Topic 5.3 and 5.4Hess’s Law and Bond Enthalpies

Hess’s LawTopic 5.3

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) = -890 H KJ

• shows three different pathways:A BA C BA D E B– enthalpy change

from reactants to products for all of these is the same

• if a series of reactions are added together, the enthalpy change for the net reaction (Hfinal) will be the sum of the enthalpy change for the individual reactions (Hind + Hind + Hind ….) – the change in enthalpy is the same whether the

reaction takes place in one step, or in a series of steps

– H is independent of the reaction pathway– depends only on the difference between the

enthalpy of the products and the reactants • H = Hproducts − Hreactants

• provides a way to calculate enthalpy changes even when the reaction cannot be performed directly

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energy in reactants

energy in products

Problem-Solving Strategy– work backwards from the final reaction, using the

reactants and products to decide how to manipulate the other given reactions at your disposal

– if a reaction is reversed the sign on ΔH is reversed– N2 (g) + 2O2 (g) → 2NO2 (g) ΔH = 68kJ

– 2NO2 (g) → N2 (g) + 2O2 (g) ΔH = - 68kJ

– multiply reactions to give the correct numbers of reactants and products in order to get the final reaction.

– the value of Δ H is also multiplied by the same integer– identical substances found on both sides of the summed

equation cancel each other out

Example 1

• Given:N2 (g) + O2 (g) 2 NO (g) DH1 = +181 kJ

2 NO (g) + O2 (g) 2 NO2 (g) DH2 = -113 kJ • Find the enthalpy change for:

N2 (g) + 2 O2 (g) 2 NO2 (g)

• DH = DH1 + DH2 = +181 kJ + (-113 kJ) = + 68 kJ

Example 2

2 2 3

2 2 2

2 N ( ) 6 H ( ) 4 NH ( ) H = 184 kJ

6 H O( ) 6 H ( ) 3 O ( ) H = +1452 kJ

g g g

g g g

2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g

ΔH = (- 184 kJ) + (+ 1452 kJ) = + 1268 kJ

Example 3• Given:

C (s) + O2 (g) CO2 (g) DH1 = - 393 kJ mol-1

H2 (g) + ½O2 (g) H2O (g) DH2 = - 286 kJ mol-1

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) DH3 = - 890 kJ mol-1

• Find the enthalpy change for: C (s) + 2H2 (g) CH4 (g)

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This equation needs to be “flipped”. The CH4 is on the wrong side of the equation

Example 3• Given:

C (s) + O2 (g) CO2 (g) DH1 = - 393 kJ mol-1

H2 (g) + ½O2 (g) H2O (g) DH2 = - 286 kJ mol-1

CO2 (g) + 2H2O (g) CH4 (g) + 2O2 (g) DH3 = + 890 kJ mol-1

• Find the enthalpy change for: C (s) + 2H2 (g) CH4 (g)

• (- 393 kJ mol-1) + (- 572 kJ mol-1 ) + (+ 890 kJ mol-1 )= - 75 kJ mol-1

12 2 - 572

Using enthalpy cycles instead…

• counter-clockwise reaction energy (-3,222) has to equal clockwise reaction energy (-3,267 + ΔH)– answer is + 45 KJ/mol

• clockwise needs to equal counter-clockwise• -109 = 52.2 + (-92.3) + ΔH• ΔH = - 68.9 kJ mol-1

• can be used to calculate the enthalpy change for a chemical reaction if we know the energy necessary to break or form bonds in the gaseous state– breaking bonds

• energy is required so enthalpy is positive (endothermic)• the molecule was stable so energy was necessary to break

apart the molecule

– forming bonds• energy is released so enthalpy is negative (exothermic)• the new molecule is more stable than the individual

atoms so energy is released

Bond Enthalpies. Topic 5.4

• a molecule with strong chemical bonds generally has less tendency to undergo chemical change than does one with weak bonds – SiO bonds are among the strongest ones that silicon forms

• it is not surprising that SiO2 and other substances containing SiO bonds (silicates) are so common

• it is estimated that over 90 percent of Earth's crust is composed of SiO2 and silicates

• we use average bond enthalpies– again, in the gaseous state– different amount of energy can be required to

break the same bond• example- methane, CH4

– if you took methane to pieces, one hydrogen at a time, it needs a different amount of energy to break each of the four C-H bonds

– every time you break a hydrogen off the carbon, the environment of those left behind changes, and the strength of the remaining bonds is affected

– therefore, the 412 kJ mol-1 needed to break C-H is just an average value, may vary, and is not very accurate

The average bond enthalpies for several types of chemical bonds are shown in the table below:

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• Bonds broken• 1 N N = 945 kJ• 3 H-H 3(435) = 1305 kJ• Total = 2250 kJ

• Bonds formed• 2x3 = 6 N-H: 6 (390) = - 2340 kJ

• Net enthalpy change• = (+ 2250) + (- 2340) = - 90 kJ (exothermic)

Calculate the enthalpy change for the reaction. Is it endo or exothermic?

N2 + 3 H2 2 NH3

Bond Enthalpy CalculationsExample 1:

H-H

Example 2

energy

course of reaction

2H2 + O2

2H2O

Working out ∆HShow all the bonds in the reactants

energy

course of reaction

2H2O

H―HH―H

O=O+

Working out ∆HShow all the bonds in the products

energy

course of reaction

H―HH―H

O=O+

H HO

H HO

Working out ∆HShow the bond energies for all the bonds

energy

course of reaction

436436

O=O+

H HO

H HO

Working out ∆HShow the bond energies for all the bonds

energy

course of reaction

436436

498+

H HO

H HO

Working out ∆HShow the bond energies for all the bonds

energy

course of reaction

436436

498+

H HO

464 464+

Working out ∆HShow the bond energies for all the bonds

energy

course of reaction

436436

498+

464 464+

464 464+

Working out ∆H

Add the reactants’ bond energies together

energy

course of reaction

464 464+

464 464+

1370

Working out ∆H

Add the products’ bond energies together

energy

course of reaction

1370

1856

Working out ∆H∆H = energy in ― energy out

energy

course of reaction

1370

1856

13701856-

+

Working out ∆H

∆H = energy in ― energy out

energy

course of reaction

1370

1856

13701856- 486

-+

Working out ∆H

∆H = energy in ― energy out

energy

course of reaction

1370

1856

∆H = -486exothermic