CHEM 101/ 105
Thermochemistry
Lect-11
Heat (q) is a form of energy (E), and is expressed in Joule or kilo Joule units. Temperature (not a form of energy)
is a measure of the direction of heat flow; heat flows spontaneously from an object at higher temperature to another
at lower temperature. Heat effects are quantitative and are used in stoichiometric calculations.
The amount of heat [ q ] necessary to raise the temperature of some >whole thing= per
degree kelvin is the Heat Capacity [ C ] of that 'whole thing'.
C = q(Joules)/ ∆ T
so
C has units of Joules/deg K
rearrange to
q = C * ∆T
where ∆ T = ( Tfinal - Tinitial )
1. Heat Capacity
Heat Capacity is a property unique to every item, even for things made from the same material. The heat capacity
of a 3 inch iron bolt will be much smaller in magnitude than the heat capacity of a 20 foot section of iron rail, even
though both are made from the same material, because they differ so much in mass.
2. Specific heat is a ratio Heat Capacity to mass., i.e., specific heat is the heat capacity for an item divided by
its mass; Sp.ht. = Heat Capacity (in Joules/deg K) / mass (in grams), and has units of [Joules / (gram-deg K )].
The specific heat of the iron bolt and rail are the same, because both heat effects are on a per gram basis. When
the product of mass and specific heat is substituted for Heat Capacity, the first expression becomes more useful:
q = ( specific heat ) ( mass) ( ∆ T )
Specific heat is an identifying and characteristic property for every type of material. Liquid water has a specific
heat of 4.18 J/gN (please remember this value and its units). Metals have much lower specific heats than water.
Consider three examples involving specific heat:
ex. 1
a.
Benzene has a specific heat of 1.72 J/gN. How many Joules of heat are required to change the
temperature of 10.0 g of benzene from 18.6 to 21.3NC? What is the heat capacity of 10.0 g of
benzene?
b.
Carbon dioxide gas has a specific heat of 0.843 J/gN. How many g of CO2 can be heated from
43.6 to 51.4NC, by supplying 100 Joules of heat energy?
c.
Chlorine gas (Cl2) has a specific heat of 0.478 J/gN. When 0.38 mole of chlorine at 18NC are
cooled so that 25.0 Joules are removed, what is the final temperature?
Important Science Fact: Each unique example of matter has its own unique specific heat value (because each
unique example of matter has a exclusive set of bonded atoms). Energies ultimately
are associated with nature and type of chemical bonds present in substances.)
Important Science Fact: The specific heat of a given individual example of matter depends upon which state or
phase it is in. The specific heat of water as a liquid is 4.18 J/g-deg, solid water (ice) has
a specific heat of 2.03 J/gN, and steam (water in the gas or vapor state) has a specific heat of 2.02 J/gN. (This is
because changing to/from one state to another involves addition/removal of heat energy.) Please remember
these values for water.
3. Thermal Effects of Phase Changes
Every item, under any condition, has a certain heat content, to which heat can be added or removed by various
processes. A Change in heat content for a process occurring under conditions of constant pressure ( q p ) is not
necessarily the same as when conditions are at constant volume ( q v ). Most laboratory processes are open to
the atmosphere and take place at a constant pressure of about 1 atm. Under usual laboratory conditions of
constant pressure ( q p ), the heat associated PER MOLE of item is called the ENTHALPY CHANGE ( ∆ H ), q
= ÄH. Some familiar changes in heat content that occur at constant pressure include phase changes, i.e.,
physical changes between solid, liquid and gas states.
(s) = (R), melting,
(R) = (g), boiling,
∆ H(fusion) ;
∆ H(vaporization) ;
(R) = (s), freezing, -
∆
∆ H(fusion) = 6000 J/mol,
H ∆ (vaporization) = 40.7 kJ/mol.
H(fusion) ; for water,
(g) = (R), condensation, - ∆ H(vaporization) .
p
Consider the following table showing normal melting and boiling temperatures (i.e., at one atmosphere
pressure) and enthalpies of phase change processes for various substances...
name
benzene
bromine
mercury
naphthalene
water
formula
normal m.p.
