Lecture 4: Standard Enthalpies
• Reading: Zumdahl 9.6
• Outline
– What is a standard enthalpy?
– Defining standard states.
– Using standard enthalpies to determine DH°rxn
Hess’ Law
• Enthalpy is a state function. As such, DH for
going from some initial state to some final state is
pathway independent.
DH for a process involving the transformation of
reactants into products is not dependent on
pathway. Therefore, we can pick any pathway to
calculate DH for a reaction.
Definition of DH°f
• We saw in the previous lecture the utility of
having a series of reactions with known
enthalpies.
•What if one could deal with combinations
of compounds directly rather than dealing
with whole reactions to determine DHrxn?
Definition of DH°f (cont.)
• Standard Enthalpy of Formation, DH°f:
“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.
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 (or 100,000 Pascal; 1 atm=101325)
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.
Definitions (cont.)
• Standard elemental states (cont.):
– Hydrogen: H2 (g) (not atomic H)
– Oxygen: O2 (g)
– Carbon: C (gr) … graphite as opposed to diamond
• We will denote the standard state using the subscript
“°”.
– Example: DH°rxn
Importance of Elements
• We will use the elemental
forms as a primary
reference when examining
compounds.
• Pictorially, for chemical
reactions we envision
taking the reactants to the
standard elemental form,
then reforming the
products.
Example: Combusion of Methane
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
Elements and DH°f
• For elements in their
standard state:
elements
DH°f = 0
• With respect to the graph,
we are envisioning
chemical reactions as
proceeding through
elemental forms. In this
way we are comparing
DH°f for reactants and
products to a common
reference (zero)
q
DH¡ rxn
reactants
products
Tabulated DH°f
• In Zumdahl, tables of DH°f are provided in
Appendix 4.
• The tabulated values represent the DH°f for
forming the compound from elements at standard
conditions.
C (gr) + O2 (g)
CO2 (g)
DHf° = -394 kJ/mol
Using DH°f to determine DH°rxn
• Since we have connected the DH°f to a common
reference point, we can now determine DH°rxn by
looking at the difference between the DH°f for
products versus reactants.
• Mathematically:
DH°rxn = SDH°f (products) - SDH° f (reactants)
How to Calculate DH°rxn
• When a reaction is reversed, DH changes sign.
• If you multiply a reaction by an integer,
DH is also multiplied (it is an extensive variable)
• Elements in their standard states are not included
in calculations since DH°f = 0.
Example
• Determine the DH°rxn for the combustion of
methanol.
CH3OH (l) + 3/2O2(g)
CO2(g) + 2H2O(l)
• DH°f in Appendix 4:
CH3OH(l)
CO2(g)
H2O(l)
-238.6 kJ/mol
-393.5 kJ/mol
-286 kJ/mol
Example (cont.)
CH3OH (l) + 3/2O2(g)
CO2(g) + 2H2O(l)
DH°rxn = SDH°f (products) - SDH° f (reactants)
= DH°f(CO2(g)) + 2DH°f(H2O(l))
- DH°f(CH3OH(l))
= -393.5 kJ - (2 mol)(286 kJ/mol)
- (-238.6 kJ)
Exothermic!
= -728.7 kJ
Another Example
• Using the following reaction:
2ClF3(g) + 2NH3(g)
N2(g) + 6HF(g) + Cl2(g)
DH°rxn = -1196 kJ
Determine the DH°f for ClF3(g)
DH°f
NH3(g) -46 kJ/mol
HF(g)
-271 kJ/mol
Another Example (cont.)
• Given the reaction of interest:
2ClF3(g) + 2NH3(g)
N2(g) + 6HF(g) + Cl2(g)
DH°rxn = -1196 kJ
DH°rxn = SDH°f (products) - SDH° f (reactants)
DH°rxn = 6DH°f(HF(g)) - 2DH°f (NH3(g))
- 2DH° f (ClF3(g))
Another Example (cont.)
DH°rxn = 6DH°f(HF(g)) - 2DH°f (NH3(g))
- 2DH°f (ClF3(g))
-1196 kJ = (6mol)(-271 kJ/mol)
- (2mol)(-46 kJ/mol)
- (2mol)DH°f (ClF3(g))
-1196 kJ = -1534 kJ - (2mol)DH°f (ClF3(g))
338 kJ = - (2mol)DH°f (ClF3(g))
-169 kJ/mol = DH°f (ClF3(g))
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 H20(l) = 75 J/mol.K
• CP H20(g) = 36 J/mol.K
Bonus Example (cont.)
(373.2 K)
H2O(l)
H2O(g)
nCP(H2O(l))DT
H2O(l)
nCP(H2O(g))DT
H2O(g)
(340.2 K)
DH = 40.66 kJ/mol
DH = ?
Bonus Example (cont.)
(373.2 K)
H2O(l)
H2O(g)
nCP(H2O(l))DT
H2O(l)
(340.2 K)
DH = 40.66 kJ/mol
nCP(H2O(g))DT
H2O(g)
DHrxn (340.2K) = DH(product) - DH(react)
= DHwater(g)
(373.2K) + nCP(H2O(g))DT
- DHwater(l) (373.2K) - nCP(H2O(l))DT
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