Topic 5-energetics SL
5.1 Exothermic and endothermic reactions
5.2 Calculation of Energy/Enthalpy Changes
5.3 Hess law
5.4 Bond enthalpy
5.1 Exothermic reactions
• Heat is produced and transferred to
surroundings
• NaOH(s) + H2O  NaOH(aq) + heat
• HCl + NaOH  NaCl + H2O + heat
Neutralisation
• Wood + O2  CO2 + H2O + heat
Combustion
Exothermic- compare explosion
Endothermic reactions
• Heat is consumed from surroundings- it gets
colder or you need to heat
• Ba(OH)2(s) + 2 NH4SCN(s) + heat 
Ba2+(aq) + 2 SCN-(aq) + 2 H2O(l) + NH3(aq)
Enthalpy, H
• H = internal energy. The total chemical energy
of a system. Some of the energy is stored in
chemical bonds.
DH = enthalpy change
• There is no “absolute zero” for enthalpy =>
enthalpy for a particular state cannot be
measured but changes in enthalpy during
reactions can be measured.
• DH = Hproducts – Hreactants
In the reaction
2 H2 (g) + O2 (g) 2 H2O (g)
Hreactants- Enthalpy of reactants
Hproducts- Enthalpy of products
Enthalpy, H
DH= Hproducts - Hreactants = - 486 kJ
2 H2 + O2
EXOTHERMIC
- 486 kJ
2 H2O
In the reaction
1/2 N2 (g) + O2 (g) NO2 (g)
DH= Hproducts - Hreactants = 33,9 kJ/mol
Enthalpy, H
NO2
33,9 kJ
1/2 N2 + O2
ENDOTHERMIC
Exothermic reaction
CH4 + 2 O2  CO2 + 2H2O + heat
Energy rich
Energy poor
• DH = (Energy poor) – (Energy rich) => negative
value
=> In Exothermic reactions: DH < 0 => Gives
more stable products
=> In Endothermic reactions DH > 0 => Gives
more reactive products.
Type of
reaction
Exothermic
Endothermic
Heat energy
change
Heat energy
evolved
Heat energy
absorbed
Temperature Sign of DH
change
Becomes
Negative (-)
hotter
Becomes
Positive (+)
colder or
energy must
be added
DHo: standard enthalpy change of
reaction
Standard conditions: p =101.3 kPa, T =298 K
Factors affecting DHo
• The nature of the reactants and products
• The amount
• Changing state involves the enthalpy change
• The temperature and pressure of the reaction
surroundings
DHfo: standard enthalpy of formation
(cf. page 8 in Data Booklet)
The enthalpy difference for the reaction when
the substance is formed from it’s elements
under standard conditions.
5.2 Calculation of Energy/Enthalpy
Changes
• Measurements:
Open calorimeter
Bomb calorimeter
Calculation of heat of solution
•
•
•
•
•
q = c . M . DT (page 1 in Data Booklet)
q= energy (J)
m = mass (g)
DT = temperature change (K)
c = specific heat capacity, different for all
substances
– E.g. 4.18 J/g*K for Water
Example
• The heat energy required to heat 50 g of
water from 20oC to 60oC is:
q = 50*4.18*(60-40) = 8364 J = 8.364 kJ
Energy and heat are always positive
Enthalpy changes
DH = -1202 kJ/mol
Exothermic
• The amount of energy released when 0.6 g of Mg
is burnt?
Mg
m
0.6 g
M
24.3 g/mol
n
0.025 mol
q=1202*0.025 = 30 kJ (energy is always positive)
Mg + ½ O2  MgO
5.3 Hess’s law
• The principle of conservation of energy states
that energy cannot be created or destroyed.
• The total change in chemical potential (enthalpy
change) must be equal to the energy gained or
lost.
• The total enthalpy change on converting a given
set of reactants to a particular set of products is
constant, irrespective of the way in which the
change is carried out.
C + ½ O2  CO
CO +½ O2  CO2
DH1=-283,0 kJ
DH2=-110,5 kJ
C + O2  CO2
DH3 = DH1+DH2= -393, 5 kJ
http://www.ausetute.com.au/hesslaw.html
http://www.mikeblaber.org/oldwine/chm1045/notes/Energy/HessLaw/Energy04.htm
5.4 Bond Enthalpies
• Break chemical bonds requires energy =>
Endothermic process
• Form chemical bonds => Exothermic process
• Approximate enthalpy change, DH, can be
calculated by looking at bonds being broken
and formed in the reaction.
Average bond enthalpies
• gaseous molecule into gaseous atoms (not
necessary the normal state)
• approx in different molecules
not so precise data, but normally within 10 %
cf. page 7 in Data Booklet
Calculate DH for the reaction:
2 H2 (g) + O2 (g) 2 H2O (g)
H-H 436 kJ/mol
H-O 464 kJ/mol
O=O 498 kJ/mol
Enthalpy
4H+2O
+ 1370 kJ
2 H2 + O2
1. The bonds of the reactants are broken, enthalpy is needed
2 mol H-H = 2* 436= 872 kJ
1 mol O=O =
498 kJ
Sum
1370 kJ is spent
Enthalpy
4H+2O
+ 1370 kJ
2 H2 + O2
- 1856 kJ
2 H2O
2.The free hydrogen and oxygen atoms form bonds to create the
products. The bond enthalpy is released
2*2 mol H-O = 4* 464= 1856 kJ is formed
Enthalpy
4H+2O
+ 1370 kJ
2 H2 + O2
+ 486 kJ
- 1856 kJ
2 H2O
3. The enthalpy of the products are 1856-1370 = 486 kJ lower than
the reactants
The excess enthalpy 486 kJ is released to the surroundings.
Exothermic reaction, DH= -486 kJ/mol
The ”extra” enthalpy needed (1370 kJ) is called ACTIVATION
ENERGy
N2 (g) + 3 H2 (g) 2 NH3 (g)
(Enthalpies involved (see data booklet page 7))
N≡N
H-H
N-H
945kJ/mol
436 kJ/mol
391 kJ/mol
DH = (bonds broken) – (bonds formed) =
(945 + 3*436) – (2*3*391) = -93 kJ/mol
Exothermic
(If using other data DHf = -92kJ/mol)
Enthalpy
6H+2N
+ 2253 kJ
N2 + 3 H2
+ 93 kJ
- 2346 kJ
2 NH3
N2 + 3 H2  2 NH3
The enthalpy of the products are 2346-2253 = 93 kJ lower than the reactants
The excess enthalpy 96 kJ is released to the surroundings.
Exothermic reaction, DH= -93 kJ/mol
1/2 N2 (g) + O2 (g) NO2 (g)
Enthalpy
2O+N
ENDOTHERMIC
+ 970,5 kJ
- 812 kJ
NO2
½ N 2 + O2
The sum of the bond enthalpies
- 93 kJ
The bonds of the reactants are broken,
enthalpy is needed
1/2 mol N≡N
1/2* 945= 472,5 kJ
1 mol O=O
= 498 kJ
Sum
970,5 kJ is spent
The free nitrogen and oxygen atoms form
bonds to create the products. The bond
enthalpy is released.
1 mol N-O = 222kJ
1mol N=O = 590 kJ
812 kJ is formed
The resulting enthalpy, 93 kJ is taken from
the surroundings.
Endothermic reaction, DH= 93 kJ/mol