Enthalpy and Hess’s Law

All chemical reactions either absorb or release energy. The science of studying
heat changes during a chemical reaction  thermochemistry.

Let’s take a typical reaction

AB + CD  AD + BC
How does this reaction happen? Break it down into steps.
AB  A + B {bonds break} energy is needed
CD  C + D {bonds break}
A + D  AD {bonds form} energy is released when bonds are formed
B + C  BC {bonds form}


For the overall reaction, the energy change depends on the relative magnitude of these
processes (the bond breaking and the bond reformation). 
1. Exothermic reaction 
heat is released to the surroundings
2. Endothermic reaction 
heat is supplied to the system by the surroundings.
3. Exothermic process  the total energy of the products is less than the total energy of
the reactants since heat was transferred to the surroundings.
4. Endothermic process 
total energy of products > total energy of reactions
(reason: heat supplied by surroundings).
Enthalpy

Most transformations (chemical and physical changes) occur under constant pressure
conditions, i.e., usually at atmospheric pressure.

Define enthalpy H  the heat content of a substance; H describes the changes in the heat
contents of the products and reactants during a constant pressure process (i.e., H = qp
where qp represents the heat evolved or absorbed by the transformation under constant
pressure conditions).

Exothermic process  heat content of system decreases, the system enthalpy
decreases (i.e., H < 0)

Endothermic process  energy of the system is increased (i.e. H > 0).
Heat
surroundings
System
An Exothermic Process
Heat
System
surroundings
An Endothermic Process

We define the enthalpy of reaction as
H = p npH (products) - r nrH (reactants)
where np = number of moles of the products formed (i.e., stoichiometric coefficients
of products in the balanced chemical reaction),
H (products) = heat content (enthalpies) of products;
nr = number of moles of the reactants (i.e., stoichiometric coefficients of reactants in
the balanced chemical reaction);
H (reactants) = heat content (enthalpies) of reactants.
Conventions for Writing Thermochemical
Equations
1. For exothermic reactions, H <0; for endothermic
reactions, H > 0.
2. Unless otherwise stated, H values are given for
processes occurring at 25C and 1 atm pressure (standard
thermodynamic temperature and pressure).
3. The physical state of each substance participating in the
process must be stated (i.e., s  solid, l  liquid, g  gas,
and aq  aqueous solutions).
4. The stoichiometric coefficients of the substances in an
equation indicate the number of moles of each substance
involved in the transformation (note: fractional
stoichiometric coefficients are permitted in
thermochemical equations). The H value given
corresponds to these specific amounts of substances.
5. If the coefficients of a chemical equation are multiplied or
divided by a certain factor, the H value must also be
multiplied or divided by that factor (i.e., enthalpy is an
extensive property).
6. When an equation is reversed, the sign of the H value is
changed but its magnitude stays the same.
Example Problems for Thermochemical conventions.
H2 (g) + ½ O2 (g)  H2O (l)
(An exothermic process, H < 0).

H = -286 kJ (for 1 mole of H2O (l)).
Reverse the reaction
H2O (l)  ½ O2 (g) + H2 (g)
H = +286 kJ (note: the magnitude is the
same, but the sign of the enthalpy is changed. It is now an endothermic process).

What if we had 2 moles of water
2 H2O (l)  O2 (g) + 2 H2 (g)
absorbed.

We can tackle problems in thermochemistry in much the same way as we tackled
stoichiometry problems.
H = 2 x (286 kJ) = 572 kJ of heat
Example
NH4NO3 is a commonly used explosive for which the decomposition reaction
NH4NO3 (s)  N2O (g) + 2 H2O (g)
is negative (H = -37.0 kJ). What is the enthalpy change when 2.50 g of NH4NO3
decomposes?
1 mole NH4 NO3
n (NH4NO3) = 2.50 g NH4NO3 *
80.04g NH4 NO3
H = n (NH4NO3) x -37.0 kJ/mole NH4NO3 = -1.16 kJ of heat released to the
surroundings
Physical states are very important

H2O (s)  H2O (l)
H2O (s)  H2O (g)
H2O (l)  H2O (g)

H = 6.0 kJ/mole H2O
H = 50.0 kJ/mole H2O
H = 44.0 kJ/mole H2O
In the above example, even though each of the reactants and products are he same chemical
substance (water), since they are in different states, they each have different heat contents.
Enthalpies (since they are energies) are temperature dependent (and slightly pressure
dependent). Therefore, when discussing various transformations, we must specify the
temperature and pressure of the transformation. Therefore, we choose 298 K and 1.00 atm
as our standard thermodynamic reference temperature and pressure.


