Enthalpies of Reactions
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.
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