THREE METHODS TO DETERMINE THE ΔH OF A REACTION
Problem: Calculate the ΔH for the reaction below:
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l)
Method #1:
Method #2:
Using Hess’s Law (you must be given two or more reactions with their ΔH values)
½N2(g) + 3/2H2(g) → NH3(g)
½N2(g) + O2(g) → NO2(g)
H2(g) + ½O2(g) → H2O(l)
ΔH = -46 kJ/mol
ΔH = +34 kJ/mol
ΔH = -286 kJ/mol
4NH3(g) → 2N2(g) + 6H2(g)
2N2(g) + 4O2(g) → 4NO2(g)
6H2(g) + 3O2(g) → 6H2O(l)
ΔH = +184 kJ/mol
ΔH = +136 kJ/mol
ΔH = -1716 kJ/mol
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l)
ΔH = -1396 kJ/mol
(x4 and reverse sign)
(x4)
(x6)
Using Heats of Formation (you must be given a chart with various ΔHf values; elements have no ΔHf)
½N2(g) + 3/2H2(g) → NH3(g)
½N2(g) + O2(g) → NO2(g)
H2(g) + ½O2(g) → H2O(l)
ΔHf = -46 kJ/mol
ΔHf = +34 kJ/mol
ΔHf = -286 kJ/mol
Use this equation: ΔH = Σ ΔHf products - Σ ΔHf reactants
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l)
[ 4(-46)
+
7(0) ] →
[ (-184) + (0) ]
→
[-184]
→
[ 4(+34)
[ (+136)
+ 6(-286) ]
+ (-1716) ]
[-1580]
ΔH = Σ ΔHf products - Σ ΔHf reactants
ΔH = (-1580) - (-184)
ΔH = -1396 kJ
Method #3:
Using average bond energies (you must be given a chart of various bond energies; less accurate)
C–H
C–C
C=C
C≡C
413 kJ/mol
347 kJ/mol
614 kJ/mol
839 kJ/mol
N–H
N≡N
N–O
N=O
391 kJ/mol
941 kJ/mol
201 kJ/mol
607 kJ/mol
H–H
O–H
O=O
C=O
432 kJ/mol
467 kJ/mol
495 kJ/mol
745 kJ/mol
Remember: reactant bonds break (requires E) and product bonds make (releases E)
ΔH = Σ E (bonds broken) - Σ E (bonds made)
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l)
[ 12(391)
[ (4692)
+ 7(495) ] → [ 4(201) + 4(607) + 12(467) ]
+ (3465) ] →
[ (804) + (2428) + (5604) ] = [ (3232) + (5604) ]
[ 8157 ] →
[ 8836 ]
ΔH = Σ E reactants - Σ E products
ΔH = (8157) - (8836)
ΔH = -679 kJ
resonance
Practice Problem
[#63 Masterton, 6th Ed.]
Thermochemistry
ASSUMPTIONS:
(1) H = (m)(ΔT)(c)
(2) 6-pk = beer + cans
(3) heat lost = heat gained
On a hot day, you take a six-pack of beer on a picnic, cooling it with ice. Each aluminum
can weighs 38.5 g and contains 340. g (12 fl. oz.) of beer. The specific heat of aluminum
is 0.902 J/g•°C. and that of beer is 4.10 J/g•°C.
(A) How much heat (J) must be absorbed from the six-pack to lower the temperature
from 25.0 to 5.0 °C. ?
H = heat lost by 6-pk = heat lost by cans + beer
H = [(38.5 g/can x 6 cans)(25.0-5.0ºC)(0.902 J/gºC)] + [(340. g/can x 6 cans)(25.0-5.0 ºC)(4.10 J/gºC)
H = [(231 g Al)(20.0ºC)(0.902 J/gºC)] + [(2040g beer)(20.0ºC)(4.10 J/gºC)]
H = [4,167 J lost by Al] + [167,280 J lost by beer]
H = 171,447 -> 3SF -> 1.71 x 105 J heat must be absorbed
(B) How much ice (g) must be melted to absorb this amount of heat?
(Hfusion = 6.02 kJ/mole = 6,020 J/mole)
(heat gained by melting ice) = (heat absorbed from 6-pk)
(Hfusion(ice) x moles ice melted) = 171,447 J
moles ice melted = 171,447 J / Hfusion(ice) = 171,447 J / 6,020 J/mole = 28.48 moles ice
28.48 moles ice x 18.0 g/mole = 513 ice must melt