Ch. 6: Thermochemistry - Ms....

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Transcript of Ch. 6: Thermochemistry - Ms....

  • Ch.6:Thermochemistry

  • Enthalpy

  • Enthalpy•  Enthalpyofformation(ΔHf)=HeatabsorbedorreleasedwhenONEmoleofcompoundisformedfromelementsintheirstandardstatesinkJ/mol– ΔHf=0forelementsinstandardstates

  • Hess’s Law •  Sometimes we cannot directly measure

    heat •  Hess’s Law allows us to determine the

    overall heat of reaction by summing the parts

  • Hess’sLaw•  Whengivenasetofreactantsandproducts,thechangeinenthalpyisthesamenomatterthenumberofstepsinbetween.

    •  Ifthereactionisreversed,thesignofΔHisalsoreversed.

    •  TheΔHisalsodirectlyproportionaltotheamountofproductsandreactants.Thereforeifyoumultiplythecoefficientsbyafactor,theΔHmustbemultipliedbythissamefactor.

  • Hess’sLaw•  AconsequenceofHess’sLawisthatreactionscanbeaddedjustlikealgebraicequations

  • Example 1 C(diamond) à C(graphite)

    Find ΔH. Is the rxn exo or endothermic? •  C(graphite) + O2 à CO2 ΔH = -393.5 kJ •  C(diamond) + O2 à CO2 ΔH = -395.4 kJ

  • Step 1 C(diamond) à C(graphite)

    •  C(graphite) + O2 à CO2 ΔH = -393.5 kJ •  C(diamond) + O2 à CO2 ΔH = -395.4 kJ

    •  If needed, flip eqs (ΔH changes sign)

  • Step 2 C(diamond) à C(graphite)

    •  C(graphite) + O2 à CO2 ΔH = -393.5 kJ •  C(diamond) + O2 à CO2 ΔH = -395.4 kJ •  If needed, multiply eqs by any factor

    (ΔH as well)

  • Step 3 C(diamond) à C(graphite)

    •  C(graphite) + O2 à CO2 ΔH = -393.5 kJ •  C(diamond) + O2 à CO2 ΔH = -395.4 kJ •  Combine!

    – Same side = add together – Opposite side = subtract

  • Ex/ Calculate ΔH for XeF2 + F2 à XeF4 •  Xe + F2 à XeF2 ΔH = -123 kJ •  Xe + 2F2 à XeF4 ΔH = -262 kJ

  • Ex/ Calculate ΔH for PCl3 + Cl2 à PCl5 •  2P + 3Cl2 à 2PCl3 ΔH = -640 kJ •  2P + 5Cl2 à 2PCl5 ΔH = -886 kJ

  • Ex/ Calculate ΔH for 2F2 + 2H2O à 4HF + O2

    •  H2 + F2 à 2HF ΔH = -542.2 kJ •  2H2 + O2 à 2H2O ΔH = -571.6 kJ

  • StandardEnthalpiesofFormation

  • Standard Enthalpies of Formation •  Standard Enthalpies of Formation (ΔHf0) = ΔH of the formation of one mole of a compound from its elements in their standard states.

  • Standard Heats of Formation •  Pure substances at standard state have a ΔHf0 = 0. – Ex/ Br2, O2, Fe, etc. – See Appendix 3 in textbook for ΔHf values

  • Ex/1 •  Calculate the standard heat of reaction

    (ΔH0) for the reaction 2NO + O2 à 2NO2

  • Ex/1 •  1st: Write the balanced equation (if not

    already given) •  (ΔH0) for the reaction 2NO + O2 à 2NO2

  • Ex/1 •  2nd: Find the ΔHf0 values for each

    molecule •  (ΔH0) for the reaction 2NO + O2 à 2NO2

    – ΔHf0 for NO = 90.37 kJ/mol – ΔHf0 for O2 = 0 (pure element) – ΔHf0 for NO2 = 33.85 kJ/mol

  • Ex/1 •  3rd: Solve for ΔHf0 products + reactants •  (ΔH0) for the reaction 2NO + O2 à 2NO2

    – ΔHf0 for NO = 90.4 kJ/mol – ΔHf0 for O2 = 0 (pure element) – ΔHf0 for NO2 = 33.85 kJ/mol

  • Ex/1 •  4th: Plug into formula

  • Ex/2: You Try •  Calculate the standard heat of reaction

    (ΔH0) for the reaction CaCO3 à CaO + CO2

  • Ex/3: You Try •  Calculate the standard heat of reaction

    (ΔH0) for the reaction 7O2 + 4NH3 à 4NO2 + 6H2O(g)

  • •  Giventhefollowingequation,whichofthefollowingstatement(s)is(are)true?

    S(s)+O2(g)àSO2(g) ΔH=-296kJ

    I.  ThereactionisexothermicII.  When0.500molsulfurisreacted,148kJofenergyis

    releasedIII.  When32.0gofsulfurisburned,2.96×105Jofenergy

    isreleaseda.  Allaretrueb.  Noneistruec.  IandIIaretrued.  IandIIIaretruee.  OnlyIIistrue

  • SampleQuestion•  Giventhefollowingequation,whichofthefollowingstatement(s)is(are)true?

    C(s)+2H2(g)àCH4(g)ΔH=-75kJ/molI.  ThereactionisendothermicII.  ThereversereactionwillhaveaΔHf=+75kJ/molIII.  TheenthalpyforH2iszeroa.  Allaretrueb.  Noneistruec.  IandIIaretrued.  IIandIIIaretruee.  OnlyIIistrue

  • Notes: •  Remember – we can always incorporate

    stoichiometry into these problems •  Don’t forget related concepts that we have

    already discussed •  Be able to explain the concepts