Thermochemistry
Objectives:
1. Describe the steps involved in a chemical reaction: reaction mechanism
2. State that breaking bonds requires energy and making bonds releases energy
3. Define:
a. Heat of …
4. Use definition of state function to validate Hess’s Law
5. Determine heat of reaction using Hess's Law
Objective 1
How does a reaction happen?
We already talked about this a little bit:
Activated Complex
Reactants
Energy
Products
Reaction Coordinate
Summary:
 How a reaction happens:
1. Energy is added
a. Called: Activation Energy, Ea
b. Used for: Breaking bonds between reactants
2. Activated complex or intermediate formed
3. New bonds are formed which releases energy
 Change in energy between reactants and products is H
In the reaction between hydrogen and chlorine,
H2 + Cl2
Energy
HCl
Reaction Coordinate
H2(g) + Cl2(g)  2HCl(g)
Objective 2
Steps in the reaction
1. The reactants are hydrogen and chlorine
2. Energy is added to break the H—H and Cl—Cl bonds
a. The bond energy of H—H = 436 kJ/mol
b. The bond energy of Cl—Cl = 242 kJ/mol
 Activation energy = 436 kJ/mol + 242 kJ/mol = 678 kJ/mol
3. Activated complex formed: single H and single Cl atoms
4. H—Cl bonds made, releasing energy
a. 2 bonds are made, so 2(431) = 862 kJ/mol
H2 + Cl2
Ea = 678 kJ/mol
Energy
862 kJ
H = -184kJ
HCl
Reaction Coordinate
A better picture: more to scale:
Ea = 678 kJ/mol
Energy
862 kJ
H2 + Cl2
HCl
H = -184kJ
Reaction Coordinate
H = energy breaking bonds – energy of making bonds
a. This is how we get the rule:
i. H < 0, exothermic
ii. H > 0, endothermic
Objective 3


Pseudonyms (other names) for H
a. Heat of Reaction: Hrxn heat produced in a chemical reaction
b. Heat of Combustion: Hcomb heat produced by a combustion
reaction
c. Heat of Neutralization: heat produced in a neutralization reaction
(when an acid and base are mixed to get water, pH = 7)
d. Heat of solution: Hsol heat produced by when something dissolves
e. Heat of Fusion: Hfus heat produced when something melts
f. Heat of Vaporization: Hvap heat produced when something
evaporates
g. Heat of Sublimation: Hsub heat produced when something sublimes
h. Heat of formation: Hf change in enthalpy that accompanies the
formation of 1 mole of compound from it’s elements (this has special
uses in chemistry…)
Objective 4
 H is a state function: it does not depend on the route, the net is still the same.
Example:
Add the following numbers:
1
2
3
4
+5
=15
2
4
5
1
+3
=15
It doesn’t matter what order they are added in, the result will be the same:
Another example: you can send your clothes to the cleaners to wash or you can wash
them at home or you can wash them at a Laundromat. It doesn’t matter how you
choose to do it, the clothes still come out clean.
A chemistry example:
a
b
c
Energy
Lets say a reaction took place in several steps as shown in the energy diagram above.
1. The net enthalpy change for step a is given by the red line
2. the net enthalpy change for step b is given by the blue line
3. the net enthalpy change for step c is given by the green line.
Hess’s Law:
 It doesn’t matter if the change takes several steps. The change in enthalpy will still be
the same as if it had taken place in one step.
 Formal Law: If a change of state occurs in stages or steps (even if only
hypothetically), the enthalpy change for the overall (net) change is the sum of the
individual enthalpy stages for the individual steps.
Hess’s Law Problems
Objective 5
 Heat of reaction can be found by: sum the heats of formation of all the products
– sum of heats of formation of all the reactants

  f
products
  f reac tan ts =Hrxn
Hf = standard enthalpy of formation. Energy required to form a compound from
its elements.
 “standard” is a term used a lot in chemistry. It usually means that the values are
experimentally determined and compared to an agreed upon reference value
 Since the Hf is given per mole, we must multiply by coefficients
______________________________________________________________________________________
Example 1:
Ca(OH)2(s) + CO2(g)  H2O(g) + CaCO3(s)
Hf
Hf
Reactants
Products
Ca(OH)2
-986.1 kJ/mol
H2O(g)
-241.8 kJ/mol
CO2
-393.5 kJ/mol
CaCO3(s)
-1206.9 kJ/mol

1.
 f
2.
f reactan ts = 1(-986.1 kJ/mol)+ 1(-393.5 kJ/mol) = -1379.6 kJ/mol
3.
  f

Hrxn = -69.1 kJ/mol, Exothermic
products
= 1(-241.8 kJ/mol) + 1(-1206.9 kJ/mol) = -1448.7 kJ/mol
products
  f reac tan ts = (-1448.7 kJ/mol) – (-1379.6 kJ/mol) = -69.1 kJ/mol
____________________________________________________________________________________________________________
Example 2: Find Hrxn for the following reaction
N2O4 + 3 CO  N2O + 3CO2
Hf
Hf
Reactants
Products
N2O4
9.7 kJ/mol
N2O
81 kJ/mol
CO
-110 kJ/mol
CO2
-393 kJ/mol
1.
 f
2.
f reactan ts =-9.7 kJ + 3(-110) kJ = -320.3 kJ/mol
3.
  f
products
= 81kJ + 3(-393 kJ/mol) = -1098 kJ/mol
products
  f reac tan ts = (-1098kJ/mol) – (-320.3 kJ/mol) = -778 kJ/mol
Hrxn = -778 kJ/mol
________________________________________________________________________
4.
Challenge Practice: Using the following reaction, find the Hf for C2H6
2C2H6 + 7O2  6H2O + 4CO2 H = -3119 kJ
Reactants
C2H6
O2
Answer: -84.5 kJ/mol
Hf
?
0
Products
H 2O
CO2
Hf
-286 kJ/mol
-393 kJ/mol