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  [16H f (CO2 )  18H f ( H 2O) ]  [2H 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