THERMOCHEMISTRY- A
By
Dr. Hisham E Abdellatef
2011-2012

THERMOCHEMISTRY
The study of heat released or required by chemical reactions or heat changes caused by chemical reactions

System: part of universe selected for the thermodynamic study.
• Surrounding: the part of the universe which surround the system.
HEAT HEAT HEAT HEAT

Heat of Reaction
• Heat of reaction : the change in energy which accompanies a chemical reaction.

H
• Endothermic reaction : the reaction which is supplied by heat (absorb heat and

H is
+ ve).
• Exothermic reaction: the reaction which is accompanied with evolution of heat
(evolve heat and

H is - ve).

System and Surroundings
In chemical reactions, heat is often transferred from the “system” to its “surroundings,” or vice versa.
• The substance or mixture of substances under study in which a change occurs is called the thermodynamic system ( or simply the system .
)
• The surroundings are everything outside of the thermodynamic system.

Exothermic
If

E < 0, E final
< E initial cellular respiration of glucose
Burning fossil fuels is an exothermic reaction

Endothermic
If

E > 0, E final
> E initial
Photosynthesis is an endothermic reaction
(requires energy input from sun)

Kinetic energy
(E
K
)
What is Energy?
Internal Energy E
Potential energy (E
P
)
Energy due to motion
Energy due to position
(stored energy)

J °g -1 C -1
• The amount of heat which required to raise the temperature of 1gof substance by 1 °C.
•
The amount of heat which required to raise the temperature of one mole of substance by 1 °C molar heat capacity = specific heat x M.wt.

• The change in energy which accompanies a chemical reaction.
• The change in enthalpy (heat evolved) when one mole of the substance is completely burned in presence of excess oxygen.
• The change in enthalpy (heat evolved) when one mole of the substance is formed from its elements their standard state. (t = 25 °C and
P = 1 atm).

• J °C -1 Specific Heat x mass
• The amount of heat which required to raise the temperature of the substance by
1 °C.
Which has the larger heat capacity?

How do we relate change in temp. to the energy transferred?
Heat capacity (J/ o C) = heat supplied (J) temperature ( o C)
Heat Capacity = heat required to raise temp. of an object by 1 o C

Specific heat capacity is the quantity of energy required to change the temperature of a 1g sample of something by 1 o C
Specific Heat
Capacity (Cs)
J / o C / g
=
=
Heat capacity
Mass
J / o C g

How much energy is transferred due to T difference?
The heat
(q)
“lost” or “gained” is related to a) sample mass b) change in T and c) specific heat capacity
Specific heat capacity = heat lost or gained by substance (J)
(mass, g)(T change, K)

Substance Spec. Heat (J/ g•K)
H
2
O 4.184
Ethylene glycol 2.39
Al glass
0.897
0.84
Aluminum

If 25.0 g of Al cool from 310 o C to 37 o C, how many joules of heat energy are lost by the Al?
Specific heat capacity = heat lost or gained by substance (J)
(mass, g)(T change, K)

If 25.0 g of Al cool from 310 o C to 37 o C, how many joules of heat energy are lost by the Al?
heat gain/lose = q = (sp. ht.)(mass)(∆T) where ∆T = T final
- T initial q = (0.897 J/ g•K)(25.0 g)(37 - 310)K q = - 6120 J
Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

1 calorie = heat required to raise temp. of 1.00 g of
H
2
O by 1.0 o C.
1000 cal = 1 kilocalorie = 1 kcal
But we use the unit called the JOULE
1 cal = 4.184 joules
James Joule
1818-1889

heat energy transferred
∆E = q + w work done by the system energy change
ة ظوُف ْحَم

heat transfer in
(endothermic), +q heat transfer out
(exothermic), -q
SYSTEM
∆E = q + w w transfer in
(+w) w transfer out
(-w)

• 1 st Law of Thermodynamics:
• Energy can neither created not destroyed , only transformed from one form to another

Exothermic Endothermic
6.2

Enthalpy Diagrams
• Values of

H are measured experimentally.
• Negative values indicate exothermic reactions.
• Positive values indicate endothermic reactions.
A decrease in enthalpy during the reaction;

H is negative.
An increase in enthalpy during the reaction;

H is positive.

Exothermic Examples
• Oxidation – wooden splint burning (giving off light, heat, CO
2
, H
2
O
• Burning H
2 in air,
• body reactions,
• dissolving metals in acid,
• mixing acid and water,
• sugar dehydration

Endothermic Examples
• Electrolysis (breaking water down into H
2
O
2 by running electricity in it)
• Photosynthesis, pasteurization, canning vegetables and
• 2 H
2
+ O
2
 2H
2
O + energy
• 4 g + 32 g  36 g 136 600 cal
• 2H
2
O + energy  2H
2
+ O
2
• 36 g 136 600 cal  4g + 32 g

Changes in Internal Energy
• If

E > 0, E final
> E initial
– Therefore, the system
absorbed energy from the surroundings.
– This energy change is called endergonic .

Changes in Internal Energy
• If

E < 0, E final
< E initial
– Therefore, the system
released energy to the surroundings.
– This energy change is called exergonic .

Thermochemical equations
• factors which affects the quantity of heat evolved or absorbed during a physical or a chemical transformation.
1. Amount of the reactants and products
2. Physical state of the reactants and products
3. Temperature
4. Pressure

Thermochemical equations
It must essentially:
1.be balanced ;
2.give the value of ΔE or ΔH corresponding to the quantities of substances given by the equation;
3.mention the physical states of the reactants and products . The physical states are represented by the symbols (s),
(L), (g) and (aq) for solid, liquid, gas and aqueous states respectively.

Example of thermochemical equation
H
2
+ ½ O
2
→ H
2
O ΔH = -68.32 Kcal
1 mole of hydrogen reacts with 0.5 mole of oxygen, one mole of water is formed and 68.32 Kcal of heats evolved at constant pressure.
But, not specify whether water is in the form of steam or liquid
H
2
H
2
(g) + ½ O
2
(g) + ½ O
2
(g)→ H
2
O (L) ΔH = -68.32 Kcal
(g)→ H
2
O (g) ΔH = -57.80 Kcal.
Effect of temperature ?????

Standard enthalpy change
ΔH°.
298 K and

Enthalpy (H) of the reaction
(Comes from Greek for “heat inside”)
• the sum of internal energy and the product of this pressure and volume.
H = E + PV
» E is the internal energy,
» P is the pressure and
» V is the volume of the system.
• It is also called heat content.
• ΔH = H product – H reactants = Hp – Hr

Calculation of ΔH from ΔE
• When the system changes at constant pressure, the change in enthalpy,

H, is

H =

(E + PV)

H =

E + P

V
At constant pressure and temperature
The enthalpy of a chemical is measured in kilojoules per mole
(kJmol -1 ).


H =

E + P

V For solid and liquid

H =

E • At constant volume
Liquids and solids are neglected in calculation of
 n
 n = moles of gaseous products - moles of gaseous reactants.

• In case of gases

H =

E + P

V (I)
ΔV =Δn x V
• Δn = no of moles of products - no of moles of reactants
PxΔV = PVxΔn (II)
• But PV = RT (for one mole of gas)
• Putting RT in place of PV in equation (II) we get
PΔV = RTΔn
• Substituting the value of P AV in equation (I) we get
ΔH = ΔE + Δn RT

or
R = 1.987 cal. (=2 cal.)
= 8.314 joules

Example 1:
• Calculate

E for the following reaction:
2 CO
(g)
+ O
2(g)
→ 2 CO
2(g)
Where

H = - 135272 cal. At 25 °C
Solution:

H =

E +
 nRT
 n = 2 -3 = -1
T(K) = 25 + 273
- 135272 cal. =

E + (- 1x2x298)

E = - 135272 + 596 = - 134676 cal.

Example 2:
Calculate

E and

H for vaporization of water at 100 °C and 1 atm., the specific heat of vaporization of water under these conditions is 540 cal/g.
Solution:
H
2
O
(l)
↔ H
2
O
(vap)
Sp. heat of vaporization = 540 cal/g. 1 mole H
2
O = 18g.

H = molar heat of vaporization = 540 x 18 at constant pressure
Endothermic reaction since

H = + 9720 cal.
 n = 1 - 0 =1

H =

E +
 nRT
9720 =

E + 1 x 2 x 373

E = 8974 cal.

