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# Topic 5.3 and 5.4 Hess’s Law and Bond Enthalpies - PowerPoint PPT Presentation

Topic 5.3 and 5.4 Hess’s Law and Bond Enthalpies. Hess’s Law Topic 5.3. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (l )  H = -890 KJ. shows three different pathways: A  B A  C  B A  D  E  B enthalpy change from reactants to products for all of these is the same.

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Topic 5.3 and 5.4Hess’s Law and Bond Enthalpies

Hess’s LawTopic 5.3

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

• 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 products change for the net reaction (

energy in reactants

Problem-Solving Strategy change for the net reaction (

• 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 change for the net reaction (

• 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 change for the net reaction (

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

Example 3 change for the net reaction (

• 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)

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

13

Example 3 change for the net reaction (

• 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

2

2

- 572

1

Using enthalpy cycles instead… change for the net reaction (

• counter-clockwise reaction energy (-3,222) has to equal clockwise reaction energy (-3,267 + ΔH)

• answer is + 45 KJ/mol

Bond Enthalpies. Topic 5.4 change for the net reaction (

• 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

• 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 tendency to undergo chemical change than does one with weak bonds 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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Bond Enthalpy Calculations bonds are shown in the table below:

Example 1:

Calculate the enthalpy change for thereaction.

Is it endo or exothermic?

N2 + 3 H2 2 NH3

• 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)

• H-H

Example 2 bonds are shown in the table below:

energy

2H2 + O2

2H2O

course of reaction

Working out ∆H bonds are shown in the table below:

Show all the bonds in the reactants

H―H

+

O=O

energy

H―H

2H2O

course of reaction

O bonds are shown in the table below:

O

H

H

H

H

Working out ∆H

Show all the bonds in the products

H―H

+

O=O

energy

H―H

course of reaction

O bonds are shown in the table below:

O

H

H

H

H

Working out ∆H

Show the bond energies for all the bonds

436

+

O=O

energy

436

course of reaction

O bonds are shown in the table below:

O

H

H

H

H

Working out ∆H

Show the bond energies for all the bonds

436

+

498

energy

436

course of reaction

O bonds are shown in the table below:

H

H

Working out ∆H

Show the bond energies for all the bonds

436

+

498

energy

436

464

+

464

course of reaction

Working out ∆H bonds are shown in the table below:

Show the bond energies for all the bonds

436

+

498

energy

436

464

+

464

464

+

464

course of reaction

Working out ∆H bonds are shown in the table below:

Add the reactants’ bond energies together

energy

1370

464

+

464

464

+

464

course of reaction

Working out ∆H bonds are shown in the table below:

Add the products’ bond energies together

1370

energy

1856

course of reaction

Working out ∆H bonds are shown in the table below:

∆H = energy in ― energy out

1370

1856

+

-

energy

1370

1856

course of reaction

Working out ∆H bonds are shown in the table below:

∆H = energy in ― energy out

1370

1856

- 486

+

-

1370

energy

1856

course of reaction

Working out ∆H bonds are shown in the table below:

∆H = energy in ― energy out

∆H = -486

exothermic

1370

energy

1856

course of reaction