Lecture 5: Standard Enthalpies

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Lecture 5 Standard Enthalpies

• Reading Zumdahl 9.6
• Outline
• What is a standard enthalpy?
• Defining standard states.
• Using standard enthalpies to determine DHrxn

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Definition of DHf

• We saw in the previous lecture the utility of
  having a series of reactions with known enthalpy
  changes.

• What if one could deal with combinations
• of compounds directly, rather than dealing
• with whole reactions, in order to determine
  DHrxn?

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Definition of DHf

• Standard Enthalpy of Formation, DHf
• The change in enthalpy that accompanies the
  formation of 1 mole of a compound from its
  elements with all substances at standard state.
• We envision taking elements at their standard
  state, and combining them to make compounds, also
  at standard state.

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What is Standard State?

• Standard State A precisely defined reference
  state. It is a common reference point that one
  can use to compare thermodynamic properties.
• Definitions of Standard State
• For a gas P 1 atm.
• For solutions 1 M (mol/L).
• For liquids and solids pure liquid or solid
• For elements The form in which the element
  exists under conditions of 1 atm and 298 K.
• Caution! Not STP (for gas law problems, 273 K)

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• Standard elemental states
• Hydrogen H2 (g) (not atomic H)
• Oxygen O2 (g)
• Carbon C (gr) graphite, not diamond
• We will denote the standard state using the
  superscript
• For example, DHrxn

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Importance of Elements

• We will use the elemental forms as a primary
  reference when examining compounds.
• Pictorally, for chemical reactions we envision
  taking the reactants apart to their standard
  elemental forms, then reforming the products.

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Example Combusion of MethaneCH4(g) 2O2(g)
CO2(g) 2H2O(l)
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Elements and DHf

• For elements in their standard state
• DHf 0
• With respect to the graph, we are envisioning
  chemical reactions as proceeding through
  elemental forms. In this way we are comparing
• DHf for reactants and products to a common
  reference state, namely
• ??????????DHf 0 for the elements.

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Tabulated DHf

• In Zumdahl, tables of DHf are provided in
  Appendix 4.
• The tabulated values represent the DHf for
  forming the compound from elements at standard
  conditions.
• C (gr) O2 (g) CO2 (g) DHf -394
  kJ/mol

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Using DHf to determine DHrxn

• Since we have connected the DHf to a common
  reference point, we can now determine DHrxn by
  looking at the difference between the DHf for
  products versus reactants.

• Mathematically
• DHrxn SDHf (products) - SDH f (reactants)

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How to Calculate DHrxn

• When a reaction is reversed, DH changes sign.
• If you multiply a reaction by an integer, then
• DH is also multiplied by the same factor (since
  it is an extensive variable)
• Elements in their standard states are not
  included in calculations since DHf 0.

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Example

• Determine the DHrxn for the combustion of
  methanol.

CH3OH (l) 3/2O2(g) CO2(g) 2H2O(l)
(From Appendix 4) DHf CH3OH(l) -238.6
kJ/mol CO2(g) -393.5 kJ/mol H2O(l)
-286 kJ/mol
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• CH3OH (l) 3/2O2(g) CO2(g) 2H2O(l)

DHrxn SDHf (products) - SDH f
(reactants) DHf(CO2(g)) 2DHf(H2O(l)) -
DHf(CH3OH(l))
-393.5 kJ - (2 mol)(286 kJ/mol) - (-238.6
kJ) -728.7 kJ
Exothermic!
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Another Example

• Using the following reaction
• 2ClF3(g) 2NH3(g) N2(g) 6HF(g) Cl2(g)
• DHrxn -1196 kJ

Determine the DHf for ClF3(g)
?????????????????????????????????????????????????D
Hf NH3(g) -46 kJ/mol HF(g) -271
kJ/mol
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• Given the reaction of interest
• 2ClF3(g) 2NH3(g) N2(g) 6HF(g) Cl2(g)
• DHrxn -1196 kJ
• And Hess Law

DHrxn SDHf (products) - SDH f (reactants)
Substituting, we have DHrxn 6DHf(HF(g)) -
2DHf (NH3(g)) - 2DH f (ClF3(g))
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DHrxn 6DHf(HF(g)) - 2DHf (NH3(g)) -
2DHf (ClF3(g)) -1196 kJ (6mol)(-271
kJ/mol) - (2mol)(-46 kJ/mol) -
(2mol)DHf (ClF3(g)) -1196 kJ -1534 kJ -
(2mol)DHf (ClF3(g)) 338 kJ - (2mol)DHf
(ClF3(g)) -169 kJ/mol DHf (ClF3(g))
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Bonus Example

• Problem 9.81 DHvap for water at the normal
  boiling point (373.2 K) is 40.66 kJ. Given the
  following heat capacity data, what is DHvap at
  340.2 K for 1 mol?

• CP H2O(l) 75 J/mol.K
• CP H2O(g) 36 J/mol.K

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Bonus Example (cont.)
(373.2 K)
H2O(l)
H2O(g)
DH 40.66 kJ/mol
nCP(H2O(g))DT
nCP(H2O(l))DT
H2O(l)
H2O(g)
DH ?
(340.2 K)
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Bonus Example (cont.)
(373.2 K)
H2O(l)
H2O(g)
DH 40.66 kJ/mol
nCP(H2O(l))DT
nCP(H2O(g))DT
H2O(l)
H2O(g)
(340.2 K)
DHrxn (340.2K) DH(product) - DH(react)
DHwater(g) (373.2K)
nCP(H2O(g))DT - DHwater(l)
(373.2K) - nCP(H2O(l))DT
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Bonus Example (cont.)
DHrxn (340.2K) DHwater(g) (373.2K)
nCP(H2O(g))DT - DHwater(l)
(373.2K) - nCP(H2O(l))DT
DHrxn (340.2K) 40.66 kJ/mol 36 J/mol.K(-33 K)
- 75 J/mol.K(-33 K)
DHrxn (340.2K) 40.66 kJ/mol 1.29 kJ/mol
41.95 kJ/mol