Example 3:
6.4g of naphthalene C
10
H
8 when burned under constant volume gave 123 KJ at
20 °C, calculate 
E and

H.
Solution: at constant volume the heat of combustion is equal to

E gave means exothermic
C
10
H
8(s)
+ 12 O
2(g)
→ 10CO
2(g)
+ 4H
2
O
(l)
6.4g at constant volume → -123KJ
128g at constant volume
→ 
E

E = = = -2460KJ = - 2460 x 10
6.4
3 J

H =

E +
 nRT
= - 2460 x 10 3 + (- 2 x 8.3 x 293)
= - 2460 X10 -3 - 4863.8 J
= - 2464863 J
= - 2464.863 KJ

Example
• The heat of combustion of ethylene at 17° C and at constant volume is -332.19 kcal. Calculate the heat of combustion at constant pressure considering water to be in liquid state (R = 2 cal.).
The chemical equation for the combustion of ethylene is
C
2
H
4
+ 3 O
2
= 2CO
2
(g) + 2H
2
O (1)
1 mole 3 moles 2moles negligible volume
No. of moles of the products = 2
No. of moles of the reactants = 4
Δn =(2-4) =-2
ΔH = ΔE + Δn RT
ΔH = -332.19 +[ 2 x I0 -3 x -2 x 290] =-333.3 kcal
Given that
ΔE=-332. 19 kcal.
T= 273+17= 290k
R=2cal=2xlO -3 kcals.

Example
• The heat of combustion of carbon monoxide at constant volume and at 17° C is -283.3 Kj. Calculate its heat of combustion at constant pressure(R= 8.314 J degree1 mole1 ).
CO(g) + ½ O
2
(g) → CO
2
(g)
1 mole ½ mole 1 mole
No. of mles of products = 1
No. of moles of reactants =1.5
n = No. of moles of products - No. of moles of reactants
=1-1.5 =-0.5
Given that :
ΔE =-283.3 kJ
T = (273+17) = 290 K.
R = 8.314 J or
8.314x10-3 KJ
ΔH = ΔE + Δn x RT
ΔH= -283.3 + (-0.5x (8.314x10
-3 ) x 290] = - 283.3-1.20 =-284.5 KJ
Heat of combustion of CO at constant pressure is -284.5 kJ.

Heat of combustion
• The change in enthalpy (heat evolved) when 1 mole of a substance is completely burnt in presence of excess oxygen.
Organic compound + O
2(g)
→ CO
2(g)
+ H
2
O
(l)
• The thermochemical equation must be balanced firstly.

Example 4:
 H for the combustion of liquid heptane C
CO
 E.
and H
2
O
(l)
7
H
Solution:
C
7
 n = 7-11 = -4
H
16(l)
+ 11O
2(g)
→
From thermochemical equation
7CO
2(g)
+ 8H
2
O
(i)
 H =  E +  nRT
 E =  H - A nRT = - 1151 x 10 -3 - (- 4x2x293)
= - 1151 x 10 3 - (-2344) cal./mol.
= -1151 x 10 3 + 2344
= -1148.656 x10 3 cal/mo into

Example 5:
The heat of combustion of benzoic acid C
CO and H
2
O
6
H
5
COOH, into at constant pressure is - 78.2
k J/mole at
27°C calculate heat of combustion at constant volume?
Solution: C
6
H
5
COOH
(S)
 H = - 78.2 x 10 3 J
 n = 7-7.5 = -0.5
+ 7.5 O
2(g)
→ 7CO
2(g)
+ 3H
2
O
 H =  E +  nRT  E =  H  nRT
 E = - 73200 - (-0.5 x 8.3 x (27 + 273)) = -81935 J
(i)

• Try to solve: 3.2 g of naphthalene
C
10
H
O
2 solid when burnt in excess gas into CO
2(g) and H under constant volume gives 1423 KJ at
20°C calculate  E and  H? (M.Wt
= 128).
2
O
Calculate ∆H of reaction?

Heat of Formation :
(Enthalpies of Formation)  H° f
• is the enthalpy change for the formation of one mole of the substance from its elements, at standard pressure (1 atm) and a specified temperature (25°C)
H
2(g)
+ ½ O
C
(graphite)
2(g)
→ H
2
O
+ 2H
2(g)
(l)
→ CH
4(g)
 H° f
 H o
= -285.8 KJ f
= -74.9 KJ

 H f o superscript o means standard state
= 25 o C and 1 atm pressure subscript f means formation from most stable elements

Example 6
:
Calculate the

H ° of the reaction
Fe
2
O
3(s)
+ 3CO
(g)
CO
(g)
→ 2Fe
(s)
= - 110.5KJ/mol
+ 3CO
2(g) given that:

H ° f

H ° f
Fe
2
O
3
CO
2(g)
= - 393.5 KJ/mol
= -822.2 KJ/mol

H ° f
Solution:

H o = (products) - (reactants).
= [(3X-393.5) + (2x0)] - [(lx-822.2) + (3x-110.5)]J
= -1180.5 + 1153.7 = -26.8 KJ.

