Thermochemistry II

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Thermochemistry II
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Thermochemical equations

• Coefficients represent MOLES of reactants and
  products.
• Amount of heat energy released or absorbed is
  proportional to the amounts of reactants and
  products
• Allows for the use of fractions as coefficients
• Phase designations must be included

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Molar Enthalpy of Formation

• ?H involved in the formation of 1 mole of a
  compound in its standard state (25C and 1atm)
• ?H0f indicates standard state enthalpy of
  formation
• ?H0f values are available in reference materials
• Stability of compounds
• A negative value indicates product is more stable
  than reactants.

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Molar Enthalpy of Combustion

• ?H involved in the combustion of 1 mole of a
  reactant in its standard state (25C and 1atm
• This equation is used by every food manufacturer
  to determine caloric content
• A bomb calorimeter measures heat transferred from
  the combustion reaction to a surrounding water
  jacket.

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Enthalpy Calculations- Hesss Law

• The total ?H in a reaction is the sum of ?H for
  each individual step.
• A process can be considered to occur in stages or
  steps, then the enthalpy change (?H) for the
  overall process can be obtained by summing the
  enthalpy changes of each individual step
• Energy changes are independent of pathway

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• Overall reaction C(s) O2(g) ? CO2(g)

DHtot DH1 DH2
-110.5kJ(-283kJ) -393.5 kJ
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• Determine the ?Hrx for the following reaction
• 3C(s) 4H2(g) ? C3H8(g) ?Hrx ?
• Given ?H
• (kJ/mol)
• 1. C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g) -2045
• 2. C(s) O2(g) ? CO2(g) -393.5
• 3. H2(g) ½ O2(g) ? H2O(g) -242

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General principles

• If a reaction is reversed, then the sign of ?H
  must be reversed.
• Multiply the coefficients of the known equations.
  When added we will end up with the desired
  equation. Multiply ?H by the same factor.

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• -1. 3CO2(g) 4H2O(g) ? C3H8(g)
  5O2(g) 2045
• 3?2. 3C(s) 3O2(g) ? 3CO2(g)
  3?(-393.5)
• 4?3. 4H2(g) 2O2(g) ? 4H2O(g) 4?
  (-242)
• 3C(s) 4H2(g) ? C3H8 (g)
• ?H rx 2045 kJ (-1180.5 kJ) (-968 kJ)
• -103.5 kJ

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Heat of Formation (DHf)

• The standard heat of formation of a compound,
  DHf, is the amount of heat absorbed or released
  when ONE mole of the compound is formed from its
  elements in their standard state.
• STANDARD STATE ()? 25C, 1 atm
• H2(g) ½ O2(g) ? H2O(l) DHf -286 kJ/mol
• H2(g) ½ O2(g) ? H2O(g) DHf -242 kJ/mol
• By definition the DHf of elements in their
  standard states is equal to ZERO. (O2(g) ?O2(g))
• DHf(Na(s)) 0
• DHf(Hg(s)) ? 0, because at 25C and 1 atm Hg is
  a liquid
• DHf(SO2(g)) ? 0, because SO2 is NOT an element
• DHf(I2(g)) ? 0, because at 25C and 1 atm I2 is
  a solid

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• If we know DHf of the reactants and products in
  a reaction, we can determine the
• DHreaction using the following equation
• DHreaction??HfPRODUCTS - ??HfREACTANTS

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• Given the following, calculate ?Hf? for the
  combustion of ethane,
• C2H6(g) 7/2 O2(g) ? 2CO2(g) 3H2O(l)

substance ?Hf? (kJ/mol)
CO2(g) -393.5
H2O(l) -286.0
C2H6(g) -84.7
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• DHreaction ??HfPRODUCTS - ?HfREACTANTS
• DHreaction sum of mol of products x
  ?HfPRODUCTS-
• sum of mol of reactants x HfREACTANTS
• 2?Hf?(CO2(g)) 3?Hf?(H2O(l))
  ?Hf?(C2H6(g)) 7/2 ?Hf?(O2(g))
• 2(-393.5 kJ) 3(-286 kJ)-(-84.7 kJ 7/2(0)
• -1560.3 kJ