The document discusses Hess's law, which states that the heat of reaction is the same whether a chemical process occurs in one or multiple steps. Specifically:
- Hess's law allows adding together multiple chemical equations to determine the enthalpy change of the overall equation.
- Two examples are provided to demonstrate calculating the enthalpy change of an overall reaction by combining individual reaction enthalpies.
- In both examples, the individual reactions are rearranged and combined to produce the overall reaction, and the enthalpy terms are summed to find the enthalpy change of the overall reaction.
Organic Name Reactions for the students and aspirants of Chemistry12th.pptx
Hess's law
1. the heat evolved or absorbed in a chemical process is
the same whether the process takes place in one or
several steps.
2. 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
3. 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.
6. 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.
7. 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
H2(g) + ½ O2(g) → H2O(g) ∆H = -241.8kJ
Example 1
8. 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
9. 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
10. 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
11. 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
12. 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
13. 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
14. 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
15. 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
16. 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)
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∆H = -125.6kJ/mol
Example 2