Example 7 :
Give the following data:
B
2
H
6(g)
+ 6 H
2
O
(l)
 H° = - 493.4 KJ
→ 2H
3
BO
3(s)
+ 6 H
2(g)
 H° f
  H° of H
3
BO f of H
2
O
3(S)
(l) is - 1088.7 KJ/mol, and is - 285.9 KJ/mol
Calculate the standard enthalpy of formation of
B
2
H
6

Solution:
= 2  H° f
H
3
B
2
H
6
- 493.4 KJ = 2x- 1088.7 KJ- [1x  H°
285.9KJ]
BO
3
- [  H° f
+ 6  H° f f
H
2
O]
B
2
H
6
+ 6X -
= - 2177.4KJ - [  H° f
B
2
H
6
- 1715.4KJ]
- 493.4KJ =- 462.0KJ  H o f
B
2
H
6
 H° f
B
2
H
6
= + 31.4KJ/mol

The equations representing the variation of heat changes of reaction with temperature are known as
Kirchoff's equations .

VARIATION OF HEAT (OR ENTHALPY)
OF REACTION WITH TEMPERATURE
KIRCHOFF'S EQUATIONS.
1. At constant volume, internal energies of the reactants and products.
ΔE=E2-E1
Differentiating this equation with respect to temperature at constant volume, we get

heat capacities
But we have already seen that heat capacities of the products and reactants
Integrating between temperature T l and T2 , we have
ΔE
2
ΔE
1
=
Δ E
2
ΔE
1
= Δ Cv (T
2
-T
1
)
(3)

2. At constant pressure
ΔH=H
2
-H
1
And finely …. ΔH
2
- ΔH
1
= ΔCp (T
2
-T
1
) (6)

Example 3 – 3
The heat of reaction
½ H
2
+ ½ Cl
2 =
HCl at 27 C° is -22.1 Kcal.
Calculate the heat of reaction at 77°C. The molar heat capacities at constant pressure at 27° C for hydrogen, chlorine and HCl are 6.82, 7.70 and 6.80 cal mol1 , respectively.
Here ½ H
2
+ ½Cl → HCl Δ H = -22.1 Kcal
Δ Cp = Heat capacities of products - Heat capacities of reactants
= 6.80[½(6.82) +½ (7.70)]
T2 = 273 + 77 =350 K
= 6.80 - 7.26 = -0.46 x 103 Kcal
T1 = 273 + 27 = 300 K
ΔH
2
ΔH
1
= ΔCp (T
2
-T
1
)
ΔH2 - (-22.1) = (-0.46 x 10 -3 ) x 50= -21.123
T2-T1= (350-300)K=50K

Example 3 – 4
• Calculate the standard heat of formation of propane (C
(g) and H
1
2
3
H
8
) if its heat of combustion is
1 . The heats of formation of CO
O (L) are - 393.5 and -285.8 KJ molrespectively.
2

C
3
H
8
(g) + 5 O
2
C (s) + O
2
(g) → 3CO
2
(g) → CO
2
(g)
(g) + 4 H
2
O (L) Δ H c
= 2220.2 KJ
ΔHf=- 393.5 KJ
H
2
(g)+½O
2
(g) →H
2
O (L) ΔHf=-285.8KJ
We should manipulate these equations in a way so as to get the required equation
3 C (s) + 4 H
2
(g) → C
3
H
8
(g) ΔH=?
Multiplying equation (ii) by 3 and equation (iii) by 4 and adding up we get
3C(s) + 3O
2
(g) → 3 CO
2
4H
2
(g) ΔH = -1180.8 KJ
(g) +2O
2
(g) → 4H
2
0(L) ΔH = -1143.2 KJ
3C(s) + 4H
2
(g) + 50
2
(g) → 3CO
2
(g) +4H
2
O(L) ΔH= - 2323.7 KJ subtracting equation (i) from equation (iv), we have
3C(s)+4H
2
(g) → C
3
H
8
(g) Δ H= -103.5 KJ

Heat of transition
• the change in enthalpy which occurs when one mole of an element changes from one allotropic form to another .
• P white
→ P
• S monoclipic red
→ S rhombic
ΔH= -1.028 kcal
ΔH=-0016 Kcal

