CAcT HomePage
Enthalpies of Reactions
Skills to develop
 Calculate the enthalpies of reactions from bond energies.
 Calculate the enthalpies of reactions from enthalpies of formation.
 Draw energy level diagrams and use them to calculate enthalpies of reactions.
Enthalpies of Reactions
Enthalpy of a reaction or energy change of a reaction
DH, is the amount of energy or heat
absorbed in a reaction. If the energy is required,
DH is positive, and if energy is
released, the DH, is negative.
_________ Products

 DH, positive for
 endothermic reaction

________ Reactants
The enthalpy can be determined by experiment, but estimates can
easily be made if
bond energies or
standard enthalpies
of formation for the reactants and products are available.
Using Formulas to Calculate DH
Due the the definitions of various types of energy related terms,
formulas for evaluating enthalpies can be very confusing. For example,
the formulas to calculate the enthalpy of a reaction depends on whether
bond energies or enthalpies of formation are available.
When standard enthalpies of formation, H_{f}^{o},
for all products and reactants are available, we have
H_{reaction} = SUM (H_{products})
 SUM (H_{reactants})
or if you prefer using symbols
H_{reaction} = SH_{products}
 SH_{reactants}
For simplicity in formulation we use H to represent
H_{f}^{o} in the above formulas.
Because bond energies are defined as the energies required to break the bonds,
positive values are usually listed whereas in reality, they are energies
released when chemical bonds are formed from respective atoms. Thus, using
the bond energies (BE) as they are given or defined, the following
formula apply:
H_{reaction} = SUM (BE_{reactants})
 SUM (BE_{products})
or if you prefer using symbols
H_{reaction} = SBE_{reactants}
 SBE_{products}
These formulas to calculate the enthalpy (heat) of a reaction can be a
very confusing and you may easily get an incorrect sign for the value.
Thus, remembering formulas is discouraged!
Well, if you use a diagram to help visualize the calculation or write
the chemical reaction equations accompanying the thermodynamic values,
you will be able to avoid the confusion. In both cases, you are applying
the principle of conservation of energy in solving the problems.
Calculate Enthalpy of Reaction from Bond Energies
Due to the principle of conservation of energy the total energy
before and after the reaction must not change. Thus, the energy of a
reaction released or absorbed in the reaction must come from the difference
in bond energies of the products and the reactants.
Example 1
The bond energy (kJ) for H_{2}, F_{2}, and HF are 436,
158 and 568 kJ
respectively, calculate the enthalpy (energy) of
the reaction,
H_{2}(g) + F_{2}(g) = 2 HF
Solution
Based on the bond energies given, we have
H_{2} ® 2H D = 436 kJ/mol
F_{2} ® 2F D = 158 kJ/mol
2H + 2F ® 2HF H = 568*2 kJ/mol
Adding all three equations and energies leads to the following
H_{2}(g) + F_{2}(g) = 2 HF
DH = 542 kJ/equation
Note that D represent bond dissociation energy, and H the
enthalpy of the reaction as written. We use DH
in the last equation to denote enthalpy of change of the overall reaction.
Discussion
Since bond energies are given, we use the monoatomic gases are the reference
level in this calculation. The energy level diagram shown below illustrates
the principle of conservatin of energy, and you are expected to have the skill
to draw such a diagram.
2 H(g) + 2 F(g)
 436+158 kJ
 
+2*568 kJ H_{2} + F_{2}
 
  DH = 542 kJ/equation
 ¯
2 HF(g)
This diagram is very similar to those of BohnHaber cycle used to
evaluate lattice energy.
Calculate Enthalpy of Reaction from Enthalpy of Formation
A similar cycle can be devised to calculate energy of a reaction when the
standard enthalpies of formation are given. We illustrate this cycle by an
an example.
Example 2
Standard enthalpies of formation are: C_{2}H_{5}OH(l) 228, CO_{2} 394,
and H_{2}O(l) 286 kJ/mol. Calculate the enthalpy of the reaction,
C_{2}H_{5}OH + 3 O_{2} ® 2 CO_{2} + 3 H_{2}O
Solution
From the definition of the enthalpy of formation, we have the following
equations and the energy changes of reactions.
C_{2}H_{5}OH(l) ® 2 C(graphite) + 3 H_{2}(l) + 0.5 O_{2}(g) H = 228 kJ/mol
2 C(graphite) + 2 O_{2}(g) ® CO_{2}(g)
H = 394*2 kJ/mol
3 H_{2}(g) + 1.5 O_{2}(g) ® 3 H_{2}O(l)
H = 286*3 kJ/mol
Adding all three equations and energies leads to the following
C_{2}H_{5}OH(l) + 3 O_{2}(g) ® 2 CO_{2}(g) + 3 H_{2}O (l)
DH = 1418 kJ/mol
Discussion
Since the standard enthalpy of formation uses the elements as the standard,
we put the elements on the top as a common level of 0 energy.
The enthalpies of formation are negative, and we have the following diagram.
Thus, the enthalpy of reaction is the difference between the
level of
C_{2}H_{5}OH +3 O_{2}
and
2 CO_{2} + 3 H_{2}O.
2 C(graphite) + 3 H_{2} + 3.5 O_{2}(g)
 
  228 kJ
 ¯
 C_{2}H_{5}OH + 3 O_{2}
394*2 286*3 
 
  DH = 394*2  286*3  (228)
  = 1418 kJ
¯ ¯
2 CO_{2} + 3 H_{2}O
A Footnote
Due to the definitions, the enthalpies of formation are negative values
whereas the bond energies are given as positive values.
Thus, calculations using these two types of data have different signs.
You can use the diagramatic method to solve these types of problems
as given in the discussions above.
Using the diagrams shown indicates clearly what should be the sign of the
results. When the reaction is from a higher level to a lower level, the
enthalpy of reaction should be negative (downward arrow). If you reverse
the direction of the reaction, you also change the sign of the
enthalpy of the reaction.
Confidence Building Questions

Calculate the heat of reaction for
2 H_{2} + O_{2} = 2 H_{2}O
Bond energies: H_{2} 436 kJ/mol, O_{2} 498, HO 463 kJ/mol
Discussion 
Note that in this problem, you are asked to calculate the heat of reaction,
but the heat of formation is 241 kJ / mol.

Calculate the heat of reaction for
H_{2} + Cl_{2} = 2 HCl
Bond energies: H_{2} 436 kJ/mol, O_{2} 498, HO 463, Cl_{2} 243, HCl 432 kJ/mol
Discussion 
If you are given the heat of the reaction, and the HH bond energy,
can you calculate the bond energy for ClCl?

Calculate the enthalpy of formation for NH_{3}.
Bond energies: H_{2} 436 kJ/mol, N_{2} 945, HN 391 kJ/mol
Another method 
Write the enthalpies of formation below their formulas in the equation, we have:
^{3}/_{2} H_{2} + ^{1}/_{2} N_{2} = NH_{3}
3*436/_{2} + 945/_{2} = 3*391 + H_{f}
From this equation, we have:
H_{f} = 3*436/2 + 945/2  3*391 = 46.5 kJ/mol
A handbook gives the enthalpy of formation of NH_{3} of 46.1 kJ/mol.
As an exercise, use this value to calculate the tripple bond energy of N_{2}.

Calculate the enthalpy of the reaction
CH_{4} + 2 O_{2} = CO_{2} + 2 H_{2}O
from the enthalpies of formation: CH_{4} 75 kJ/mol, CO_{2} 394,
and H_{2}O(l) 286 kJ/mol.
Discussion 
The value calculated is the enthalpy of combustion of methane
©cchieh@uwaterloo.ca