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Hess’ Law Starter: Why is it easy to measure enthalpies of combustion but often difficult to measure enthalpies of formation?

2.3 8 Hess' Law

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Page 1: 2.3 8 Hess' Law

Hess’ Law

Starter: Why is it easy to measure

enthalpies of combustion but often

difficult to measure enthalpies of

formation?

Page 2: 2.3 8 Hess' Law

Problem

• How would you measure the enthalpy

change for the formation of methane from

graphite and hydrogen gas?

– problem: graphite doesn’t react with

hydrogen gas to form methane

• C(graphite) + 2H2(g) CH4(g)

Page 3: 2.3 8 Hess' Law

Hess’s Law

StartFinish

Both lines accomplished the same result,

they went from start to finish.

Net result = same.

Page 4: 2.3 8 Hess' Law

Hess’s Law

• When reactants are converted to products,

the change in enthalpy is the same

whether the reaction takes place in one

step or a series of steps.

Page 5: 2.3 8 Hess' Law

Remember

• P4(s) + 3O2(g) P4O6(s) ΔH = -1640.1 kJ

• Transform the equation so that P4O6(s) is on the

reactant side.

• P4O6(s) P4(s) + 3O2(g) -ΔH = 1640.1 kJ

• Note: if the equation switches, the sign of ΔH

does as well.

Page 6: 2.3 8 Hess' Law

Example

• What is the change in enthalpy for the

formation of methane from graphite and

hydrogen gas?

– problem: graphite doesn’t react with

hydrogen gas to form methane

• C(graphite) + 2H2(g) CH4(g)

Page 7: 2.3 8 Hess' Law

Solution

• Find reactions that do occur that link

graphite to methane through a series of

steps.

Page 8: 2.3 8 Hess' Law

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 formation

Page 9: 2.3 8 Hess' Law

Calculate the enthalpy change for the hydrogenation of ethene

Enthalpy of reaction from bond enthalpies

Page 10: 2.3 8 Hess' Law

Calculate the enthalpy change for the hydrogenation of ethene

Enthalpy of reaction from bond enthalpies

DH2 1 x C=C bond @ 611 = 611 kJ

4 x C-H bonds @ 413 = 1652 kJ

1 x H-H bond @ 436 = 436 kJ

Total energy to break bonds of reactants = 2699 kJ

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Calculate the enthalpy change for the hydrogenation of ethene

Enthalpy of reaction from bond enthalpies

DH2 1 x C=C bond @ 611 = 611 kJ

4 x C-H bonds @ 413 = 1652 kJ

1 x H-H bond @ 436 = 436 kJ

Total energy to break bonds of reactants = 2699 kJ

DH3 1 x C-C bond @ 346 = 346 kJ

6 x C-H bonds @ 413 = 2478 kJ

Total energy to break bonds of products = 2824 kJ

Applying Hess’s Law DH1 = DH2 – DH3 = (2699 – 2824) = – 125 kJ

Page 12: 2.3 8 Hess' Law

Sample calculation

Calculate the standard enthalpy change for the following reaction, given that the

standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286,

+33 and -173 kJ mol-1 respectively; the value for oxygen is ZERO as it is an element

2H2O(l) + 4NO2(g) + O2(g) ———> 4HNO3(l)

By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be...

PRODUCTS REACTANTS

[ 4 x DHf of HNO3 ] minus [ (2 x DHf of H2O) + (4 x DHf of NO2) + (1 x DHf of O2) ]

DH°r = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0

ANSWER = - 252 kJ

Enthalpy of reaction from enthalpies of formation

DH = DHf of products – DHf of reactants

Page 13: 2.3 8 Hess' Law

Enthalpy of reaction from enthalpies of combustion

If you burned all the products you should get the same amounts of oxidation products

such a CO2 and H2O as if you burned the reactants.

Enthalpy of combustion is an exothermic process

Page 14: 2.3 8 Hess' Law

Enthalpy of reaction from enthalpies of combustion

Sample calculation

Calculate the standard enthalpy of formation of methane; the standard enthalpies of

combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1 .

C(graphite) + 2H2(g) ———> CH4(g)

By applying Hess’s Law ... The Standard Enthalpy of Reaction DH°r will be...

REACTANTS PRODUCTS

[ (1 x DHc of C) + (2 x DHc of H2) ] minus [ 1 x DHc of CH4]

DH°r = 1 x (-394) + 2 x (-286) - 1 x (-890)

ANSWER = - 76 kJ mol-1

DH = DHc of reactants – DHc of products

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Determine the heat of reaction for the reaction:

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Using the following sets of reactions:

N2(g) + O2(g) 2NO(g) DH = 180.6 kJ

N2(g) + 3H2(g) 2NH3(g) DH = -91.8 kJ

2H2(g) + O2(g) 2H2O(g) DH = -483.7 kJ

Page 16: 2.3 8 Hess' Law

Determine the heat of reaction for the reaction:

C2H4(g) + H2(g) C2H6(g)

Use the following reactions:

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) DH = -1401 kJ

C2H6(g) + 3.5 O2(g) 2CO2(g) + 3H2O(l) DH = -1550 kJ

H2(g) + 0.5 O2(g) H2O(l) DH = -286 kJ

Consult your neighbor if necessary.