Hess’s Law
For applying the concept of Hess’s Law, it is important to understand that enthalpy is a state function; therefore, the enthalpy change for a chemical reaction is independent from the path by which the products are formed. In fewer words, multi-step processes can be expressed as one-step processes if all of the individual steps are taken in account. Hess’s Law states that the enthalpy change for a given reaction equals the sum of the enthalpy values for each individual step. For example, consider the formation of carbon monoxide from graphite and oxygen:
2C(graphite) + O2(g) ® 2CO(g)
This is, really, a two-step process, in which graphite and oxygen form carbon dioxide, which further decomposes into carbon monoxide and oxygen gas:
2C(graphite) + 2O2(g) ® 2CO2(g)
2CO2(g) ® 2CO(g) + O2(g)
Each individual step has an enthalpy value. In the first step, DH = -393.5 kJ per mole of CO2 formed. Notice that in this equation, 2 moles of CO2 are formed; therefore, the enthalpy value must be multiplied by two (DH = -393.5 kJ x 2 = -787 kJ).
Enthalpy values are usually obtained from tables that show the enthalpy value for each selected equation. Sometimes, the equation that we are looking for may not appear in the direction that we expect but as a reverse reaction as well. If this is the case, we just flip the equation and change the sign for DH. For instance, the decomposition of CO2 into CO and O2 may appear as follows:
2CO(g) + O2(g) ® 2CO2(g); DH = -566 kJ
So, we just reverse the equation and change the sign:
2CO2(g) ® 2CO(g) + O2(g); DH = +566 kJ
In the end, we have:
2C(graphite) + 2O2(g) ® 2CO2(g) DH1 = (-393.5kJ) x (2)
2CO2(g) ® 2CO(g) + O2(g) DH2 = (+566.0kJ)
2C(graphite) + O2(g)® 2CO(g) DH3 = (-221.0kJ)
Sample Problem:
Hydrazine, N2H4, is a colorless liquid used as a rocket fuel. What is the enthalpy change for the process in which hydrazine is formed from its elements?
N2(g) + 2H2(g) ® N2H4(l)
Use the following reactions and enthalpy changes:
(a) N2H4(l) + O2(g) ® N2(g) + 2H2O(l); DH = -622.2 kJ
(b) H2(g) + ½ O2(g) ® H2O(l); DH = -285.8 kJ
Solution:
First, look at the general equation for the formation of the desired product. See what things are on the reactant’s side and what things are on the product’s side. Then, set your other two equations in such a way that reactants and products are on the side where you want them.
For (a), you want N2H4 to be on the right side and N2 on the left side. So, we flip the equation and change the sign:
N2(g) + 2H2O(l) ® N2H4(l) + O2(g); DH = +622.2 kJ
For (b), you want H2 to be on the reactant’s side (right where it is). Also, want O2 and H2O to be on opposite sides in relation to the same elements on the previous equation, so at the end they may all cancel out. Notice, however, that 2 moles of H2 are needed to react in order to form the desired product; also, 2 moles of water and 1 mole of O2 are needed to cancel out with such elements on the opposite side of the previous equation. Everything needs to be multiplied by 2, including DH:
(2) x (H2(g) + ½ O2(g) ® H2O(l); DH = -285.8 kJ)
= 2H2(g) + O2(g) ® 2H2O(l); DH = -571.6 kJ
At the end, we have:
N2(g) + 2H2O(l) ® N2H4(l) + O2(g) DH1 = +622.2 kJ
2H2(g) + O2(g) ® 2H2O(l) DH2 = -571.6 kJ
N2(g) + 2H2(g) ® N2H4(l) DH3 = +50.6 kJ
Exercises:
- Ammonia will burn in the presence of a platinum catalyst to produce nitric oxide, NO.
4NH3(g) + 5O2(g) ® 4NO(g) + 6H2O(g)
What is the heat of reaction at constant pressure? Use the following thermochemical equations:
N2(g) + O2(g) ® 2NO(g); DH = 180.6 kJ
N2(g) + 3H2(g) ® 2NH3(g); DH = -91.8 kJ
2H2(g) + O2(g) ® 2H2O(g); DH = -483.7 kJ
(Answer: -906.3kJ)
- Compounds containing carbon-carbon double bonds, such as ethylene, C2H4, add hydrogen in a reaction called hydrogenation.
C2H4(g) + H2(g) ® C2H6(g)
Calculate the enthalpy change for this reaction, using the following combustion data:
C2H4(g) + 3O2(g) ® 2CO2(g) + 2H2O(l); DH = -1401 kJ
C2H6(g) + 7/2O2(g) ® 2CO2(g) + 3H2O(l); DH = -1550 kJ
H2(g) + ½ O2(g) ® H2O(l); DH = -286 Kj
(Answer: -137kJ)