5. ThermoChemistry

Hess’ Law

Since enthalpy change, ΔH, is a state function, we can go to products via several routes and the enthalpy change will still be the same.

Another way to say this is: the enthalpy change in a reaction does not depend on the reaction pathway.  This allows us to define any path we choose to get from reactants to products and, as long as we keep track of the enthalpy changes for each step, we will be able to calculate the overall enthalpy change for the process.

Let’s look at the example of ethane burning in excess oxygen to produce carbon dioxide and water.  We have two possibilities: we make a series of reactions (1,2,3) that gets us to the final products and keep track of the enthalpy changes for each step; we simply burn the ethane in a calorimeter and measure the enthalpy change directly (labelled T).  These two possibilities are diagrammed below where the vertical axis represents enthalpy.

We can consider the reaction to occur in one complete process [T] or in three distinct steps [1], [2] and [3]. In either case, the total enthalpy change should be the same.

[1] Ethane is heated to drive off H2.
C2H6 image C2H4 + H2
Δr1 = +136.2 kJ
[2] Ethene is burned in excess O2.
C2H4 + 3 O2 → 2 CO2 + 2 H2O
Δr2 = -1410.8 kJ
[3] H2 is burned in excess O2.
H2 + 1/2 O2 → H2O
Δr3 =  -285.8 kJ
[T] overall reaction is sum [1]+[2]+[3]
C2H6 + 3.5 O2 → 2 CO2 + 3 H2O
ΔrT = -1560.4 kJ

The example here may not seem to have been practical at first glance.   After all, why set up three separate calorimetry experiments when we could have just done the overall experiment in one step?  Well, it serves as a good demonstration of the principle.  Additionally, we can use this idea to calculate enthalpies changes for reactions where direct measurement is impractical.  Also, we can use this method to calculate the expected enthalpy change of a reaction using tabulated values without the need to do any experiments.

Example: Use the following standard enthalpies of combustion to calculate the standard enthalpy change for the formation of methane.

The standard formation reaction for methane is:

C(s) + 2 H2(g) → CH4(g)    ΔrH° = ?

This will be our target reaction.  We need to use the Standard Combustion Enthalpies listed here.

1)ΔcombH°(C) = -393.5 kJ
2)ΔcombH°(H2) = -285.8 kJ
3)ΔcombH°(CH4) = -890.4 kJ

Knowing the standard format for the combustion reaction (1 mol of reactant burns to completion in oxygen to produce fully oxidized products; CO2 and/or H2O in this case) we can easily write down the three reactions.

1)  C(s) + O2(g) → CO2(g)Δr = -393.5 kJ/mol
2) H2(g) + 1/2 O2(g) → H2O(g)Δr = -285.8 kJ/mol
3) CH4(g) + 2 O2(g) → CO2(g) + 2H2O(l)Δr = -890.4 kJ/mol
Now lets add up the three reactions to give the target reaction
1)  C(s) + O2(g) → CO2(g)Δr = -393.5 kJ/mol
2) 2 ×[ H2(g) + 1/2 O2(g)  H2O(g)Δr = -285.8 kJ/mol]
3) CO2(g) + 2H2O(l) → CH4(g) + 2 O2(g)
Δr = +890.4 kJ/mol
    C(s) + 2 H2(g)  → CH4(g)ΔrHº  = -74.7 kJ/mol

To get to the desired final reaction, we needed to use equation 2 twice (needed to use up 2 hydrogens). Thus, we also have added the enthalpy for that reaction twice (note the square brackets extends around the equation and the enthalpy value.  We also reversed the direction of equation 3 (multiplied all coefficients by -1, and hence multiplied the enthalpy change by -1) Whatever we do to the equation, we must also do the Δ value.

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