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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 5.4 Hess’s Law and Bond Enthalpies

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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


  • 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


  • 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

8


energy in products

energy in reactants


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

    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)

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

    13


    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

    2

    2

    - 572

    1


    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


    Bond Enthalpies. Topic 5.4

    • 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 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:

    20


    Bond Enthalpy Calculations

    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

    energy

    2H2 + O2

    2H2O

    course of reaction


    Working out ∆H

    Show all the bonds in the reactants

    H―H

    +

    O=O

    energy

    H―H

    2H2O

    course of reaction


    O

    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

    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

    O

    H

    H

    H

    H

    Working out ∆H

    Show the bond energies for all the bonds

    436

    +

    498

    energy

    436

    course of reaction


    O

    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

    Show the bond energies for all the bonds

    436

    +

    498

    energy

    436

    464

    +

    464

    464

    +

    464

    course of reaction


    Working out ∆H

    Add the reactants’ bond energies together

    energy

    1370

    464

    +

    464

    464

    +

    464

    course of reaction


    Working out ∆H

    Add the products’ bond energies together

    1370

    energy

    1856

    course of reaction


    Working out ∆H

    ∆H = energy in ― energy out

    1370

    1856

    +

    -

    energy

    1370

    1856

    course of reaction


    Working out ∆H

    ∆H = energy in ― energy out

    1370

    1856

    - 486

    +

    -

    1370

    energy

    1856

    course of reaction


    Working out ∆H

    ∆H = energy in ― energy out

    ∆H = -486

    exothermic

    1370

    energy

    1856

    course of reaction


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