Standard Enthalpy of Formation
We can use:
1) q=mcDT to experimentally determine an energy
value from a reaction,
2) Hess’ Law to theoretically determine an energy
value for a reaction from a series of reactions, and
3)
standard enthalpy of formation values to
theoretically determine an energy value for any
hypothetical reaction.
A standard enthalpy of formation is an experimentally
determined value that involves the forming of 1 mole
of a substance from its elements in their standard
states (SATP conditions).
Eg: H2(g) + 1/2O2(g)  H2O(l) DHof= -285.8 kJ/mol
By definition all elements in their “usual” states have
DHof= 0 kJ/mol
Eg:
O2(g)  O2(g)
DHof= 0 kJ/mol
Using Standard Molar Enthalpies (DHf’)
For any chemical reaction:
DH = SDHofproducts - SDHofreactants
Eg:
Ans.
Calculate the molar enthalpy of combustion of
methane.
Chemical reaction is:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
DHo = SDHofproducts - SDHofreactants
DHo = S(DHofCO2(g) +2DHofH2O(l))- S(DHofCH4(g) + 2DHofO2(g))
DHo=(1 mol x -393.5 kJ/mol + 2 mol x -285.8 kJ/mol) (1 mol x -74.4 kJ/mol + 2 mol x 0 kJ/mol)
DHo= (-965.1 kJ) – (-74.4 kJ)
DHo= -890.7 kJ
Since 1 mol of methane in chemical reaction,
DHocomb = -890.7 kJ/mol
We could also generate our own given equations and
use Hess’ Law to rearrange them in order to reach our
target equation.
Eg: For target: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Derive:
1) H2(g) + 1/2O2(g)  H2O(l) DHof= -285.8 kJ/mol
2) C(s) + O2(g)  CO2(g)
DHof= -393.5 kJ/mol
3) C(s) + 2H2(g)  CH4(g)
DHof= -74.4 kJ/mol
4) O2(g)  O2(g)
DHof= 0 kJ/mol
____________________________________________
1)
2)
3)
4)
X2
X1
X-1
X2
(-285.8 x2) = -571.6kJ
(-393.5 x1) = -393.5kJ
(-74.4 x-1) = 74.4kJ
(0 x2)
= 0 kJ
o
DH = -890.7kJ