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Lesson
Hess’s Law
IB Chemistry Power Points
Topic 05
Energetics
www.pedagogics.ca
Great thanks to
JONATHAN HOPTON & KNOCKHARDY PUBLISHING
www.knockhardy.org.uk/sci.htm
Some taken from
ENTHALPY
CHANGES
“The overall enthalpy change of a chemical process is
independent of the path taken”
The enthalpy change going from A to B can be found by
adding the values of the enthalpy changes for the reactions
A to X, X to Y and Y to B.
ΔHr = ΔH1 + ΔH2 + ΔH3
HESS’S LAW
Dissolving solid sodium
hydroxide in water
This process produces sodium and
hydroxide ions ie. NaOH (aq)
solution.
1. NaOH(s) + H2O  NaOH (aq) + H2O ΔH1
Consider three reactions
Reacting the sodium hydroxide
solution with a hydrochloric acid
solution = neutralization
Na+ OH-
2. NaOH (aq) +HCl (aq)  NaCl (aq) + H2O ΔH2
Alternatively - add solid
sodium hydroxide directly to
hydrochloric acid solution.
H3O+
Cl-
3. NaOH(s) + HCl (aq)  NaCl (aq) + H2O ΔH3
Recap:
1. NaOH(s) + H2O  NaOH (aq) + H2O ΔH1
2. NaOH (aq) +HCl (aq)  NaCl (aq) + H2O ΔH2
3. NaOH(s) + HCl (aq)  NaCl (aq) + H2O ΔH3
Show that equation 1 plus equation 2 is the same as equation 3
What conclusion about enthalpy can be made?
Equation 1 + 2
1. NaOH(s) + H2O  NaOH (aq) + H2O ΔH1
2. NaOH (aq) +HCl (aq)  NaCl (aq) + H2O ΔH2
3. NaOH(s) + HCl (aq)  NaCl (aq) + H2O ΔH3
Combined – equations 1 and 2 give equation 3
Since the equation 1 then 2 sequence starts and
finishes with the same reactants and products as
equation 3, Hess’s Law says ΔH3 = ΔH1 + ΔH2
Represented as an enthalpy level diagram
NaOH (aq) +HCl (aq)
ΔH3
NaOH (s) reactants + HCl (aq)
products NaCl (aq) + H2O
ΔH1
ΔH2
+ H2O - 42 kJ/mol
- 57 kJ/mol
- 99 kJ/mol ΔH3
ΔH1
ΔH2
Represented as an enthalpy cycle
+ H2O
NaOH (aq)
ΔH3
NaOH (s) + HCl NaCl (aq) + H2O
ΔH1
+ HCl
ΔH2

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2016 topic 5.2 hess's law

  • 1. Lesson Hess’s Law IB Chemistry Power Points Topic 05 Energetics www.pedagogics.ca
  • 2. Great thanks to JONATHAN HOPTON & KNOCKHARDY PUBLISHING www.knockhardy.org.uk/sci.htm Some taken from ENTHALPY CHANGES
  • 3. “The overall enthalpy change of a chemical process is independent of the path taken” The enthalpy change going from A to B can be found by adding the values of the enthalpy changes for the reactions A to X, X to Y and Y to B. ΔHr = ΔH1 + ΔH2 + ΔH3 HESS’S LAW
  • 4. Dissolving solid sodium hydroxide in water This process produces sodium and hydroxide ions ie. NaOH (aq) solution. 1. NaOH(s) + H2O  NaOH (aq) + H2O ΔH1 Consider three reactions
  • 5. Reacting the sodium hydroxide solution with a hydrochloric acid solution = neutralization Na+ OH- 2. NaOH (aq) +HCl (aq)  NaCl (aq) + H2O ΔH2
  • 6. Alternatively - add solid sodium hydroxide directly to hydrochloric acid solution. H3O+ Cl- 3. NaOH(s) + HCl (aq)  NaCl (aq) + H2O ΔH3
  • 7. Recap: 1. NaOH(s) + H2O  NaOH (aq) + H2O ΔH1 2. NaOH (aq) +HCl (aq)  NaCl (aq) + H2O ΔH2 3. NaOH(s) + HCl (aq)  NaCl (aq) + H2O ΔH3 Show that equation 1 plus equation 2 is the same as equation 3 What conclusion about enthalpy can be made?
  • 8. Equation 1 + 2 1. NaOH(s) + H2O  NaOH (aq) + H2O ΔH1 2. NaOH (aq) +HCl (aq)  NaCl (aq) + H2O ΔH2 3. NaOH(s) + HCl (aq)  NaCl (aq) + H2O ΔH3 Combined – equations 1 and 2 give equation 3 Since the equation 1 then 2 sequence starts and finishes with the same reactants and products as equation 3, Hess’s Law says ΔH3 = ΔH1 + ΔH2
  • 9. Represented as an enthalpy level diagram NaOH (aq) +HCl (aq) ΔH3 NaOH (s) reactants + HCl (aq) products NaCl (aq) + H2O ΔH1 ΔH2 + H2O - 42 kJ/mol - 57 kJ/mol - 99 kJ/mol ΔH3 ΔH1 ΔH2
  • 10. Represented as an enthalpy cycle + H2O NaOH (aq) ΔH3 NaOH (s) + HCl NaCl (aq) + H2O ΔH1 + HCl ΔH2