1. Hess' Law Learning Goals:
Will be able to apply Hess’s Law to determine the enthalpy change of chemical equations.
Will be able to write target equations from word equations

2. Hess’ Law Start Finish Path independent
Both lines accomplished the same result, they went from start to finish.
Net result = same.

3. Hess’s Law states….
The enthalpy change (ΔH) in a chemical reaction – from reactants to products – is the same whether the conversion occurs in one step or several steps.
ie. ΔH is independent of the path taken

4. Hess’s Law Rules
The value of the ΔH for any reaction can be written in steps equals the sum of the values of ΔH for each of the individual reactions ∆Htarget = ∑ ∆Hunknowns
If a chemical reaction is reversed then the sign of ΔH changes.
If changes are made to the coefficients of a chemical equation (i.e. to balance it) then the value of ΔH is altered in the same way.

5. Example 1:
What is the enthalpy change for the formation of nitrogen monoxide from its elements?
N2(g) +O2(g) → NO(g) ΔH=?
Given the following known reactions:
½N2(g) +O2(g) → NO2(g) ΔH= + 34kJ
NO(g) + ½ O2(g) → NO2(g) ΔH= - 56kJ
Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction.
and.. the H values must be treated accordingly.

6. Example 2:
Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values:
2CO2 (g) +H2O(g) →C2H2 (g) + 5/2 O2 (g)
C2H2 (g) +2H2 (g) →C2H6 (g) ΔH=-94.5kJ
H2O(g)  H2 (g) + ½ O2 (g) ΔH=71.2kJ
C2H6 (g) +7/2O2 (g) →2CO2 (g) +3H2O(g) ΔH=-283kJ

7. Example 3:
How much energy can be obtained from the roasting of 50.0kg of zinc sulphide ore?
ZnS (s) + 3/2 O2 (g)  ZnO(s) + SO2(g)
ZnO(s)  Zn(s) + ½ O2(g) ΔH= 350.5kJ
S(s) + O2(g)  SO2(g) ΔH=-296.8kJ
ZnS(s)  Zn(s) + S (s) ΔH= 206.0kJ

8. Found in more than one place, SKIP IT (its hard).
Goal:
4NH3(g) O2(g)  4NO(g) + 6H2O(g)
Using the following sets of reactions:
(1) N2(g) + O2(g)  2NO(g) H = kJ
(2) N2(g) + 3H2(g)  2NH3(g) H = kJ
(3) 2H2(g) O2(g)  2H2O(g) H = kJ
4NH3  2N H2 H = kJ
NH3:
(2)(Reverse and x 2)
Found in more than one place, SKIP IT (its hard).
O2 :
NO:
(1) (Same x2)
2N O2  4NO H = kJ
H2O:
(3)(Same x3)
6H O2  6H2O H = kJ

9. Found in more than one place, SKIP IT.
Goal:
4NH3(g) O2(g)  4NO(g) + 6H2O(g)
4NH3  2N H2 H = kJ
NH3:
Reverse and x2
Found in more than one place, SKIP IT.
O2 :
NO: x2
2N O2  4NO H = kJ
6H O2  6H2O H = kJ
H2O: x3
Cancel terms and take sum.
H = kJ kJ + (-1451kJ)
+ 5O2
+ 6H2O
H = kJ
4NH3 
4NO
Is the reaction endothermic or exothermic?

10. Consult your neighbor if necessary.
Determine the heat of reaction for the reaction:
TARGET  C2H4(g) + H2(g)  C2H6(g) H = ?
Use the following reactions:
(1) C2H4(g) O2(g)  2CO2(g) H2O(l) H = kJ
(2) C2H6(g) + 7/2O2(g)  2CO2(g) H2O(l) H = kJ
(3) H2(g) /2O2(g)  H2O(l) H = -286 kJ
Consult your neighbor if necessary.

11. Determine the heat of reaction for the reaction:
Goal: C2H4(g) + H2(g)  C2H6(g) H = ?
Use the following reactions:
(1) C2H4(g) O2(g)  2CO2(g) H2O(l) H = kJ
(2) C2H6(g) + 7/2O2(g)  2CO2(g) H2O(l) H = kJ
(3) H2(g) /2O2(g)  H2O(l) H = -286 kJ
C2H4(g) :use 1 as is C2H4(g) O2(g)  2CO2(g) H2O(l) H = kJ
H2(g) :# 3 as is H2(g) /2O2(g)  H2O(l) H = -286 kJ
C2H6(g) : rev # CO2(g) H2O(l)  C2H6(g) + 7/2O2(g) H = kJ
C2H4(g) + H2(g)  C2H6(g) H = -137 kJ

12. Used for reactions that
Summary:
Used for reactions that
cannot be determined
experimentally
Summary:
H is independent of
the path taken