∆H( fusion )
normal b.p.
∆ H(vaporization)
C6 H6
Br 2
Hg
C 10 H 8
H2 O
5
-7
-39
80
0
9.84 kJ / mole
10.8
2.33
19.3
6.00
80
59
357
218
100
30.8 kJ / mole
29.6
59.4
43.3
40.7
ex. 2
Write thermochemical
equations for:
a. boiling of bromine, b. condensation of naphthalene,
c. freezing of benzene, d. boiling of Hg.
ex. 3
How many Joules would be required to convert 50.00 g of liquid mercury to mercury vapor,
at its boiling temperature?
ex. 4
When 40.0 g of steam at 125EC are mixed with 125 g of ice at -15EC,
what is the final state and final temperature of the mixture?
(apple problem)
Important Science Fact: Phase changes occur at constant temperature. Energies of phase changes are
expressed in Joules or kilo Joules per mole and require interpretation of
( ±) sign.
HISTORY: addition of HEAT energy to a quantity of WATER
Thermochemistry
of Water
sp.ht.=2.02 J/g-deg
Gas Phase
(@ 1.0 atm. pressure)
(liq)
+/- 40.7 kJ/mole
phase change
(c)
Liquid Phase
temperature C
100
sp.ht.=4.18 J/g-deg
(s)
0
(liq)
(b)
(a)
phase change
+/- 6000 J/mole
Solid Phase
sp.ht.=2.03 J/g-deg
time
(gas)
(d)
(e)
(a)
(b)
(c)
(d)
(e)
ice (solid phase), only ONE phase present, the relation
q = (sp.ht. solid)(mass)(Tf-Ti) can be applied.
PHASE CHANGE: solid and liquid phases are both present and in equilibrium
at a constant melting point temperature.
TWO phases are present and temperature is constant
q = (sp.ht.)(mass)(Tf-Ti) cannot be used.
Must use ∆H( fusion ) (on a mole basis) and furnish sign.
liquid (liquid phase), only ONE phase present, the relation
q = (sp.ht. liquid)(mass)(Tf-Ti) can be applied.
PHASE CHANGE: liquid and gas phases are both present and in equilibrium
at a constant boiling point temperature.
TWO phases are present and temperature is constant
q = (sp.ht.)(mass)(Tf-Ti) cannot be used.
Must use ∆H( vapor .) (on a mole basis) and furnish sign.
steam (gas phase), only ONE phase present, the relation
q = (sp.ht. gas)(mass)(Tf-Ti) can be applied.
4. Differentiating between SYSTEM and SURROUNDINGS.
Consider the following questions:
When heat is added to something, where does it come from? When lost by something, where does it go?
A fundamental law of thermodynamics states that the
energy of the universe is constant.
Consequently, the "something" of major interest to the experimenter it is labeled as the "system", and
everything else in the universe (except the system) is labeled as the "surroundings".
A scheme for defining energy terms and algebraic signs is based on the delineation of a system from its
surroundings. So, when
direction of heat flow is FROM system TO surroundings
then sign of heat ( q ) for the system is
change in temperature of system is
term used for the system process is
negative (-),
system loses heat
Tfinal is LOWER than Tinitial
EXOthermic
corresponding sign of heat for surroundings is
change in temperature of surroundings is
term used for process w/r to surroundings is
positive (+),
surroundings gain heat
Tfinal is HIGHER than Tinitial
ENDOthermic
When direction of heat flow is FROM surroundings TO system - then all of the above is reversed.
5. Calorimetry A common method used to measure heat gain/loss in thermochemistry is to immerse the system
in a known quantity of liquid water, and measure (indirectly) the heat gain/loss for the system, via
the corresponding temperature change of the water surrounding it. The mass of surrounding water is
easily determined, and its temperature change readily measured, so the calculation of heat transferred
between system and surrounding water is given by:
q(surrounding water) = (4.18 J/gN) x (mass H2O (surr) ) x ( ∆ T H2O (surr)) = q (calorimeter)
Recognize if heat is transferred from the system to the surrounding water, then the process w/r to the system is
exothermic, the system will have lost heat (q sys is negative), and surrounding water will have gained heat and
thereby experience an increase in its temperature.
Thus, in calorimetry, heat effects of system and surrounding always have same magnitude, but opposite sign, so
q( system) = −q( surroundings )
ex. 5
When 12.78 g of a metal at 87.6EC is placed in 51 mL of liquid water at 17.4EC, the highest temperature
reached by the water and metal is 18.1EC. What is the specific heat of this metal?
A laboratory reactor that does NOT allow volume to change is called a bomb calorimeter. Reactions in bomb
calorimeters are constant volume processes. (SHOW a bomb calorimeter) Energies involved in constant volume
processes are given symbols of q v (per gram) and E (TOTAL ENERGY per mole).
E is always associated with constant volume processes, as for example measured in bomb calorimeters.
ex. 6
A 1.374 g sample of an organic compound was sealed in a "bomb calorimeter" and immersed in a liquid
water calorimeter at a temperature of 18.32 C. The sample was ignited and burned, and the final
temperature of the liquid water in the calorimeter was 21.53 C. (In a separate experiment the Heat
Capacity of the calorimeter, i.e., the surroundings, was measured as 10.12 kJ/N.) If the molecular weight
of the organic compound is 109 g/mol, what is the TOTAL ENERGY (symbol E) for this combustion?
Clearly, chemical reactions cause changes in heat content. Buildings are Aheated@ by energy released when
organic compounds are combusted. What quantity of heat is involved in a reaction? The quantity involved has to
be the difference between heat content of products, and heat content of reactants. So,
∆ E( reaction) = [ ∑ ∆ E( products ) ] − [ ∑ ∆ E(reactants ) ]
(at constant volume)
For processes occurring under conditions of constant pressure and temperature, (i.e., volume changes, but not P
and T) the quantity of energy involved (per mole) is defined as ENTHALPY (symbol H). In a manner similar to the
above expression for ∆E , an expression for ∆H is …
∆ H( reaction) = [ ∑ ∆ H( products ) ] − [ ∑ ∆ H( reactants) ]
For example, given that the reaction P(s) + 3/2 Cl2(g) = PCl3(g)
EXOTHERMIC process can be represented as:
(at constant P and T)
has an enthalpy ( ∆H ) of -287.0 kJ/mol, this
∆H( reaction) = [ heat content of product(s) ] - [ heat content of reactants ] = - 287.0 kJ/mol , or
∆ H( reaction) = [ ∆ H( PCl 3 ) ] − [ ∆ H( P ) + 3 / 2 ∆ H( Cl2 ) ] = -287 kJ / mole
This Rx will be used to define a standard enthalpy of formation, which is represented by
∆H f
6. Enthalpies of Formation
Enthalpy values for chemical substances cannot be determined without some
reference point, just as locations on Earth cannot be specified without some reference point - for example the
ZERO meridian line established at Greenwich, England. If ZERO meridian were located in Paris, then EACH point
on Earth would have different coordinates, BUT differences between the coordinates of any TWO points on Earth
would still be the same. In an analogous manner only differences in enthalpies , ∆H , between two substances
are informative in chemical processes. Between what two substances? The difference is always between
PRODUCTS and REACTANTS.
A special consideration applies when the product is a single compound, and the reactants are all elements that
form the compound.
∆H is then called the enthalpy of formation and is given the symbol ∆Hof .
The superscript
ZERO is used to specify standard state conditions (1 atm and 25 deg.C).
The reference point in chemical thermodynamics comes from defining the enthalpy for elements in their standard
state as ZERO. Consider the consequences of this definition:
Phosphorus trichloride is formed from the elements phosphorus and chorine according to the reaction,
P (s)
+ 3/2 Cl 2 (gas) =
P Cl 3 (gas)
and the heat effects involved with formation of one mole of the compound under a constant pressure of one atm and at a temperature of 25 deg.C - can be readily measured. As noted above, 287 kJ of heat energy are liberated
for every mole of PCl 3 formed.
