1. Hess’s Law

2. Hess’s law states:
If a reaction can be written as a series of steps that add up to the overall reaction, the ΔH for the overall reaction will be the sum of the ΔH’s for each of the steps.
If we know ΔH for some reactions, we can manipulate them (reverse the order, multiply or divide by a factor) to add up to the desired overall equation.

3. Why do we need it?
A science reference will only include ΔH for some types of reactions (like formation). We can use the reactions included in the reference to find ΔH for reactions not included in the reference.
The ΔH for some reactions are very difficult to measure experimentally.

4. Why do we need it?
From the AP course description!
Students should understand these Hess’s law ideas: When a reaction is reversed, the sign of the enthalpy of the reaction is changed; when two (or more) reactions are summed to obtain an overall reaction, the enthalpies of reaction are summed to obtain the net enthalpy of reaction.

5. Use these hints for doing problems with Hess’s Law.
The reactions that add up to the overall reaction will not necessarily be given in the correct direction or with the correct number of moles.
Use these hints for doing problems with Hess’s Law.
1. Use the overall reaction as a guide.
2. If you reverse the reaction, change the sign of H.
3. If you multiply the reaction by a constant, do the same to H.

6. calculate H for the reaction
Examples
1. Given the following data:
S(s) + 3/2 O2(g)  SO3(g) H = kJ/molrxn
2 SO2(g) + O2(g)  2 SO3(g) H = kJ/molrxn
calculate H for the reaction
S(s) + O2(g)  SO2(g)

7. calculate H for the reaction
Examples
2. Given the following data:
2 O3(g)  3 O2(g) H = -427 kJ/molrxn
O2(g)  2 O(g) H = +495 kJ/molrxn
NO(g) + O3(g)  NO2(g) + O2(g) H = -199 kJ/molrxn
calculate H for the reaction
NO(g) + O(g)  NO2(g)

8. Examples 3. Given the following data:
C2H2(g) + 2H2(g)  →  C2H6(g)                        ΔH  = kJ/molrxn
H2O(g)  →  H2(g) + ½ O2 (g)                           ΔH  = 71.2 kJ/molrxn
C2H6(g) +  7/2 O2(g)  →  2CO2(g)  +  3H2O(g)   ΔH  = -283 kJ/molrxn
calculate H for the reaction
2CO2(g)  +  H2O(g)  →  C 2H2(g) +  5/2 O2(g)

9. Examples 4. Given the following data:
P4(s) + 6 Cl2(g)  4 PCl3(g) H = kJ/molrxn
P4(s) + 5 O2(g)  P4O10(s) H = kJ/molrxn
PCl3(g) + Cl2(g)  PCl5(g) H = kJ/molrxn
PCl3(g) + ½ O2(g)  Cl3PO(g) H = kJ/molrxn
calculate H for the reaction
P4O10(s) + 6 PCl5(g)  10 Cl3PO(g)