Rxns

Consider the formation of a molecule of 2 in the gas

phase. (By in the gas phase we mean that there are no

interactions between 2 units. (This is an idealization

achieved only at low .)

() +2()→ 2()

1. We take the most stable elemental form at 1 bar and

the specified temperature (usually 298.15 K) as the

REFERENCE state for the energy state function

. That is, the = ≡ 02. The combustion of diamond (to 2) yields almost

2 kJ/mole more energy than does graphite at STP.

This means that at modest pressures diamond has

a higher enthalpy than graphite, i.e. diamond is

“metastable” wrt graphite at STP. However, the

barrier is so large that diamonds will easily out-

last human-kind and therefore we can consider dia-

monds a stationary state.

3. A reaction with a -ve ∆ ≡[(P)− (R)] meansthat the enthalpy is lower for the products, i.e. the

reaction is energetically “down-hill”. This type of

reaction is called exothermic, and as energy must

be conserved, some energy will come “out”. If your

holding the beaker (a diabatic vessel), your hand

will feel the energy released (from the molecular as-

sociations sector into molecular kinetic energy), as

the reaction proceeds.

4.3 Hess’s law

4.3.1 Overview

1. Hess’s law is no more than a reminder that enthalpy

is a state function.H = 0 =

P .

2. For a specified reaction,

+ → + ; ∆ =P

∆.

The stoichiometric coefficients being +ve for

products and -ve for reactants. This sign conven-

tion achieves the ∆ ≡ P− R Think about it thisway: The energy (heat) released and enthalpy are on

opposite sides of an equation that balances energy

as well atoms. Therefore energy balanced reaction

expressions would read

+ → + + or

31

∆ + + → +

Energy Q ∆h Name

out +ve -ve exothermic

in -ve +ve endothermic

3. The value of ∆ only makes sense in the context

of a reaction with specified coefficients. (The sub-

script is telling you, the reader, to look for the

balanced equation that the writer is referring to.) In

other words, if an alternate reaction0 is specified as

2+ 2 → 2 + 2 then

∆0 = 2∆

Definition 1 The subscript refers to “one unit

of the reaction” with the stoichiometric coefficients

as written. This means, any quantity such as

∆ has an implied reaction with specified coef-

ficients. The change (∆) refers to taking the sto-

ichiometric coefficient number of moles of R and

converting them into P.

4. Hess’s law is most often applied to get the standard

(P= 1 bar, T

= 298.15 K) enthalpy of reac-

tion, which would read, for a rxn statement with

specified stoichiometric coefficients ,

∆ =P

∆.

Definition 2 The standard enthalpic state ( =

= 0) is taken (for convenience) to be that of the

most stable elemental form at STP.

Definition 3 If the reaction produces a single mole

of product ( = 1) from the elements in their stan-

dard states then ∆=∆

5. Consider this example.

Note that there are 4 measurable enthalpies, 3

of which would be standard enthalpies (if P=1 bar,

T=298.15 K) of formation:

∆ [2] = −286 [

]∆ [] = −635 [

]

and ∆ [()2] = −985 [

]

Therefore, for the rxn 2 + →, we have∆ = ∆

[()2]−(∆ [2]+∆

[])

= −64 []

4.3.2 Bond Enthalpies (energies)

One can get enthalpies corresponding to average bond

energies. These bond enthalpies are extremely valuable

in building up (in a ticker-toy fashion) estimates of the

enthalpies of molecules1. These energies are useful for

systems with no delocalized bonding. Values for organ-

ics are far more accurate than those for inorganics. Using

tables like those given in the text on pg. 69 (also found

on the WWW) are the starting point. To help under-

stand what bond enthalpies are, what they are not and

how to use them, consider the following examples.

1. The enthalpy of 4 is 1664 kJ/mole less than that

of the separated atoms.

(a) Therefore, − ≡ 166344

= 416

(b) The energy required to dissociate one H from

4 is 427 .

∆ = +427 .

2. Review example 4.1 in the text, − = 4635

kJ/mole.

3. Do problem 4.30, pg 78. ....

4. Finally, consider the enthalpy of formation

22(), i.e.

2( ) + 2()→ 22() ;∆

Enthalpy diagram for 22 All values in

1The utility of bond enthalpies is that much (but certainly not

all) of chemistry is “local”.

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The value of∆ (22()) can be estimated from

the following data.

