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