THERMOCHEMISTRY

Part IV: Hess's Law

State function | Hess's Law | Applications of Hess's Law | Exercises |

 

You far you have learned how to determine the enthalpy change of a chemical reaction using calorimetry or to calculate the enthalpy change from the standard enthalpies of formation of products and reactants. In this module you will learn how to determine the reaction enthalpy of an overall process from the enthalpy changes of individual reactions.

Basic definitions:

State function: The value of a state function is independent on how the state was achieved, in other words the path is not important (such as the altitude of a mountain climber). Other examples for state functions include temperature and volume. Likewise the enthalpy of a system is a state function, since the change of enthalpy for an overall physical process is independent of it's path. The same applies to chemical reaction and is summarized as
Hess's Law:

The overall reaction enthalpy is the sum of the reaction enthalpies of the steps into which the reaction can be divided. Examples:

An example is the highly exothermic reaction of hydrogen with oxygen to form water (applications: space shuttle, fuel cells)

H2(g) + 1/2 O2(g) è H2O(l) DH = -285 kJ/mol

This process can be divided into smaller steps such as
    1.     H2(g) + 1/2 O2(g) è H2O(g)            DH = -241 kJ/mol

    2.     H2O(g) è H2O(l)                            DH = -44 kJ/mol
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    3.     H2(g) + 1/2 O2(g) è H2O(l)            DH = -285 kJ/mol

The sum of the two steps is the overall reaction. Likewise, the sum of the reaction enthalpy is the reaction enthalpy for the overall process:

DH reaction = DH reaction1 + DH reaction 2


Enthalpy Diagram:

 An example is the oxidation of carbon to form carbon monoxide:

CO(s) + 1/2 O2(g) è CO2                    DH = ?

Use indirect method:

            1. C(s) + 1/2 O2(g) è CO(g)                DH1 = -110 kJ

2. CO(g) + 1/2 O2(g) è CO2 (g)         DH2 = ?

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add both equations :         C(s) + O2(g) è CO2(g) DHoverall = -393 kJ

Solve for DH2:

        DH = DHoverall - DH1 = -393 kJ-(-110kJ) = -283 kJ

Applications of Hess's Law include reactions for which it is not always possible to measure the reaction enthalpy directly. An indirect method can be used to find the enthalpy for a particular step. An example is the reaction of ethene (produced from fossil fuels) and water to form ethanol alcohol:

C2H4(g) + H2O(l) è C2H5OH (l)                DH = ?

In this case we have to devise a sequence of reactions ("invent" some reactions) that allow us to determine the enthalpy for above reaction indirectly. In this class we will focus on just combustion reactions as "invented" reactions.

Scheme:

                1. Select one of the reactants and write a balanced combustion reaction

                                                C2H4(g) + 3 O2(g) è 2 CO2 (g) + 2 H2O(l)                DH = -1411 kJ

            2. Select one of the products and write a balanced combustion reaction
       

      C2H5OH (l) + 3 O2(g) è 2 CO2(g) + 3 H2O (l)            DH = -1367 kJ

      3. Reverse equation number (2) so that C2H5OH becomes a product, reverse sign for reaction enthalpy as well.

      2 CO2(g) + 3 H2O (l) è C2H5OH (l) + 3 O2(g)            DH = +1367 kJ

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      4. Add reaction (1) and (3); add reaction enthalpies for reaction (1) and (3):

C2H4(g) + H2O(l) è C2H5OH (l)                                DH = -44 kJ
 You just found the enthalpy for the overall reaction!

Exercise: Devise a similar scheme to determine the reaction enthalpy for these reactions:

(a) C2H2 (g) + H2O(l) è CH2OH-CH2OH            DH = ?

(b) C2H2(g) + H2(g) è C2H6                                DH=?
 
 


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