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Hess

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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. – PowerPoint PPT presentation

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Title: Hess


1
Hesss Law
Start
Finish
A State Function Path independent.
Both lines accomplished the same result, they
went from start to finish. Net
result same.
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
Goal
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
4NH3 ? 2N2 6H2 ?H 183.6 kJ
NH3
Reverse and x 2
Found in more than one place, SKIP IT (its
hard).
O2
NO
x2
2N2 2O2 ? 4NO ?H 361.2 kJ
H2O
x3
6H2 3O2 ? 6H2O ?H -1451.1 kJ
4
Goal
4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g)
4NH3 ? 2N2 6H2 ?H 183.6 kJ
NH3
Reverse and x2
Found in more than one place, SKIP IT.
O2
NO
x2
2N2 2O2 ? 4NO ?H 361.2 kJ
H2O
x3
6H2 3O2 ? 6H2O ?H -1451.1 kJ
Cancel terms and take sum.
5O2
6H2O
?H -906.3 kJ
?
4NH3
4NO
Is the reaction endothermic or exothermic?
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
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
7
Summary enthalpy is a state function and is path
independent.
8
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