For the reaction,
C2H5OH (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (g)
H (1) = H (products) - H (reactants)
= 2 x H [CO2 (g)] + 3 x H [H2O (g)] – 1 x H [C2H5OH (l)]
= -992.95 kJ/mol C2H5OH
(1)

For the reverse reaction.
2 CO2 (g) + 3 H2O (g) 3 O2 (g) + C2H5OH (l) (2)
H (2) = 1 x H [C2H5OH (l)] - 3 x H [H2O (g)] – 2 x H [CO2 (g)] = + 992.95 kJ

We again note that H1 = H2 (i.e., the values for the heat content difference are exactly
the same in magnitude, the sign is merely reversed).
Hvalues and Hess’s Law


In order to calculate the enthalpy changes for physical and chemical transformations, we can
resort to tabulated values of reaction enthalpies. These reaction enthalpies have largely been
measured directly with a calorimeter (which we will discuss later). However, we may also
estimate the enthalpies (or heat contents) of chemical systems using an indirect method.

Let’s look at the following example.
For the reactions
CO (g) + ½ O2 (g)  CO2 (g)
H2 (g) + ½ O2 (g)  H2O (g)
H (1) = -111 kJ
H (2) = -242 kJ

We are interested in the enthalpy change for the following process.
CO (g) + H2O (g)  CO2 (g) + H2 (g)

If we reverse the second reaction
H2O (g)  H2 (g) + ½ O2 (g)

The sign of the reaction enthalpy is changed. Now let’s add the third reaction to the first
reaction
CO (g) + ½ O2 (g)  CO2 (g)
H2O (g)  H2 (g) + ½ O2 (g)
CO (g) + H2O (g)  CO2 (g) + H2 (g)
H (4)
H (3) = +242 kJ
Which is the desired enthalpy change!!


To calculate the H value, H (4), all we need to do is add the enthalpy changes for
reactions 3 and 1
i.e., H (4) = H (3) + H (1) = +242 kJ + (-111 kJ)
= +131 kJ.
We have just seen an example of Hess’s Law. Hess’s Law  the heat changes
accompanying a given reaction is the same whether the reaction occurs in a single step or in
many steps.
NOTES ON HESS’S LAW PROBLEMS



General Rule: We should arrange a series of reactions such that when the reactions
are added together, all non-essential reactions go away! We may have to reverse
some reactions and multiply others to get what we want!
The major utility of Hess’s Law is in calculating the enthalpy changes of reactions
that would be difficult to measure.
Note: according to the law of conservation of energy, energy cane neither be created
nor destroyed in an ordinary chemical reaction. Hess’s law tells us that we will never
get more energy (or less energy for that matter) from a chemical reaction.
Another Example
Given the following series of reactions,
N2O4 (g)  2 NO2 (g)
NO (g) + ½ O2 (g)  NO2 (g)
H (1) = 58 kJ
H (2) = -56 kJ
Calculate H for the process
2 NO (g) + O2 (g)  N2O4(g)
Solution
Multiply reaction 2 by 2
2 NO (g) + O2 (g)  2 NO2 (g)
Reverse reaction 1.
2 NO2 (g) N2O4 (g)
H (3) = 2 x H (2)
= -56 kJ x 2
= -112 kJ
H (4) = -58 kJ
Add reaction 3 to reaction 4
2 NO (g) + O2 (g)  2 NO2 (g)
2 NO2 (g) N2O4 (g)
2 NO (g) + O2 (g)  N2O4(g)
H (5) = H (3) + H (4) = -112 kJ + -58 kJ
= -170 kJ.