(a) The energy released to form 22 from

atomic elements. This energy is a sum of bond

enthalpies2.

2() + 2()→ 22();

−[− + 2−] =−[172 + 2(261)] = −695[

]

(b) The value of ∆ for 2,

2()→ 2() ;∆ = 2434[

]

(c) The energy required to take atoms of from

the reference state to the gas phase.

2()→ 2() ;∆ = 2 ∗ 2271[

]

(d) Adding the above (a+b+c) we estimate,

∆ (22()) = 25 []

a result that relies on estimates of the bond enthalpies.

Remark 1 Note this is inaccurate as it involves the

difference between large numbers with a resulting small

value. This is a situation always to be avoided if at all

possible. This points to WHY we choose the reference

states to be the stable forms at a convenient T and P

rather than say the separated atoms. Let me elaborate

on this point. Consider the molecules O2 and N2. The

enthalpies associated with these bonds are 494 and 942

kJ/mole, respectively. (The binding energy of N2, the

triple bond, is one of the strongest molecular bonds.) If

we chose our reference state to be the separated atoms,

the molecular enthalpies would be H=(O2) = -494

and H=(N2) = -942 kJ/mole below the separated

atoms. In general, the bonded forms are hundreds of

kJ (100 kJ/mole = 1.03 ev/unit), in association energy,

below the separated atoms. As we are almost always in-

terested in the enthalpy difference between bonded forms,

it is convenient to take our reference as a bonded form.

This largely avoids taking the difference between large

numbers.

Remark 2 An analogy: Say we are interested in the

gravitational potential energy of objects in this room. We

need a reference from which to measure the height of the

object. We could take this reference as the center of the

earth, the sidewalk outside, the floor level of the lab, or

the level of the bench top we are doing the experiment on.

If we choose the later we can carry the ruler in our hands

and it can be made cheaply with precision markings. If

we choose any of the others, we have to have a long (very

2Some of these values have large uncertainties. For example,

another source for the Se-Cl gives 243 rather than 261 kJ/mole.

long indeed for the first ones in the list) and these rulers

will be very hard (expensive) to make accurately and easy

to use.

Remark 3 We are free to choose the zero of energy

variables as we please! However we must always be sure

our zero is communicated to all potential users. This is

why we all agree to use the same “smart” choices of the

reference states.

4.4 ∆( )

To determine how the enthalpy for a reaction changes

with temperature ∆( ), all we need to appreciate

is that in changing the temperature all the enthalpies

(those of the reactants R and products P) change inaccordance with their heat capacities. The individual

enthalpies increase with T (C is a strictly positive quan-

tity) but ∆( ) can increase or decrease with T.

Consider my favorite reaction: + → +

∆(2) = ∆(1)+R 21[() + ()− ()− ()]

It is convenient (for me) to define,

[] = C ≡ (P)−(R)=PP −

PR

Then we just have

∆(2) = ∆(1) +R 21C( )

Note that if, as is generally the case, the individual

heat capacities are represented as polynomials, C( )can be represented as a polynomial, derived by just sum-

ming terms of like order, with + signs for products and

− for reactants. (Don’t forget the stoichiometric coef-ficients!)

Remark 4 You should now understand why heat ca-

pacities are fit to readily integrated standard functional

forms (such as polynomials.)

Exercise 4.4.1 In the figure above, two of four cases

are drawn. Draw the other two. For which cases can the

sign of the enthalpy change with increasing temperature?

33

4.5 Experimental ∆ and ∆

Data from:

1. Calorimetry - The heat released by the reaction is

captured in calorimeter which is ideally adiabati-

cally separated from the remainder of the universe.

2. Spectroscopic studies

3. And most importantly, from other ∆ and

∆ values using Hess’s law. Here you have to

be confident of the values for the other “legs”. This

has been done for tabulated values by measuring all

legs and checking for consistency, (i.e. going in a

complete cycle and ending up with no change.)

4. QMical calc (isodesmic)

4.5.1 Calorimetry

Lets discuss the three most common types of calorimetry.

1. Constant-pressure calorimetry is the simplest

type and can be used if there are no gases or volatile

species involved. As (in the ideal experiment) no

heat is lost to the outside, the heat given up by the

reaction is captured by the calorimeter, ∆ =

However, the process that actually occurs (A)

is not the reaction enthalpy change at a fixed tem-

perature (as heating or cooling) can occur during

the reaction. Therefore to get the reaction enthalpy

∆, we need to break down the real process into

two steps.

