1. Thermochemistry
Chapter 6
Dr. Ali Bumajdad

2. Study of heat change in a chemical reaction
Chapter 6 Topics
Thermochemistry
Study of heat change in a chemical reaction
Nature and Type of Energy
Energy Changes in Chemical Reaction
Intro. to Thermodynamics: 1st Law, Wark, Heat, Enthalpy
Calorimetry, Specific Heat, Heat Capacity
Sstandard enthalpy of formation and reaction
Hess Law
Enthalpy of solution
Dr. Ali Bumajdad

3. Energy is the capacity to do work
Work = Fd
Radiant energy comes from the sun (also called solar energy)
Thermal energy is the energy associated with the random motion of atoms and molecules
Chemical energy is the energy stored within the bonds of chemical substances
Nuclear energy is the energy stored within the collection of neutrons and protons in the atom
Potential energy is the energy object position relative to other objects (e.g. gravity result in pot. E to objects)
Kinetic Energy is the energy of motion
Unit = J = Kgm2s-2
Energy convert from one form to another (Law of Conservation of Energy)

4. Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that are at different temperatures.
Q) Water cup in the fridge, what is the direction of heat transfer?
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
900C
400C
greater thermal energy

5. Thermochemistry is the study of heat change in chemical reactions.
System is the specific part of the universe that is of interest in the study
Surrounding is every things except the thing we are studying
Open
system
mass & energy
Exchange:
Closed
energy
Isolated
nothing
Type:

7. 2H2 (g) + O2 (g) 2H2O (l) + energy
Exothermic process is any process that gives off heat or transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g) H2O (l) + energy
H2O (g) H2O (l) + energy
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
energy + H2O (s) H2O (l)
energy + 2HgO (s) Hg (l) + O2 (g)

8. ‘The study of the interconversion of heat and other kinds of energy’
Thermodynamics
‘The study of the interconversion of heat and other kinds of energy’
State functions properties determined by the current state of the system, regardless of how that condition was achieved
e.g. energy
, pressure, volume, temperature
DE = Efinal - Einitial
Work and Heat
Not State Function
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

9. First law of thermodynamics:
1) energy can be converted from one form to another, but cannot be created or destroyed
DEsystem + DEsurroundings = 0
DEsystem = -DEsurroundings OR
C3H8 + 5O CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings 2)
Heat transfer to (+) or from (-) the system
Work done on (+) or by (-) the system
W=-PV
Change in internal energy of a system
DEsys = q + w
(1)
(2)

10. Combustion reaction release heat
q=-ve
Combustion reaction release heat
Vaporization absorbed heat
q=+ve
Condensation release heat
q=-ve
Freezing release heat
q=-ve
Melting absorbed heat
q=+ve

11. Work is not a state function!
Work Done On the System
w = F x d
w = -P DV
DV > 0
-PDV < 0
wsys < 0
P x V = x d3 = F x d = w F d2
pressure
vol
Work is not a state function!
Dw = wfinal - winitial
initial
final

12.  Work is not state Function
Q) A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm?
1L atm =101.3 J
w = -P DV
(2)
(a)
DV = 5.4 L – 1.6 L = 3.8 L
P = 0 atm
W = -0 atm x 3.8 L = 0 L•atm = 0 joules
(b)
DV = 5.4 L – 1.6 L = 3.8 L
P = 3.7 atm
w = -3.7 atm x 3.8 L = L•atm
w = L•atm x
101.3 J
1L•atm
= J
 Work is not state Function

15. Enthalpy and the First Law of Thermodynamics
Enthalpy (H) used to quantify heat flow into or out of a system in a process that occurs at constant pressure.
DE = q + w
At constant volume:
w = 0 and DE = qv
At constant pressure:
DH = qp and w = -PDV
DE = DH - PDV
DH = DE + PDV
(3)
State Function

16. DH = H (products) – H (reactants)
Enthalpy of Reactions
DH = H (products) – H (reactants)
(4)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0

17. Is DH negative or positive?
H2O (s) H2O (l)
System absorbs heat
Endothermic
DH > 0, H =+ve
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
H2O (s) H2O (l)
DH = 6.01 kJ/mol

18. Is DH negative or positive?
System gives off heat
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
Exothermic
DH < 0, H =+ve
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
DH = kJ/mol

19. Thermochemical Equations
H2O (s) H2O (l)
DH = 6.01 kJ/mol
1) If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s)
DH = kJ/mol
 Hforward =-Hbackward
2) If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
H extensive property
2H2O (s) H2O (l)
DH = 2 x 6.01 = 12.0 kJ/mol
3) The physical states of all reactants and products must be specified in thermochemical equations
H2O (s) H2O (l)
DH = 6.01 kJ/mol
H2O (l) H2O (g)
DH = 44.0 kJ/mol
H H2O(g) > H H2O(l) > H H2O(l)

