Any property that does not depend on history is a State Function. Enthalpy, and Energy are State Functions. This is a terribly useful, but fairly subtle, observation. Consider the following reactions (all taking place at 298 K)
Reaction #1: CH4 + O2 = H2CO + H2O ; DHrxn = Q1
Reaction #2: H2CO = CO + H2 ; DHrxn = Q2
Reaction #3: 2 CO + O2 = 2 CO2 ; DHrxn = Q3
Reaction #4: 2 H2 + O2 = 2 H2O ; DHrxn = Q4
Suppose that I have run all three of the following
reactions and measures their constant pressure heat output, i.e. the Enthalpy change
for the reaction. But, suppose I want the enthalpy change for the combustion
reaction that, for some reason, cannot be run in the laboratory:
Reaction #5: CH4 + 2 O2 = CO2 + 2 H2O ; DHrxn = Q5The state properties and in particular the enthalpy of the carbon dioxide and water formed by this reaction are no different than that formed in reactions #1-4, so we can use what we know about them to learn about #5. In particular, let's combine some of the reactions to produce a net reaction. If we simply allow the formaldehyde formed in reaction #1 to decompose in reaction #2, the net (sum) of the reactions will be
Reaction #1+2: CH4 + O2 = CO + H2 + H2O ; DHrxn = Q1 + Q2The heat of the net reaction will simply be the sum of the heats of reaction 1 and 2. Lets add this reaction to reaction #3
#1+2+3: CH4 + 2 O2 + CO = H2 + H2O + 2 CO2 ; DHrxn = Q1 + Q2 + Q3This isn't quite what we want because there is an extra CO on the left and an extra H2 on the right. But lets add up reaction 1 plus reaction 2 plus one half of reaction 3:
#1+2+(1/2)3: CH4 + 3/2 O2 = H2 + H2O + CO2 ; DHrxn = Q1 + Q2 + (1/2)Q3Now we are on the right track. The combustion reaction is actually a combination of all of the reactions numbered 1 through four, but reactions 3 and 4 must be scaled (divided by two) to make the net reaction equal to the combustion reaction written as reaction 5. So:
#1+2+(1/2)3 + (1/2)4: CH4 + 2 O2 = 2 H2O + CO2 ;
DHrxn = Q5 = Q1 + Q2 + (1/2)Q3 + (1/2)Q4
We can find the heat of a reaction
that we have never actually run in the laboratory if we take advantage
of the fact that the Enthalpy of chemical
species is a state function and is therefore independent of the Path of
the reaction!
Here is even a different way of calculating the heat of combustion of methane:
Hess's Law: If a change of state occurs
in stages or steps (even if only hypothetically), the enthalpy change for
the overall (net) change is the sum of the individual enthalpy stages for
the individual steps.
{This statement is of course true for any state function,
not just enthalpy}
Here is another example of the calculation of the net enthalpy change for a reaction from related heats of reaction. Consider the following reactions:
Reaction #6 : C(graphite) + O2 (gas) = CO2 (gas) DHrxn = -393.5 kJ/mol Reaction #7 : H2 (gas) + 1/2 O2 (gas) = H2O(l) DHrxn = -285.8 kJ/mol Reaction #8 : C3H8 (gas)+ 5 O2 (gas) = 3 CO2 + 4 H2O(l) DHrxn = -2220.1 kJ/mol Reaction #9 : 3 C(graphite) + 4 H2 (gas) = C3H8 (gas) DHrxn = ??The heat of formation of propane gas is the heat of reaction of equation #9. This is a difficult reaction to run, but the heat of the combustion reaction (#8) is easy to measure in a Calorimeter. Together with the known heats of formation of CO2 and Water, we can calculate the heat of reaction of equation #9 without actually running the reaction at all.
How can one construct the reaction we want to know about from the ones for which we have thermochemical data?
3*{ C(graphite) + O2 (gas) = CO2 (gas) }
+ 4*{ H2 (gas) + 1/2 O2 (gas) = H2O(l) }
- { C3H8 (gas)+ 5 O2 (gas) = 3 CO2 + 4 H2O(l) }
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3 C(graphite) + 4 H2 (gas) = C3H8 (gas)
What is the Heat of Formation of propane, anyway? answer