Hess`s Law PPT
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Transcript Hess`s Law PPT
the heat evolved or absorbed in a chemical process is the
same whether the process takes place in one or several
steps.
if two or more chemical equations can be added
together to produce an overall equation, the sum of the
enthalpy equals the enthalpy change of the overall
equation.
This is called the Heat of Summation, ∆H
Analogy for Hess's Law
There is an old Chinese proverb which
says: There are many ways to the top of
a mountain, but the view from the top
is always the same.
Develop an analogy for soccer
and scoring a goal.
Develop an analogy for soccer
and scoring a goal.
Hess’s Law
Read through the whole question
Plan a Strategy
Evaluate the given equations.
Rearrange and manipulate the equations so that they
will produce the overall equation.
Add the enthalpy terms.
Example 1
H2O(g) + C(s) → CO(g) + H2(g)
Use these equations to calculate the molar enthalpy
change which produces hydrogen gas.
C(s) + ½ O2(g) → CO(g)
H2(g) + ½ O2(g) → H2O(g)
∆H = -110.5kJ
∆H = -241.8kJ
H2O(g) + C(s) → CO(g) + H2(g)
Use these equations to calculate the molar enthalpy
change which produces hydrogen gas.
C(s) + ½ O2(g) → CO(g)
∆H = -110.5kJ
H2O(g) → H2(g) + ½ O2(g) ∆H = +241.8kJ
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C(s) + H2O(g) → H2(g) + CO(g) ∆H=+131.3kJ
Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change
which produces butane gas.
C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= -2657.4kJ/mol
C(s) + O2(g) → CO2(g)
∆H= -393.5kJ/mol
H2(g) + ½O2(g) → H2O(g)
∆H= -241.8kJ/mol
Read through the whole question
Plan a Strategy
Evaluate the given equations.
Rearrange and manipulate the equations so that they will produce the overall
equation.
Add the enthalpy terms. REWRITE THE CHANGES.
Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change
which produces butane gas.
C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= -2657.4kJ/mol
C(s) + O2(g) → CO2(g)
∆H= -393.5kJ/mol
H2(g) + ½O2(g) → H2O(g)
∆H= -241.8kJ/mol
Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change
which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol
C(s) + O2(g) → CO2(g)
∆H= -393.5kJ/mol
H2(g) + ½O2(g) → H2O(g)
∆H= -241.8kJ/mol
Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change
which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol
4(C(s) + O2(g) → CO2(g))
∆H= 4(-393.5kJ/mol)
H2(g) + ½O2(g) → H2O(g)
∆H= -241.8kJ/mol
Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change
which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol
4C(s) + 4O2(g) → 4CO2(g) distribute the 4
∆H= 4(-393.5kJ/mol)
H2(g) + ½O2(g) → H2O(g)
∆H= -241.8kJ/mol
Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change
which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol
4C(s) + 4O2(g) → 4CO2(g) distribute the 4
∆H= 4(-393.5kJ/mol)
5(H2(g) + ½O2(g) → H2O(g)) distribute the 5
∆H= 5(-241.8kJ/mol)
Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change
which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol
4C(s) + 4O2(g) → 4CO2(g)
∆H= 4(-393.5kJ/mol)
5H2(g) + 2½O2(g) → 5H2O(g)
∆H= 5(-241.8kJ/mol)
Example 2
4C(s) + 5H2(g) → C4H10(g)
Use these equations to calculate the molar enthalpy change
which produces butane gas.
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol
4C(s) + 4O2(g) → 4CO2(g)
∆H= 4(-393.5kJ/mol)
5H2(g) + 2½O2(g) → 5H2O(g)
∆H= 5(-241.8kJ/mol)
_____________________________________________________
∆H = -125.6kJ/mol