Enthalpies of Reactions
You can find the overall enthalpy of a reaction by subtracting
the enthalpy at the beginning of the reaction from the enthalpy
at the end of the reaction:
DH = Hfinal - Hinitial
which is virtually the same as saying
DH = Hproducts - Hreactants
under the thermodynamic standard states 25ºC (298K), 1
atm, and 1 M. Try using this equation in a problem.
Example
Calculate the DH for the following:
3Al(s) + 3NH4ClO4(s)
Al2O3(s) +
AlCl3(s) + 3NO(g) +
6H2O(g)
given the following values:
| Substance |
 (kJ/mol) |
| NH4ClO4(s) |
-295 |
| Al2O3(s) |
-1676 |
| AlCl3(s) |
-704 |
| NO(g) |
90 |
| H2O(g) |
-242 |
| Al(x) |
0 (since it’s an element) |
Explanation
[1(-1676) + 1(-704) + 3(90) + 6(-242)] -
[3(0) + 3(-295)] = -2677 kJ
This is an exothermic reaction since DH is
negative. Don’t forget to multiply values by coefficients since
each coefficient represents the number of moles of each substance!
Try another one:
Example
Find the DHf of
C6H12O6(s) using
the following information:
C6H12O6(s) +
6O2(g) 
6CO2(g) +
6H2O(l) + 2800 kJ
| Substance |
 (kJ/mol) |
| CO2(g) |
-393.5 |
| H2O(l) |
-285.8 |
Explanation
[6(-393.5) + 6(-285.8)] - [1(x)
+ 6(0)] = -2800 kJ
Now solve the above for
x and you have
your answer! The value 2800 is negative because this reaction is
exothermic. Oxygen is considered an element in its free state, so
it is assigned a value of zero. (All diatomic molecules are assigned
zeros for the same reason.) After solving for
x,
you get
for glucose.