∆H(oreaction) = - 287.0 kJ/mol
In terms of enthalpy differences between products and reactants, this expression may be recast as…
∆ H(oreaction) = [ ∆ H(oPCl 3 ) ] − [ ∆ H(oP ) + 3 / 2 ∆ H(oCl2 ) ] = - 287.0 kJ/mole
Application of the reference point regarding enthalpy of formation of elements gives the expression…
∆ H(oreaction) = [∆ H(oPCl3 ) ] − [0 + 3 / 2 x 0] = - 287.0 kJ / mole
which gives the value for the enthalpy of formation of phosphorus trichloride
∆ H(oreaction) = ∆ Hof = - 287.0 kJ/mol
A table of enthalpy of formation values is listed in all general chemistry texts. The usefulness of values of
∆Hof
to the working chemist is remarkable, and they are most often applied in the following relation:
∆ H(oreaction) = [∑ ∆ Hof (products) ] − [∑ ∆ H(of ( reactants) ]
ex. 7
Write thermochemical equations for the enthalpy of formation,
∆Hof , of the following compounds
carbon dioxide, magnesium carbonate, ethyl alcohol, ammonium perchlorate
ex. 8
When 25.0 g of barium carbonate is heated to form barium oxide and carbon dioxide
(a) write a balanced thermochemical equation for the process.
(b) how many joules of heat will be involved,
per mole?
for 25.0 g barium carbonate?
ex. 9
Determine
∆H(oreaction) for combustion of ethane, C2H6(g) , to carbon dioxide and liquid water.
Another practical application for the above expression, is to find
measurement of enthalpy for a thermochemical process (i.e.,
∆Hof values for a specific substance, given
∆H(oreaction) ), and when ∆Hof values are known for
all remaining substances present.
ex. 10 Combustion of 1.243 g of dimethyl ether at constant pressure, C2H6O (g), liberates 39.3 kJ of
heat.
o
Determine ∆H f of dimethyl ether. (given qp = 39.3 kJ for 1.243 g cmpd.)
7. The algebra of thermochemical equations: Hess' Law Thermochemical equations apply to both physical and
chemical changes, and they obey all rules of algebra. Consider the next three thermochemical equations which
are "algebraic" variations of the same chemical reaction…
Mg(s) + 2 O2(g) = MgO(s)
∆HoRx = - 601.7 kJ,
MgO(s) = Mg(s) + 2 O2(g)
2 Mg(s) + O2(g) = 2 MgO(s)
ex. 11 Given
∆HoRx = - 1203.4 kJ
P(s) + 3/2 Cl2(g) = PCl3(g)
∆HoRx = -287 kJ
P(s) + 5/2 Cl2(g) = PCl5(g)
∆HoRx = -374.9 kJ
Determine ÄH for the process,
ex. 12 Given
∆HoRx = + 601.7 kJ
2 N2(g) + 2 O2(g) =
PCl3(s) + Cl2(g) = PCl5(g).
NO(g)
∆HoRx = +90.2 kJ
NH4NO3(s) = 2 N2(g) + NO(g) + 2 H2O(R)
∆HoRx = -115.7 kJ
H2(g) + 2 O2(g) = H2O(R)
∆HoRx = -285.8 kJ
Calculate the enthalpy of formation for ammonium nitrate (s).
Answers to Problems Worked Out in Class
ex. 1
ex.2
ex. 3
a.
q = (sp.ht.)(mass)( Ä T )
b.
c.
+100 Joule = (0.843 J / g-deg)(? gram)(51.4 - 43.6 deg)
? gram = 15.2
 71 g 
first convert from mole to gram: ? g = 0.38 mole 
 = 27 gram
1 mole 
-25.0 Joule = (0.478 J / g-deg)( 27 gram)(? Tf - 18)
Tfinal = 18 - 1.94 = 16.06 deg.C
a.
Br
b.
C10H8 (gas) = C10H8 (liquid) ÄH = - 43.3 kJ/mol
c.
C6H6 (liquid) = C6H6 (solid) ÄH = - 9.84 kJ/mol
d.
Hg (liquid) = Hg (gas)
2 (liquid)
= Br 2 (gas)
? q = ( 1.72 J / g-deg)(10.0 g)(21.3 - 18.6 deg) = 52 Joule
ÄH = + 29.6 kJ/mol
ÄH = + 59.4 kJ/mol
Note: temperature DOES NOT CHANGE during phase transitions,
so can=t use q =(sp.ht)(mass)(Ä T) .
Work directly off the fact that enthalpy always has units of J/mole or kJ/mole, and scale kJ for
given masses that are not equal to the mass of one mole. .
1mole 
Convert from mass to mole:? mole = 50.00 g Hg 
 = 0.249 mole
 201 g 
Convert from listed value of enthalpy of vaporization, as 59.4 kJ / one mole, to mole value of
problem
 59.4 kJ 
? kJ = 0.249 mole 
? kJ = 14.8 kJ

 1 mole 
ex. 5
a.
system: metal
T(initial) = 87.6 deg.C
T(final) = 18.1 deg.C
mass = 12.78 gram
sp.ht. = ?
metal LOST heat,
process is exothermic
can=t determine sp.ht. metal b/c heat loss ( -q ) is not known
b.
surroundings: water
T(initial) = 17.4 deg.C
T(final) = 18.1 deg.C
mass = 51 gram
sp.ht. = 4.18 J/g-deg
water GAINED heat,
process is endothermic. find q(water)
? q(surroundings) = ( 417
. J / g − deg) (51g )(181
. − 17.4 deg) = + 149.2 Joule
c.
The magnitude of heat gained by surroundings (water) must be the same as heat lost by
system (metal) but has opposite sign in order to conserve energy. So endothermic for
surroundings must mean exothermic for system, or
q(system) = - q(surroundings)
d.
now find sp.ht. metal:
q(metal) = - (+ 149.2 Joule) = - 149.2 Joule
q(metal) = (sp.ht. metal) (mass metal) [ T(final) - T(initial) metal ]
-149.2 J = ( ? sp.ht. ) ( 12.78 g ) ( 18.1 - 87.6 deg)
sp.ht.metal = 0.168 J/g-deg
(note: sp.ht. cannot have a negative value. Why?)
ex. 6
a.
system:
a chemical reaction
mass cmpd. = 1.372
ÄT(system) really not applicable here,
b.
surroundings:
calorimeter
T(initial) = 18.32 T(final) = 21.53
heat capacity : 10.12 kJ/deg.
This is a constant volume process ( q v )
temperature of calorimeter increased, therefore it gained heat energy. Calorimeter
process is endothermic. How much heat was gained by the calorimeter?
q v (calorimeter) = (calorimeter heat capacity) [T(final) - T(initial) calorimeter]
q v (surr) = (10.12 kJ / deg ) ( 21.53 - 18.32 deg ) = + 32.5 kJ
q v (reaction) = - q v (calorimeter); q v (reaction) = - 32.5 kJ
c.
q v (reaction) = - 32.5 kJ (for 1.374 g cmpd)
scale up for one mole cmpd (109 g)
 − 32.5 kJ 
? q v (reaction per mole) = 109 g / mole 
 = - 2580 kJ/mole
 1.374 g 
Use upper case thermochemical symbols for per mole quantities. B./c this process
took place in a bomb calorimeter it is a constant volume process,
so q v (on a per mole base) gives the TOTAL ENERGY (E)
Ä E = - 2580 kJ/mole
ex. 7
ex. 8
a.
C (s)
b.
Mg
c.
2C
d.
N2 (g)
a.
balanced chemical equation:
b.
=
CO2 (g)
(s)
+
C (s)
+
3/2 O2 (g)
=
MgCO3 (s)
∆ Hf0 = - 1113 kJ/mole
(s)
+
3 H2 (g) +
1/2 O2 (g)
=
C2H5OH (liq)
∆ Hf0 = - 278 kJ/mole
+
2 H2 (g)
+
1/2 Cl2 (g)
+
=
NH4ClO4 (s)
∆ Hf0 = - 295 kJ/mole
BaCO3 (s) = BaO (s) + CO2 (g)
∆ H o (reaction) = ?
