Enthalpy
C6H12O6(s) + 6O2 (g) --> 6CO2 (g) + 6H2O(l) + 2803 kJ
2C57H110O6 + 163O2 (g) --> 114 CO2(g) + 110 H2O(l) + 75,520 kJ
The heat lost or gained by a system undergoing a process at
constant pressure is related to the change in ENTHALPY (DH)
of the system
DH = Hfinal - Hinitial = qp
Note: DH is a state function
Most physical and chemical changes take place under the
constant pressure of the Earth’s atmosphere.
For an exothermic process at constant pressure DH < 0
For an endothermic process at constant pressure DH > 0
DH < 0
DH > 0
Enthalpy, H is defined as
H = E + PV
For a system undergoing pressure-volume work, the
change in enthalpy in the system is
DH = DE + DPV
If the pressure is a constant external pressure, Pext,
DH = DE + PextDV
From the 1st law: DE = q + w
For constant P: DH = qp + w + PextDV
Since w = - Pext DV
DH = qp - Pext DV + PextDV = qp
If volume is held constant, no pressure-volume work can
be done on that system or by that system.
work = - PextDV = 0 at constant volume
DE = q + w,
For constant volume processes, DE = qv
A constant pressure calorimeter measures DH
A constant volume calorimeter (like a bomb calorimeter)
measures DE.
Relationship between DH and DE
C(s) + 1/2 O2(g) --> CO(g)
DH = -110.5 kJ
Determine the change in internal energy accompanying this
reaction.
DH = DE + D (PV)
DE = DH - D (PV)
D(PV) = D (nRT) = RT(Dng)
DE = DH - RT(Dng) or
DH = DE + RTDng
For this reaction Dng = 0.5 mol; hence DE = -111.7 kJ
For reactions that do not involve gases; DH ≈ DE
Enthalpy of Physical Change
Phase transitions involve change in energy
Vaporization is endothermic; condensation exothermic.
Since phase transitions typically take place at constant
pressure, the heat transfer is the change in enthalpy.
DHvap = Hvapor - Hliquid
liquid --> gas
DHfusion = Hliquid - Hsolid
solid --> liquid
DHfreezing = - DHfusion
DHforward = - DHreverse
DHsublimation = DHfusion + DHvaporization
Heating curve of water
H2O(s) --> H2O(l)
DH = 6.01 kJ
H2O(l) --> H2O(g)
DH = 40.7 kJ
Enthalpy of Chemical Change
The enthalpy change for a chemical reaction is given by
DH = SH(products) - SH(reactants)
For example:
2H2(g) + O2 (g) --> 2H2O(g)
DH = - 483.6 kJ
Thermochemical reaction
In a chemical reaction, the enthalpy change during the
reaction indicates whether the reaction releases energy or
consumes energy.
If DH < 0, the reaction releases heat and is EXOTHERMIC
If DH > 0, the reaction absorbs heat and is ENDOTHERMIC
The magnitude of DH for a reaction is directly proportional
to the amount of reactants consumed by the reaction.
CH4(g) + 2O2(g) --> CO2(g) + 2H2O(l)
DH = -890 kJ
Calculate the amount of heat that would be released when
4.50 g of CH4(g) is burned in an oxygen atmosphere at
constant pressure.
4.50g CH4 => 0.28 mole CH4
=> (0.28 mol CH4)(-890 kJ/mol CH4) = -250 kJ
The enthalpy change for a reaction is equal in magnitude
but opposite in sign to DH for the reverse reaction
CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l) DH = -890 kJ
CO2(g) + 2H2O(l) --> CH4(g) + 2 O2(g) DH = +890 kJ
The enthalpy change for a reaction depends on the phase of
the reactants and products.
1) CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)
DH = -890 kJ
2) CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(g)
DH = -802 kJ
3) 2H2O(g) --> 2H2O(l)
DH = -88kJ
Standard Reaction Enthalpies
CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)
DH = -890 kJ
CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(g) DH = -802 kJ
In order to compare DH’s accompanying reactions need to
define a standard state
Standard State: reactant and products in their pure forms
at a pressure of 1 atm (or 1 bar).
A solute in a liquid solution is in its standard state when its
concentration is 1 mol L-1
While temperature is not part of the definition of the
standard state, standard reaction enthalpies are usually
reported for a temperature of 298.15 K.
Standard Reaction Enthalpies (DHo): enthalpy change
accompanying a reaction when reactants in their standard
states change to products in their standard states.
CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)
DHo = -890 kJ
Hess’s Law
If a reaction is carried out in a series of steps, DH for the
reaction will be equal to the sum of the enthalpy change for
the individual steps.
The enthalpy of combustion of C to CO2 is -393.5 kJ/mol, and
the enthalpy of combustion of CO to CO2 is -283.0 kJ/mol CO.
(1) C(s) + O2 (g) --> CO2 (g)
DH = -393.5 kJ
(2) CO(g) + 1
2 O2 (g) --> CO2 (g) DH = -283.0 kJ
Calculate the enthalpy change of combustion of C to CO
(3) C(s) + 1
2 O2 (g) --> CO (g)
DH = ?
C(s) + O2 (g) --> CO2 (g)
DH = -393.5 kJ
CO2 (g) --> CO(g) + 1
2 O2 (g)
DH = 283.0 kJ
C(s) + 1 O2 (g) --> CO (g)
2
DH = -110.5 kJ
Standard Enthalpy of Combustion
Change in enthalpy per mole of a substance that is burned
in a combustion reaction under standard conditions
Calculate the mass of propane that you would need to burn
to obtain 350 kJ of heat which is just enough energy to heat
I L of water from room temperature (20oC) to boiling at sea
level. Assume that all the heat generated is absorbed by the
water.
C3H8(g) + 5 O2(g) --> 3 CO2(g) + 4H2O(l)
DHo = -2220 kJ
Combustion of 1 mole of propane generates 2220 kJ of heat
Mass or propane required =
(350 kJ) (1 mole C3H8 /2220kJ) (44.09 g/mol) = 6.95 g
If you were to use butane instead of propane in the previous
example, how much butane (in grams) would you need?
2 C4H10(g) + 13 O2(g) --> 8 CO2(g) + 10 H2O(l)
DHo = -5756 kJ
Answer: 7.07 g
If you were to use ethanol instead of propane in the previous
example, how much ethanol (in grams) would you need?
C2H5OH(l) + 3 O2(g) --> 2 CO2(g) + 3 H2O(l)
Answer: 11.8 g
DHo = -1368 kJ
Enthalpies of Formation
The enthalpy of formation, DHf, or heat of formation, is
defined as the change in enthalpy when one mole of a
compound is formed from its stable elements.
The standard enthalpy of formation (DHfo) of a compound is
defined as the enthalpy change for the reaction that forms 1
mole of compound from its elements, with all substances in
their standard states.
2C(s) + 1/2 O2(g) + 3 H2 (g) --> C2H5OH(l)
DHfo = -277.69 kJ
Standard enthalpies of formation of substance can be used
to determine standard reaction enthalpies.
The standard enthalpy of formation of the most stable form of
an element under standard conditions is ZERO.
DHfo for C(graphite), H2(g), O2(g) are zero
The stoichiometry for formation reactions indicate the
formation of 1 mole of the desired compound, hence
enthalpies of formation are always listed as kJ/mol.