5 3 and 5 4
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Transcript of 5 3 and 5 4
Lesson Hess’s Law
Bond Enthalpies
IB Chemistry Power Points
Topic 05Energetics
www.pedagogics.ca
Great thanks toJONATHAN HOPTON & KNOCKHARDY PUBLISHING
www.knockhardy.org.uk/sci.htm
Some taken from
ENTHALPYCHANGES
“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 = ΔH
1 + ΔH
2 + ΔH
3
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-
H3 O + Cl -
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?
Represented as an enthalpy level diagram
NaOH (aq) +HCl (aq) ΔH
3
NaOH (s) + HCl (aq)
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
Theory Imagine that, during a reaction, all the bonds of reacting species are broken
and the individual atoms join up again but in the form of products. The
overall energy change will depend on the difference between the energy
required to break the bonds and that released as bonds are made.
energy released making bonds > energy used to break bonds ... EXOTHERMIC energy used to break bonds > energy released making bonds ... ENDOTHERMIC
Enthalpy of reaction from bond enthalpies
Step 1 Energy is put in to break bonds to form separate, gaseous atomsStep 2 The gaseous atoms then combine to form bonds and energy is released
its value will be equal and opposite to that of breaking the bonds
Applying Hess’s Law ΔHr = Step 1 + Step 2
Calculate the enthalpy change for the hydrogenation of ethene
Enthalpy of reaction from bond enthalpies
DH2
1 x C=C bond @ 611 = 611 kJ4 x C-H bonds @ 413 = 1652 kJ1 x H-H bond @ 436 = 436 kJ
Total energy to break bonds (reactants) = 2699 kJ
DH3
1 x C-C bond @ 346 = 346 kJ6 x C-H bonds @ 413 = 2478 kJ
Total energy to break bonds (products) = 2824 kJ
DH = bonds broken – bonds made = (2699 – 2824) = – 125 kJ