Heats or reaction from heats of formation

7.8 DH_rxn from DH_f° and Hess's Law
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DH_f° = standard enthalpy of formation: the enthalpy change that accompanies the formation of 1 mole of compound from the elements with all substances in their standard states. DH_f° are intensive, not extensive, by virtue of the fact that they are based on the formation of 1 mole of product!

for water (at 25 °C):
H₂(g) + 1/2O₂(g) ---> H₂O(l) DH_f° = -286 kJ/mol

for water (at 110 °C):
H₂(g) + 1/2O₂(g) ---> H₂O(g) DH_f° = -241 kJ/mol

standard state: the most common state of matter for a given substance at a specified temperature.

    gases: 1 atm and the temperature of interest
    solutes: 1 molar and the temperature of interest
    substances: either solid or liquid, depending
                        on the temperature
    elements: most common allotrope at the
                        temperature of interest

examples: H₂O at 25°C is a liquid
H₂O at 200 °C is a gas, and at -10 °C is a solid.
At room temperature oxygen is most commonly found as gaseous O₂, while nitrogen is gaseous N₂. Hg is a liquid, while Na is a solid.
Note that DH_f° depend on the state of the product (gas, liquid, or solid) since heat is typically goven off when a solid forms from a liquid and when a liquid forms from a gas.

DH_f° = for an element is DEFINED as 0 kJ/mole.
    Why? Because of the definition: one mole of
    element formed from the elements would
    necessarily have 0 enthalpy change.

DH_f° are tabulated conveniently in reference manuals
    of chemistry. Most substances are easily found
    because of alphabetization and conventions for
    formulating compounds.

How are they useful?

Hess's Law can be used to define another way of
determining DH_rxn° from DH_f°.

DH_rxn° = Sn_pDH_f°(products) - Sn_pDH_f°(reactants)

where n_p= coefficient before product or reactant in the balanced chemical equation.

What is the "physical" significance, and why does this work?

-H is a "state function" and it does not matter how we get to point B from point A to calculate DH. One way of getting from reactants to products would be to take each compound and convert it back into elements. Then take the elements and convert them into products. Since DH_f° is defined as the enthalpy change associated with the formation of 1 mole of compound from the elements, the decomposition of the reactants into elements is the exact reverse of this process. The DH associated with this step is -DH_f°.

Example:

Calculate DH for the conversion of 1 mole of gaseous water into 1 mole of liquid water at 25°C.

At the top of this page we stated that DH_f° for H₂O(l) was -286 kJ/mol, while DH_f° for H₂O(g) was -241 kJ/mol.

(i) H₂(g) + 1/2O₂(g) ---> H₂O(l) DH_f° = -286 kJ/mol
(ii) H₂(g) + 1/2O₂(g) ---> H₂O(g) DH_f° = -241 kJ/mol

If we take (i) - (ii) and subtract the common elements
from both the left and right sides we are left with:

H₂O(g) ---> H₂O(l)
DH = [-286 -(-241)] = -45 kJ/mol

DH_f° are found in appendix D and some appear below.

substance	DH_f° (298, kJ/mol)	substance	DH_f° (298, kJ/mol)
CO(g)	-110.5	HBr(g)	-36.40
CO₂(g)	-393.5	HI(g)	26.48
CH₄(g)	-74.81	H₂O(g)	-241.8
C₂H₂(g)	226.7	H₂O(l)	-285.8
C₂H₄(g)	52.26	H₂S(g)	-20.63
C₂H₆(g)	-84.68	NH₃(g)	-46.11
C₃H₈(g)	-104.7	NO(g)	90.25
C₄H₁₀(g)	-125.6	N₂O(g)	82.05
CH₃OH(l)	-238.7	NO₂(g)	33.18
C₂H₅OH(l)	-277.7	N₂O₄(g)	9.16
HF(g)	-271.1	SO₂(g)	-296.8
HCl(g)	-92.31	SO₃(g)	-395.7

Example: calculate the heat of combustion of 1 mole of methane, CH₄?

1st) balance the reaction:

CH₄(g) + 2O₂(g) ---> CO₂(g) + 2H₂O(l)

2nd) find DH_f° in the table for each reactant/product

CH₄(g) + 2O₂(g) ---> CO₂(g) + 2H₂O(l)
-74.81 0 -393.5 -286 kJ/mol

3rd) use Hess's Law to compute DH_rxn

DH_rxn = -393.5 + 2(-285.8) -(-74.81) - 0
DH_rxn = -890.3 kJ

Notice that while DH_f° are intensive, DH_rxn are extensive and depend on the amount of "stuff" we are "reacting."

How much heat is absorbed when 4 moles of water and 2 moles of CO₂ are converted into oxygen and methane (the reverse of the combustion of methane)...

B: 4H₂O(l) + 2CO₂(g) ---> 4O₂(g) +2CH₄(g)

DH_B = -2DH_combustion = +1780 kJ/mol

One can also calculate DH_f° from DH_comb which is actually how most of these values were actually arrived at.

For instance, if DH_comb for acetylene (C₂H₂) = -1299.7 kJ/mol
what is DH_f°(C₂H₂)?

1) balanced reaction is:

C₂H₂(g) + O₂(g) ---> 2CO₂(g) + H₂O(l)
DH_comb = -1299.7 kJ/mol

2) Hess's Law gives (for DH_comb):

DH_f°(CO₂) H₂O O₂
-1299.7 = 2(-393.5) + (-286) - 0 - DH_f°(C₂H₂)

rearranging we get...

DH_f°(C₂H₂) = 2(-393.5)+(-286)+1299.7
DH_f°(C₂H₂) = 226.7 kJ/mol

Example: Chapter 7, #61

The table of heats of formation above provides the information necessary to solve the problem.

2 Cl₂(g) + 2 H₂O(l) ----> 4 HCl(g) + O₂(g)
DH_f° 0 -286 kJ/mol -92.31 kJ/mol 0

DH_rxn° = Sn_pDH_f°(products) - Sn_pDH_f°(reactants)

= 4 x (-92.31) - 2 x (-286) kJ
= (-369.24 + 572) kJ
= +204 kJ

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Click here to move on to Chapter 9.