Thermochem Notes Set-3 (Std. Enthalpy of Formation)
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Transcript Thermochem Notes Set-3 (Std. Enthalpy of Formation)
Thermochemistry
Standard Enthalpies of Formation
Standard Enthalpy of Formation
∆Hf°
change in enthalpy that accompanies the
formation of one mole of a compound from
its elements in standard states
° means that the process happened under
standard conditions so we can compare
more easily
Standard States
For a COMPOUND:
for gas: P = 1 atm
pure liquid or solid state
in solution: concentration is 1 M
For an ELEMENT:
form that it exists in at 1 atm and 25°C
O: O2(g) K: K(s)
Br: Br2(l)
Writing Formation Equations
always write equation where 1 mole of
compound is formed (even if you must use
non-integer coefficients)
NO2(g):
½N2(g) + O2(g) NO2(g)
∆Hf°= 34 kJ/mol
CH3OH(l):
C(s) + 2H2(g) + ½O2(g) CH3OH(l)
∆Hf°= -239 kJ/mol
Using Standard Enthalpies of
Formation
ΔHreaction Σn pΔHf(products) Σn rΔHf(reactants)
where
n = number of moles of products/reactants
∑ means “sum of”
∆Hf° is the standard enthalpy of formation for
reactants or products
∆Hf° for any element in standard state is zero so
elements are not included in the summation
Using Standard Enthalpies of
Formation
since ∆H is a state function, we can use
any pathway to calculate it
one convenient pathway is to break
reactants into elements and then
recombine them into products
Using Standard Enthalpies of
Formation
H H
f ( CO2 )
H
f ( H 2O )
H
f ( CH 4 )
0
Using Standard Enthalpies of
Formation
Example 1
Calculate the standard enthalpy change
for the reaction that occurs when ammonia
is burned in air to make nitrogen dioxide
and water
4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l)
break them apart into elements and then
recombine them into products
Example 1
Example 1
4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l)
can be solved using Hess’ Law:
(1) 4NH3(g) 2N2(g) + 6H2(g)
(2) 7O2(g) 7O2(g)
(3) 2N2(g) + 4O2(g) 4NO2(g)
(4) 6H2(g) + 3O2(g) 6H2O(l)
-4 ∆Hf°NH3
0
4 ∆Hf°NO2
6 ∆Hf°H2O
Look up:
-46 kJ/mol
34 kJ/mol
-286 kJ/mol
46kJ
34kJ
286kJ
H 4mol (
) 0 4mol (
) 6mol (
)
mol
mol
mol
H 1396kJ
values are in Appendix 3: p. A8-A12
Example 1
can also be solved using enthalpy of formation
equation:
ΔH
reaction
ΔH
reaction
Σn pΔH
f(products)
[4(34) 6(286)] [4(46) 0]
H 1396kJ
Σn r ΔH
f(reactant s)
Example 2
Calculate the standard enthalpy change for the
following reaction:
H 0f
2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s)
= 0 kJ/mol
ΔH
reaction
-826 kJ/mol
Σn pΔH
-1676 kJ/mol
f(products)
0 kJ/mol
Σn r ΔH
f(reactant s)
ΔHreaction [1(1676) 2(0)] [2(0) 1(826)]
ΔHreaction 850.kJ
Example 3
Compare the standard enthalpy of
combustion per gram of methanol with per
gram of gasoline (it is C8H18).
Write equations:
2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(l)
2C8H18(l) + 25O2(g)16CO2(g) + 18H2O(l)
Example 3
Calculate the enthalpy of combustion per mole:
HCH
[
2
H
4
H
]
[
2
H
3OH
f (CO2 )
f ( H 2O )
f (CH 3OH ) ]
ΔH
CH 3OH
1454 kJ for 2 moles CH3OH
HC 8H18 [16H f (CO2 ) 18H f ( H 2O) ] [2H f (C8 H18 ) ]
ΔH
C8H18
1.09 10 kJ for 2 moles C8H18
4
Example 3
Convert to per gram using molar mass:
1 mol CH 3OH
1454 kJ
kJ
22.7
2 mol CH 3OH
32.0 g
g
1.09 10 kJ 1 mol C8 H18
kJ
47.8
2 mol C8 H18
114.2 g
g
4
so octane is about 2x more effective