Hess's law of constant heat summation
If a reaction is carried out in a series of steps, ΔH for the reaction is the sum of
ΔH’s for each of the steps.
A →B + q1
B → C + q2
C → Z + q3
Δ H
1
= q1
Δ H
2
= q2
Δ H
3
= q
3
A →z + Q
1
ΔH
1
= -Q
1
The total evolution of heat = q1 +q2 +q3 =Q2
According to Hess's law Q1= Q2

We can find

H subtracting

H
(a)
(b) by from

H
(c)

Hess’s Law
& Energy Level Diagrams
Forming CO
2 can occur in a single step or in a two steps.
∆H total is the same no matter which path is followed.

Hess’s Law
& Energy Level Diagrams
Forming H
2
O can occur in a single step or in a two steps.
∆H total is the same no matter which path is followed.

Illustrations of Hess's law
• (1) Burning of carbon to CO
2
1 st way:
C(S) + 0
2
KJ)
(g) → CO
2
(g) ΔH = -94.05kcal (-393.50
2 nd way:
C(s)+½O
2
(g) →CO(g) ΔH =-26.42 kcal and
CO
2
(g) + ½ O
2
(g) → CO
2
(g) ΔH = -67.71 kcal
Overall change C (s) + O kcal
2
(g) →CO
2
(g) ΔH = -94.13

Illustrations of Hess's law
(2) Formation of sodium by Hydroxide from Na
1 st way:
2 Na (s) + ½ O
2
(g) → Na
2
O (s) ΔH = -100 kcal
Na
2
O + H
2
O (L) →2 NaOH (aq) Δ H = -56 kcal
2 Na (s) + H
2
O (L) + ½ O
2
(g) → 2 NaOH (aq) ΔH = - 156 Kcal
2 nd way
2 Na (s) + 2 H
2
O (L) → 2 NaOH (aq) + H
2
H
2
(g) + ½ O
2
(g) → H
2
(g) ΔH = - 88 Kcal
O (g) ΔH = - 68.5 Kcal
2 Na (s) + H
2
O (L) + ½ O
2
(g) → 2 NaOH (aq) ΔH = - 156 Kcal

Calculate the heat of hydrogenation of acetylene
1) C
2
H
2(g)
+ 2 H
2(g)
→ C
2
H
6(g)
At 25°C and 1 atm., given that;
 H
1
= ?
2) 2C
2
3) 2C
2
H
2(g)
H
6(g)
+ 5 O
2(g)
4) H
2(g)
+ 7O
+ ½O
2(g)
2(g)
→ 4CO
→ 4CO
2
2(g)
→ H
2
O
(l)
+ 2H
2
O
(l)
+ 6H
2
O
(I )
 H
2
 H
3
 H
4
= - 2602 KJ
=-3123KJ
= -286KJ

Solution:
5) C
2
H
2(g)
+ 5/2 O
2(g)
→
 H
5
2CO
2(g)
+ H
2
O
(l)
= -2602/2 = -1301KJ
2 يلع ) 2 ( ةمسقب
6) 2H
2(g)
+ O
2(g)
→ 2H
2
 H
6
O
(I)
2
= 2 (-286KJ) = - 572 KJ
يف 4 برضب
7) 2CO
2(g)
+ 3H
2
O
(l)
→ 2C
2
H
6(g)
+ 7/2 O
2(g)
2 يلع ) 3 ( ةمسق و سكع
 H7 = + 3123/2 = +1561KJ

Adding equation 5,6 and 7 gives
C
2
H
2(g)
2CO
2(g)
+ 2H
2(g)
+ 3H
2
O
+ 7/2O
2(g)
(|)
+ C
2
H
6(g)
+ 2CO
2
+ 3H
+ 7/2 O
2(g)
2
O
(I)
→
Canceling elements that are the same on both sides, we get
C
2
H
2(g)
+ 2H
2(g)
→ C
2
H
6(g)
 H
1
=  H
5
+  H
6
+  H
7
= (-1301 KJ) + (-572 KJ) + (1561 KJ) = -312 KJ

Calorimetry
• We measure heat flow using calorimetry .
• A calorimeter is a device used to make this measurement.
• A “coffee cup” calorimeter may be used for measuring heat involving solutions.
A “bomb” calorimeter is used to find heat of combustion ; the
“bomb” contains oxygen and a sample of the material to be burned.

Bomb Calorimeter

Coffee-cup calorimeter.