1. Hess’s Law Hess’s law allows you to determine the energy of chemical reaction without directly measuring it. The enthalpy change of a chemical process is equal to the sum of the enthalpy changes of all the individual steps that make up the process.

2. Either way, you get to the finish.
Hess’s Law
The enthalpy change of a physical or chemical process is independent of the pathway of the process and the number of intermediate steps in the process.
Finish
Start
Either way, you get to the finish.

3. Most reactions occur in more than one step or could hypothetically occur in more than one step.
We focus on the OVERALL reaction based on the balanced chemical equation.

4. and.. the H values must be treated accordingly.
Determine the heat of reaction for the reaction:
4NH3(g) O2(g)  4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g)  2NO(g) H = kJ
N2(g) + 3H2(g)  2NH3(g) H = kJ
2H2(g) + O2(g)  2H2O(g) H = kJ
Hint: The three reactions must be algebraically manipulated to add up to the desired reaction.
and.. the H values must be treated accordingly.

5. 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:
N2(g) + O2(g)  2NO(g) H = kJ
N2(g) + 3H2(g)  2NH3(g) H = kJ
2H2(g) O2(g)  2H2O(g) H = kJ
4NH3  2N H2 H = kJ
NH3:
Reverse and x 2
Found in more than one place, SKIP IT (its hard).
O2 :
NO: x2
2N O2  4NO H = kJ
H2O: x3
6H O2  6H2O H = kJ

6. 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
H2O: x3
6H O2  6H2O H = kJ
Cancel terms and take sum.
+ 5O2
+ 6H2O
H = kJ
4NH3 
4NO
Is the reaction endothermic or exothermic?

7. . Determine the heat of reaction for the reaction:
C2H4(g) + H2(g)  C2H6(g)
Use the following reactions:
C2H4(g) O2(g)  2CO2(g) H2O(l) H = kJ
C2H6(g) + 7/2O2(g)  2CO2(g) H2O(l) H = kJ
H2(g) /2O2(g)  H2O(l) H = -286 kJ .

8. Determine the heat of reaction for the reaction:
Goal: C2H4(g) + H2(g)  C2H6(g) H = ?
Use the following reactions:
C2H4(g) O2(g)  2CO2(g) H2O(l) H = kJ
C2H6(g) + 7/2O2(g)  2CO2(g) H2O(l) H = kJ
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

9. Goal:
Achieve the Target equation by manipulating several sub-equations.
The enthalpy change of the target equation will be the sum total of the enthalpy changes for each sub-equation.

10. Rule #1 If you reverse an equation the ∆H must switch signs.
Na (s) + ½ Cl2(g) NaCl (s) ∆H = -411 kJ
NaCl (s)  Na (s) + ½ Cl2(g) ∆H = 411 kJ

11. Rule #2
If you need to multiply or divide any part of the equation, the whole equation must follow the same operation, including ∆H.
Na (s) + ½ Cl2(g) NaCl (s) ∆H = -411 kJ
2 Na (s) + Cl2(g) 2 NaCl (s) ∆H = -822 kJ

12. Practice A + B  AB ∆H1 = 20 kJ AB + B  AB2 ∆H2 = 50 kJ
What is the ∆H for the overall reaction
A + 2 B  AB2 ∆H3 = ?
∆H3 = 70 kJ

13. For example, suppose you are given the following data:
Could you use these data to obtain the enthalpy change for the following reaction? 3

14. If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third. 3

15. Example Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ
Calculate DHº for this reaction

16. Practice:

17. Practice: