|Standard Heats of Formation and Hess's Law|
|The preceding sections have lead to this point.
= [Ho products] - [Ho reactants]
|What goes into the products and reactants brackets are the standard
heats of formation. We'll begin by the defining the term standard state.
Any substance in its most stable physical form (gas, liquid, or solid)
at 25oC and under a pressure of 1 atm is said to be in its standard
state. The element oxygen for example, is in its standard state when its
exists as a molecule of O2 -not as atoms and not as molecules
of O3. The element carbon is in its standard state when it exits
as graphite - not as diamond - at 25oC. Diamond is also a form
of carbon, but it is actually slightly less stable than graphite.
|The quantities that we'll use from now on to compute values for Ho
are called the standard enthalpies of formation of standard heats
of formation. The standard heats of formation of a compound Hfois
the amount of heat absorbed or evolved when one mole of the compound is
formed from its elements in their standard states. Thus, the thermochemical
equation for the formation of one mole of liquid water from oxygen and hydrogen
in their standard states is
H2(g) + ½O2(g) ------> H2O(l)
Hfo = -285.8 kJ/mol
|The standard enthalpy change for this reaction, that is, the enthalpy
change at 25oC and 1 atm, is called the standard heat of formation
of liquid water. This is a point that often causes confusion among students.
Each of the following equations, for example involves the formation of CO(g)
C(s) + O2(g) ----> CO2(g)
CO(g) + ½O2(g) ----> CO2(g)
2 C(s) + 2 O2(g) ----> 2 CO2(g)
|However, only the first involves just the elements as reactants
and the formation of just one mole of CO2. In the second
equation, one of the reactants is a compound, carbon monoxide, and in the
third, two moles of CO2 are formed. Only the first equation, therefore,
has a standard enthalpy change that we identify as Hfo.
Your databook has tables of standard heats of formation for a
variety of substances. They are listed alphabetically and in some cases not
logically. There are also standard heats of formation values for the organic
|***************** IMPORTANT **********************
The Hfo for any element in its standard state is zero. (0 kJ/mol)
|This makes sense if you think about it. There would be no enthalpy
change if you "form an element in its standard state from itself."
|Using Hess's Law with Standard Heats of Formation|
|The Ho for any reaction must be the difference between
the total enthalpies of formation of the products and those of the reactants.
Any reaction can be generalized by the equation
aA + bB + ..... -----> nN + mM + .....
|where a, b, n, m, etc, are the coefficients of substances A, B, N,
M, etc. The value of Ho can be found using Hess's law equation.
Ho = [nHfo(N) + mHfo(M)
+ ...] - [aHfo(A) + bHfo(B) +
|Putting this into a more useful form we get:
Ho = [sum of the Hfo of products]-[sum of
the Hfo of reactants]
|Some chefs keep baking soda, NaHCO3, handy to put out grease
fires. When thrown on the fire, baking soda partly smothers the fire, and
the heat decomposes it to give CO2, which further smothers the
flame. The equation for the decomposition of NaHCO3 is
2 NaHCO3(s) -------> Na2CO3(s) + H2O(g)
|Calculate the ΔHo for this reaction in kilojoules
ΔHo = sum of products - sum of reactants
= [ Na2CO3(s) + H2O(g) + CO2(g) ] - [ (2)NaHCO3(s)]
|Look up the values in the databook tables for each substance.
Make sure the physical states are identical.
= [-1130.7 -241.8 -393.5 kJ/mol ]-[(2)(-950.8 kJ/mol)]
= -1766 kJ/mol - (-1901.6 kJ/mol)
= +135.6 kJ/mol
Under standard conditions, the reaction is endothermic by 135.6
1 cal = 4.184 J or 1 kcal = 4.184 kJ
In kilocalories the reaction is endothermic by 135.6 kJ/mol
which is 32.41 kcal/mole.
|Example Problem #2|
|What is the ΔHo in kilojoules for the combustion
of 1 mol of ethanol, C2H5OH(l), to form gaseous carbon
dioxide and gaseous water?
|First write and balance the combustion equation.
C2H5OH(l) + 3 O2(g) -----> 2 CO2(g) + 3 H2O(g)
Hess's law for this equation is:
= [(2)CO2(g) + (3)H2O(g)] - [C2H5OH(l)
The reaction for the combustion of ethanol is exothermic by 1235.3 kJ/mol.