.

(a) The actual process (A) is R(1)→ P(2) is thesum of:

(b) The desired step (B), the conversion of reac-

tants to productsR→ P at a constant (initial)temperature R(1)→ P(1)

(c) The heating of the products to the final

temperature (C): P(1)→ P(2) ∆ =R 21[(P) +] Clearly one needs a sep-

arate experiment to determine (P) +

(d) The desired ∆ = ∆− ∆

(e) However if really in an adiabatic enclosure

∆ = −∆

(f) ∴ the object of the experiment is really to get∆ so that the

Rin (c) can be done.

2. Bomb calorimetry (Constant V, ∆ = )

is used when there are volatile or gaseous species in

the reaction. As dV=0, ∆ = is the measured

quantity and ∆ must be calculated by adding the

change in PV energy.

Overview

While the process by which accurate values can be

obtained is quite involved, the basic process can be

analyzed with the aid of another triangle diagram,

this time referring the changes in the internal en-

ergies (rather than H) between various thermody-

namic states.

(a) The actual process (A) R(1)→ P(2) can bedecomposed into:

(b) A heating step (B),

∆ =R 21[(R) + ]

(again requires one or more separate experi-

ments) and

34

(c) The desired internal energy change is (C). If

the calorimeter is really adiabatic,

∆ = ∆ −∆ = −∆

(d) Again the object of the experiment is really to

get ∆ so that theRin (b) can be done.

(e) Corrections must be applies for the energy in-

put from ignition and stirring. To get stan-

dard state values requires an additional set of

corrections3.

Step-by-step Bare-bones logic.

(a) Determine the heat capacity of the calorimeter

(C = all parts including the water in

inner bath, not a molar value.) Measure the

change in temperature ∆ when you dump

in a known amount of energy. This can be

done by i) passing a known current through a

known resistor ( = = 2 = 2)

or ii) using a reference reaction for which the

heat released (under constant volume condi-

tions) is already known.

C)=

∆

[ ])=−∆

∆

(b) run reaction and measure temperature increase

in the heat bath ⇒ ∆exp(c) Calculate exp = [

] ·∆exp[K]

(d) Calculate the ∆ = exp and the molar value

by dividing by the number of moles of product

∆ = ∆

(e) Calculate ∆ =∆ +∆( )

∆( ) =∼ 0 if only l s

˜∆( ) if g involved

This procedure illustrates the “first-pass” logic

but cannot give the standard reference values.

To get these, the previously mentioned (Wash-

burn) corrections must be made (see footnote)

for which you need to know the pressure of the

“bomb” and the EoS of the gases involved.

3. Differential Scanning Calorimetry (DSC) is

a simple and fast method of determining the en-

thalpies of condensed phase (s-s or s-l) exother-

mic phase transitions (PTs). The way to view this

process in as follows:

3Search WEB for “Washburn corrections”. These procedures

correct for: a) Compression of the condensed phase and gas phase

species to 1 atm pressure, b) Enthalpy of solution and of dilution

of the product gases in the aqueous phase (e.g. HCl, 1:600), c)

Vaporization of water because of the heat liberated by the reaction,

and d) energy changes associated with non-isothermal reactions.

(a) The experimental and reference sample are

each connected to heaters which try to keep

the temperatures of the two samples the same.

(That is, the temperatures of both samples

start at 298 K and heat at a rate of, say, 5

C/min.

(b) The power [] ( = · = 2) is mea-

sured for both samples. There will be a power

difference (∆ = −) due to the

fact that the samples will have different heat

capacities. (The sample with the larger heat

capacity will require more power to heat.) i)

If the heat capacities are different but constant

∆ will be a constant. ii) If the heat capac-

ity of the sample grows linearly while the heat

capacity of the reference is constant, ∆

will increase linearly. If they both increase lin-

early, ∆ will increase linearly with the dif-

ference in slopes of the heat capacities.

(c) Or iii) when a PT occurs in the sample (one

must not occur in the reference), much more

heat will have to be dumped into the sam-

ple than in the reference. The total heat re-

quired to execute the PT is the excess en-

ergy, which is calculated from the excess power

times the time = ∆ =R[∆] with

∆ = ∆ −∆

(d) As DSC is employed for condensed phases

∆ ≈ ∆

Fig. of DSC

35