20. P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ/mol
Q) How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ/mol
= 6470 kJ

22. A Comparison of DH and DE
2Na (s) + 2H2O (l) NaOH (aq) + H2 (g) DH = kJ/mol
DE = DH - PDV
At 25 0C, 1 mole H2 = 24.5 L at 1 atm
and VVgas
PDV = 1 atm x 24.5 L = 2.5 kJ
DE = kJ/mol – 2.5 kJ/mol = kJ/mol
E > H
Some of Internal energy released used to do gas expansion work, this is why
For reactions involving gases E slightly different than H
For reactions do not involve gases E  H (V is very small)

23. Calorimetry, Specific Heat, Heat Capacity
Calorimetry The measurement of heat flow
Specific heat (s) is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
q = m x s x DT s= q
mT
(5)
J/g.°C
Heat capacity (C) is the amount of heat (q) required to raise the temperature of a given quantity (m) by one degree Celsius.
q = C x DT C= q T
(6)
J/°C
C = m x s
(7)
From (5) and (6):

25. Q) How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = J/g • 0C
q = m x s x Dt s= q
mT
(5)
J/g.°C
= -34,000 J

26. q = m x s x Dt

27. (1) Constant-Volume Calorimetry (Good for combustion reaction)
Heat produced by combustion and absorbed by surrounding (calorimeter)
qsys = qcal + qrxn
qsys = 0
qrxn = - qcal
qcal = Ccal x DT
qrxn = -Ccal x DT
(8)
For Reaction at Constant V
DH = qrxn
DE = qrxn
No heat enters or leaves (Adiabatic)

28. Molar heat of combustion: Heat of combustion of one mole
qrxn = -Ccal x Dt
(8b) n
molar

29. (2) Constant-Pressure Calorimetry
Heat produced (or absorbed) by reaction and absorbed (or produced) by the surrounding (solution)
Endo.
qsys = qsolution + qrxn
Exoth.
qsys = 0
qrxn = - qsolution
qcal = Ccal x DT
(8)
qrxn = - Ccal x DT
Reaction at Constant P
DE  qrxn
DH = qrxn
No heat enters or leaves! (Adiabatic)

31. qrxn = -Ccal x Dt
(8b) n
molar

32. s= q mT J/g.°C =79kJ Cmolar= q nT n or
Sa. Ex.5.6a: How much heat is needed to warm 250g of water
from 22°C to 98°C. (water specific heat =4.184 J/g.°C)
q = m x s x DT s= q
mT
(5)
J/g.°C
=79kJ
Sa. Ex.5.6b: What is the molar heat capacity of water?
Cmolar= q
nT
C = m x s
(7) n or

33. qwater = m x s x DT =-qPb
qPb = m x s x DT

34. Reaction is exothermic
qrxn = - qsolution
when density volume = mass
mol of NaOH = MNaOH VNaOH
mol of HCl = MHCl VHCl
=-56.2kJ/mol

35. Standard Enthalpy of Formation (H0f)
(It is the heat change when 1 mole of a compound formed from
its elements at 1 atm)
It is a reference point for all enthalpy expressions (similar to the sea level for measuring heights)
The standard enthalpy of formation of any element in its most stable form is zero
DH0 (O2) = 0 f
DH0 (C, graphite) = 0 f
DH0 (O3) = 142 kJ/mol f
DH0 (C, diamond) = 1.90 kJ/mol f

37. Hrxn= Hstep1 + Hstep2+ Hstep3+……
Standard Enthalpy of Reaction (H0 )
rxn
(It is the enthalpy of a reaction carried out at 1 atm)
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
nDH0 (products) S
mDH0 (reactants)
(9)
Hess’s Law
When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. Enthalpy is a state function It doesn’t matter how you get there, only where you start and end.
If a reaction is carried out in steps, H will equal the sum of the enthalpy changes for the individual steps.
Hrxn= Hstep1 + Hstep2+ Hstep3+……
(10)

38. C (graphite) + 1/2O2 (g) CO (g)
CO (g) + 1/2O2 (g) CO2 (g)
C (graphite) + O2 (g) CO2 (g)

39. Q) Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ
rxn +
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ
rxn
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = (2x-296.1) = 86.3 kJ
rxn

40. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Q) Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04, for CO2 is and for H2O is kJ/mol
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
(9)
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[ 12x– x–285.8 ] – [ 2x49.04 ] = kJ
-6535 kJ
2 mol
= kJ/mol C6H6

41. Hrxn= Hstep1 + Hstep2+ Hstep3+……
(10)

42. Hf°Fe2O3(s)=-822.2kJ/mol, Hf°Al2O3(s)=-1669.8 kJ/mol (9)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
(9)
Divide by 2 to have the molar value then divide by
the molar mass of Al to get it kJ/g

43. Enthalpy of solution (Hsoln)
The heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
DHsoln = Hsoln - Hcomponents

44. The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol