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 (ΔHf
0) = 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 ΔHf
0 = 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 ΔHf
0 values for each molecule
• (ΔH0) for the reaction 2NO + O2 à 2NO2 – ΔHf
0 for NO = 90.37 kJ/mol – ΔHf
0 for O2 = 0 (pure element) – ΔHf
0 for NO2 = 33.85 kJ/mol
Ex/1 • 3rd: Solve for ΔHf
0 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
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