Transcript Hess’s Law

Hess’s Law
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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. 2
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) +
NH3:
O2 :
NO:
H2O:
O2(g)  2H2O(g)
H = -483.7 kJ
Reverse and x 2 4NH3  2N2 + 6H2 H = +183.6 kJ
Found in more than one place, SKIP IT (its hard).
x2
x3
2N2 + 2O2  4NO
6H2 + 3O2  6H2O
H = 361.2 kJ
H = -1451.1 kJ
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Goal:
4NH3(g) + 5O2(g) 
4NO(g) + 6H2O(g)
NH3: Reverse and x2 4NH3  2N2 + 6H2 H = +183.6 kJ
O2 : Found in more than one place, SKIP IT.
NO: x2
2N2 + 2O2  4NO
H = 361.2 kJ
H2O: 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 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
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Summary:
enthalpy is a state function
and is path independent.
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Standard Enthalpies of formation:
The standard enthalpy of formation is a measure of the energy
released or consumed when one mole of a substance is created
under standard conditions from its pure elements.
The symbol of the standard enthalpy of formation is ΔHof.
Δ = A change in enthalpy
o = A degree signifies that it's a standard enthalpy change.
f = The f indicates that the substance is formed from its elements
An important point to be made about the standard enthalpy
of formation is that when a pure element is in its standard
form its standard enthalpy formation is zero.
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The equation for the standard enthalpy change
of formation, shown below, is commonly used:
ΔHoreaction=∑ΔHof(products)−∑ΔHof(Reactants)
This equation essentially states that the
standard enthalpy change of formation is equal
to the sum of the standard enthalpies of
formation of the products minus the sum of
the standard enthalpies of formation of the
reactants.
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Given a simple chemical equation with the variables A, B and C representing
different compounds:
A+B⇋C
and the standard enthalpy of formation values:
•ΔHfo[A] = 433 KJ/mol
•ΔHfo[B] = -256 KJ/mol
•ΔHfo[C] = 523 KJ/mol
the equation for the standard enthalpy change of formation is as follows:
ΔHreactiono = ΔHfo[C] - (ΔHfo[A] + ΔHfo[B])
ΔHreactiono = (1 mol)(523 kJ/mol) - ((1 mol)(433 kJ/mol) + (1 mol)(-256 kJ/mol))
Because there is one mole each of A, B and C, the standard enthalpy of
formation of each reactant and product is multiplied by 1 mole, which
eliminates the mol denominator:
ΔHreactiono = 346 kJ
The result is 346 kJ, which is the standard enthalpy change of formation for the
creation of variable "C".
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Calculate the ΔHreactiono for the formation of NO2(g).
NO2(g) is formed from the combination of NO(g) and O2(g) in the following
reaction:
2NO(g)+O2(g)⇋2NO2(g)
To find the ΔHreactiono, use the formula for the standard enthalpy change of
formation:
ΔHoreaction=∑ΔHof(products)−∑ΔHof(Reactants)
The relevant standard enthalpy of formation values from the data table are as
follows:
•O2(g): 0 kJ/mol
•NO(g): 90.25 kJ/mol
•NO2(g): 33.18 kJ/mol
Plugging these values into the formula above gives the following:
ΔHreactiono= (2 mol)(33.18 kJ/mol) - ((2 mol)(90.25 kJ/mol) + (1 mol)(0 kJ/mol))
ΔHreactiono =-114.14kJ
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Homework:
Page 324 #1-6
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Thermodynamic Quantities of Selected Substances @ 298.15 K
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