Chemistry 121:
Notes:
Standard Enthalpy of Formation
The change in enthalpy that accompanies the formation of one mole of a compound in
its standard state from the elements in their standard states. 
-See chart
-Elements in their standard state have a Hf= 0.0 kJ/mol
The importance of tabulated fis that enthalpies for many reactions can be
calculated using these numbers.
Example: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)
Since enthalpy is a state function we can choose any convenient pathway from
reactants to products and then sum the enthalpy changes along the chosen pathway.
In this pathway for the combustion of methane, the reactants are first taken apart in
reactions (a) and (b) to form the constituent elements in their standard states, which are
then used to assemble the products in reactions (c) and (d).
In reaction A where methane is taken apart into its elements, its just the reverse of the
formation of methane
C(s) + 2H2(g) -> CH4(g) f= -75 kJ/mol
So the sign changes to f= +
75kJ
In reaction B, oxygen is already an element in its standard state so no change is
needed this f= 0.0 kJ
In reactions C and D, we use the elements formed in reactions A and B to form the
products. Note that the reaction C is simply the formation reaction for carbon dioxide.
C(s) + O2(g) -> CO2(g) f= -394 kJ/mol
Reaction D is the formation of water:
H2(g) + 1/2 O2(g) -> H2O(l) f= -286
kJ/mol
However, since 2 moles are required in the balanced equation, f= -286 kJ/mol x 2=
-572 kJ
Having completed the pathway from reactants to products, we can calculate the rxn
rxn = A +B +C +D
=
+75kJ + 0kJ + (-394kJ) + (-572kJ)
= -891kJ
The enthalpy change for a given reaction can be calculated by subtracting the
enthalpies of formation of the reactants from the enthalpies of formation of the productsremembering to multiply the enthalpies of formation by the integers required by the
balanced equation.
A chemical Equation can be express as a
sum of formation reactions.
∆Horeaction= ∑nHOf(Products)- ∑nHOf(reactants)
Steps:
1. Write the balanced equation for the reaction
2. Obtain the molar heats of formation for all reactants and products
3. Multiply the molar heat of formation by the molar coefficient for each reactant and
product.
H=nHf
4. Sum the heats of formation for the reactants
and products separately.
5. Calculate the molar heat of reaction for the substance specified. ∆H/#moles (n)
Example: Calculate the Hrxn CO for the reaction of carbon monoxide gas with oxygen to
form carbon dioxide gas.
Step 1:Write the balanced equation for the reaction
2CO(g) + O2(g) --> 2CO2(g)
Step 2: Obtain the molar heats of formation for all reactants and products
Hf  CO(g)= -110.5kJ/mol
Hf  O2(g) = 0 kJ/mol
Hf  CO2(g) = -393.5 kJ/mol
Step 3: Multiply the molar heat of formation by the molar coefficient for each reactant
and product. H=nHf
nHf 2mol x -110.5kJ/mol kJ/mol=-221.0 kJ
1mol x 0 kJ/mol= kJ
2mol x -393.5 kJ/mol=-786.0 kJ
Step 4: Sum the heats of formation for the reactants and products separately.
Reactant sum:
Product sum:
-221.0kJ
-786.0 kJ
Step 5: Calculate the molar heat of reaction for the substance specified. ∆H/#moles (n)
∆Horeaction= ∑nHOf(Products)- ∑nHOf(reactants)
= -786.0 - (-221.0kJ)
= -565.0 kJ / 2mol
HoCO =-282.5 kJ/mol
In respiration, glucose is oxidized by oxygen to produce carbon dioxide gas, liquid
water, and energy. What is the energy released when 18.0g of glucose is consumed
assuming products are at standard conditions?