1. 1
Hess’s Law
Start
Finish
A State Function: Path independent.
Both lines accomplished the same
result, they went from start to finish.
Net result = same.
2. 2
Determine the heat of reaction for the reaction:
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g) 2NO(g) ∆H = 180.6 kJ
N2(g) + 3H2(g) 2NH3(g) ∆H = -91.8 kJ
2H2(g) + O2(g) 2H2O(g) ∆H = -483.7 kJ
Hint: The three reactions must be algebraically
manipulated to sum up to the desired reaction.
and.. the ∆H values must be treated accordingly.
3. 3
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g) 2NO(g) ∆H = 180.6 kJ
N2(g) + 3H2(g) 2NH3(g) ∆H = -91.8 kJ
2H2(g) + O2(g) 2H2O(g) ∆H = -483.7 kJ
Goal:
NH3:
O2 :
NO:
H2O:
Reverse and x 2 4NH3 2N2 + 6H2 ∆H = +183.6 kJ
Found in more than one place, SKIP IT (its hard).
x2 2N2 + 2O2 4NO ∆H = 361.2 kJ
x3 6H2 + 3O2 6H2O ∆H = -1451.1 kJ
4. 4
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)Goal:
NH3:
O2 :
NO:
H2O:
Reverse and x2 4NH3 2N2 + 6H2 ∆H = +183.6 kJ
Found in more than one place, SKIP IT.
x2 2N2 + 2O2 4NO ∆H = 361.2 kJ
x3 6H2 + 3O2 6H2O ∆H = -1451.1 kJ
Cancel terms and take sum.
4NH3
+ 5O2 4NO + 6H2O ∆H = -906.3 kJ
Is the reaction endothermic or exothermic?
5. 5
Determine the heat of reaction for the reaction:
C2H4(g) + H2(g) C2H6(g)
Use the following reactions:
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) ∆H = -1401 kJ
C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(l) ∆H = -1550 kJ
H2(g) + 1/2O2(g) H2O(l) ∆H = -286 kJ
Consult your neighbor if necessary.
6. 6
Determine the heat of reaction for the reaction:
Goal: C2H4(g) + H2(g) C2H6(g) ∆H = ?
Use the following reactions:
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) ∆H = -1401 kJ
C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(l) ∆H = -1550 kJ
H2(g) + 1/2O2(g) H2O(l) ∆H = -286 kJ
C2H4(g) :use 1 as is C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) ∆H = -1401 kJ
H2(g) :# 3 as is H2(g) + 1/2O2(g) H2O(l) ∆H = -286 kJ
C2H6(g) : rev #2 2CO2(g) + 3H2O(l) C2H6(g) + 7/2O2(g) ∆H = +1550 kJ
C2H4(g) + H2(g) C2H6(g) ∆H = -137 kJ