Law for States - International Law, Constitutional Law, Public Law
Hess's law
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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 =
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Transcript of Hess's law
1
Hess’s Law
StartFinish
A State Function: Path independent. Both lines accomplished the same result, they went from start to finish. Net result = same.
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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.
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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
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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?
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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.
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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 kJC2H6(g) : rev #2 2CO2(g) + 3H2O(l) C2H6(g) + 7/2O2(g) H = +1550 kJ
C2H4(g) + H2(g) C2H6(g) H = -137 kJ
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Summary:enthalpy is a state functionand is path independent.
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Standard Enthalpies of formation:
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Thermodynamic Quantities of Selected Substances @ 298.15 K