3
= [ ( - 582 ) + ( - 394 ) ] - [ - 1219 ] = + 243 kJ/mole
for 25.0 g barium carbonate
b.
2 O2 (g)
o = [∆ H o
o
o
∆H Rx
f ( BaO) + ∆H f (CO ) ] − [ ∆H f ( BaCO ) ]
∆ H o (reaction)
a.
Hf0 = - 393.5 kJ/mole
O2 (g)
2
ex. 9
∆
+
? kJ = 25.0 g
balanced chemical equation:
C2H6 (g) + 7/2 O2 (g) = 2 CO2 (g)
 +243 kJ  1 mole 
 1 mole   197 g 


+ 3 H2O (liq)
= + 30.8 kJ
∆ H o (reaction) = ?
∆H(oreaction ) = [ 2∆H of ( CO2 ) + 3∆H of ( H2O ) ] − [ ∆H of (C2H6 ) + ( 7 / 2) ∆H of ( O2 ) ]
= [ 2 ( - 394) + 3 ( - 286) ] - [ ( - 85) + ( ZERO) ] = - 1561 kJ/mole
ex. 10 a.
balanced chemical equation:
C2H6O (g) + 3 O2 (g) = 2 CO2 (g)
b.
∆ H o (reaction) = ?
combustion of 1.243 g (at constant pressure) liberates 39.3 kJ, so q p = -39.3 kJ
convert to ÄH o (reaction)
c.
+ 3 H2O (liq)
?
∆ Ho =
??? = - 1454 kJ/mole
∆H(oreaction ) = [ 2∆H of ( CO2 ) + 3∆H of ( H2O ) ] − [ ∆H of ( C2 H6O) + 3∆H of (O2 ) ]
- 1454 kJ/mole = [ 2 ( - 394) + 3 ( - 286) ] - [ ? ∆H f (C H O ) + 3 ( ZERO) ]
2 6
o
∆H of (C2H6O )
ex. 11 given:
(eq 1.)
(eq 2.)
find
= - { ( - 1454 ) - [ - 788 - 858 ] } = - 192 kJ/mole
P (s) + 3/2 Cl2 (g) = PCl3 (g)
∆ Hf o = - 287 kJ/mole
P (s) + 5/2 Cl2 (g) = PCl5 (g)
∆ Hf o = - 375 kJ/mole
∆ H o (reaction) for
PCl3 (s) + Cl2 (g) = PCl5 (g)
alter given chemical equations and associated enthalpies to yield desired reaction.
PCl3
add this to (eq 2.)
P (s) + 5/2 Cl2 (g) = PCl5
(g)
∆ Hf o = - 375 kJ/mole
PCl3 (s) + Cl2 (g) = PCl5
(g)
∆ H o (reaction) = - 88 kJ/mole
net equation is:
ex. 12 given:
(eq 1.)
(eq 2.)
(eq 3.)
find:
reverse (eq 2.)
(g)
= P (s) + 3/2 Cl2 (g)
∆ Hf o = + 287 kJ/mole
reverse (eq 1.)
1/2 N2 (g) + 1/2 O2 (g) = NO (g)
NH4NO3 (s) = 1/2 N2 (g) + NO (g) + 2 H2O (liq)
H2 (g) + 1/2 O2 (g) = H2O (liq)
∆ Hf o = + 90 kJ/mole
∆ Hf o = - 116 kJ/mole
∆ Hf o = - 286 kJ/mole
N2 (g) + 2 H2 (g) + 3 / 2 O2 (g) = NH4NO3 (s)
∆ H o (reaction) = ???
1/2 N2 (g) + NO (g) + 2 H2O (liq) = NH4NO3 (s)
∆ Hf o = + 116 kJ/mole
add (eq 1.)
1/2 N2 (g) + 1/2 O2 (g) = NO (g)
∆ Hf o = + 90 kJ/mole
add TWO (eq 3.)
2 H2 (g) + O2 (g) = 2 H2O (liq)
∆ Hf o = - 572
net equation is:
N2 (g) + 2 H2 (g) + 3 / 2 O2 (g) = NH4NO3 (s)
∆ H of
kJ/mole
= - 366 kJ/mol