Bond Enthalpy

L.O.:Explain exothermic and endothermic reactions in terms of enthalpy changes associated with the breaking and making of chemical bonds. Define and use the term average bond enthalpy. Calculate an enthalpy change of reaction from average bond enthalpies.

• H-H (g) →H(g) + H(g) ∆H = + 436 KJ mol-1

The bond enthalpy is the change that takes place when breaking one mole of a given bond in the molecules of a gaseous species.

Average bond enthalpy is the average enthalpy change that takes place when breaking by homolytic fission 1 mole of a given type of bond in the molecules of a gaseous species.

Theory Imagine that, during a reaction, all the bonds of reacting species are brokenand the individual atoms join up again but in the form of products. Theoverall energy change will depend on the difference between the energyrequired 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 enthalpiesEnthalpy of reaction from bond enthalpies

Step 1 Energy is put in to break bonds to form separate, gaseous atoms

Step 2 The gaseous atoms then combine to form bonds and energy is releasedits value will be equal and opposite to that of breaking the bonds

Applying Hess’s Law r = Step 1 + Step 2

H = ∑ (bond enthalpies of bonds broken) - ∑ (bond enthalpies of bonds made)

H = Σ(bonds broken) − Σ(bonds formed)

(worth 1 mark)

Page 134. Q1-3

Enthalpy of reaction from bond enthalpiesEnthalpy of reaction from bond enthalpies

= bond enthalpies – bond enthalpies of reactants of products

= bond enthalpies – bond enthalpies of reactants of products

Alternative view

Step 1 Energy is put in to break bonds to form separate, gaseous atoms.

Step 2 Gaseous atoms then combineto form bonds and energy is released; its value will be equaland opposite to that of breaking the bonds

r = Step 1 - Step 2

Because, in Step 2 the route involvesgoing in the OPPOSITE DIRECTIONto the defined change of bond enthalpy,it’s value is subtracted.

SUM OFTHE BOND ENTHALPIES OF THE REACTANTS

REACTANTS

PRODUCTS

ATOMS

SUM OFTHE BOND ENTHALPIES OF THE PRODUCTS

Calculate the enthalpy change for the hydrogenation of ethene

Enthalpy of reaction from bond enthalpiesEnthalpy of reaction from bond enthalpies

Calculate the enthalpy change for the hydrogenation of ethene

Enthalpy of reaction from bond enthalpiesEnthalpy of reaction from bond enthalpies

2 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 of reactants = 2699 kJ

Calculate the enthalpy change for the hydrogenation of ethene

Enthalpy of reaction from bond enthalpiesEnthalpy of reaction from bond enthalpies

2 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 of reactants = 2699 kJ

3 1 x C-C bond @ 346 = 346 kJ6 x C-H bonds @ 413 = 2478 kJ

Total energy to break bonds of products = 2824 kJ

Applying Hess’s Law 1 = 2 – 3 = (2699 – 2824) = – 125 kJ

If you formed the products from their elements you should need the same amounts of every substance as if you formed the reactants from their elements.

Enthalpy of formation tends to be an exothermic process

Enthalpy of reaction from enthalpies of formationEnthalpy of reaction from enthalpies of formation

Step 1 Energy is released as reactantsare formed from their elements.

Step 2 Energy is released as productsare formed from their elements.

r = - Step 1 + Step 2

or Step 2 - Step 1

In Step 1 the route involves going in the OPPOSITE DIRECTION to the defined enthalpy change, it’s value is subtracted.

SUM OFTHE ENTHALPIES OF FORMATION OF

THE REACTANTS

REACTANTS

PRODUCTS

ELEMENTS

SUM OFTHE ENTHALPIES OF FORMATION OF THE PRODUCTS

Enthalpy of reaction from enthalpies of formationEnthalpy of reaction from enthalpies of formation

= f of products – f of reactants = f of products – f of reactants