Molar Enthalpies of Formation
From UCDavis chemwiki
By now chemists have measured the enthalpy changes for so many reactions that it would take several large volumes to list all the thermochemical equations. Fortunately Hess' law (which we will discuss next) makes it possible to list a single value, the standard enthalpy of formation `DeltaH_(f)^(O)`, for each compound. The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substance is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C).
For example, if we know that
`DeltaH_(f) [H_2O(l)] = –285.8 kJ mol^(-1)`,
we can immediately write the thermochemical equation
`H_2(g) + (1/2)O_2(g) -> H_2O(l)`
`DeltaH_m = –285.8 kJ mol^(-1)` (1)
The elements `H` and `O` appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off per mole of `H_2O(l)` formed. Equation (1) must specify formation of 1 mol `H_2O(l)`, and so the coefficient of `O_2` must be `1/2`.
In some cases, such as that of water, the elements will react directly to form a compound, and measurement of the heat absorbed serves to determine `DeltaH_f`. Quite often, however, elements do not react directly with each other to form the desired compound, and `DeltaH_f` must be calculated by combining the enthalpy changes for other reactions. A case in point is the gas acetylene, `C_2H_2`.
`2C(s) + H_2(g) -> C_2H_2(g)`
`DeltaH_m = 227.0 kJ mol^(-1)`
was valid. Since it involves 1 mol `C_2H_2` and the elements are in their most stable forms, we can say that `DeltaH_(f) [C_2H_2(g)] = 227.0 kJ mol^(-1)`.
One further point arises from the definition of `DeltaH_f`. The standard enthalpy of formation for an element in its most stable state must be zero. If we form mercury from its elements, for example, we are talking about the reaction
`Hg(l) -> Hg(l)`
Since the mercury is unchanged, there can be no enthalpy change, and `DeltaH_(f) = 0 kJ mol^(-1)`.
Some Standard Enthalpies of Formation at 25°C.
| Compound | `DeltaH_(f) "/"kJmol^(-1)` | `H_(f) "/"kcalmol^(-1)` | Compound | `DeltaH_(f) "/"kJmol^(-1)` | `DeltaH_(f) "/"kcalmol^(-1)` |
| AgCl(s) | –127.068 | –30.35 | H2O(g) | –241.818 | –57.79 |
| AgN3(s) | +620.6 | +148.3 | H2O(l) | –285.8 | –68.3 |
| Ag2O(s) | –31.0 | –7.41 | H2O2(l) | –187.78 | –44.86 |
| Al2O3(s) | –1675.7 | –400.40 | H2S(g) | –20.63 | –4.93 |
| Br2(l) | 0.0 | 0.00 | HgO(s) | –90.83 | –21.70 |
| Br2(g) | +30.907 | +7.385 | I2(s) | 0.0 | 0.0 |
| C(s), graphite | 0.0 | 0.00 | I2(g) | +62.438 | +14.92 |
| C(s), diamond | +1.895 | +0.453 | KCl(s) | –436.747 | –104.36 |
| CH4(g) | –74.81 | –17.88 | KBr(s) | –393.798 | –94.097 |
| CO(g) | –110.525 | –26.41 | MgO(s) | –601.7 | –143.77 |
| CO2(g) | –393.509 | –94.05 | NH3(g) | –46.11 | –11.02 |
| C2H2(g) | +226.73 | +54.18 | NO(g) | +90.25 | +21.57 |
| C2H4(g) | +52.26 | +12.49 | NO2(g) | +33.18 | +7.93 |
| C2H6(g) | –84.68 | –20.23 | N2O4(g) | +9.16 | +2.19 |
| C6H6(l) | +49.03 | +11.72 | NF3(g) | –124.7 | –29.80 |
| CaO(s) | –635.09 | –151.75 | NaBr(s) | –361.062 | –86.28 |
| CaCO3(s) | –1206.92 | –288.39 | NaCl(s) | –411.153 | –98.24 |
| CuO(s) | –157.3 | –37.59 | O3(g) | +142.7 | +34.11 |
| Fe2O3(s) | –824.2 | –196.9 | SO2(g) | –296.83 | –70.93 |
| HBr(g) | –36.4 | –8.70 | SO3(g) | –395.72 | –94.56 |
| HCl(g) | –92.307 | –22.06 | ZnO(s) | –348.28 | –83.22 |
| HI(g) | +26.48 | +6.33 |
Standard enthalpies of formation for some common compounds are given in the table above. These values may be used to calculate `DeltaH_m` for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example.
EXAMPLE 8 Use standard enthalpies of formation to calculate `DeltaH_m` for the reaction
`2CO(g) + O_2(g) -> 2CO_2(g) `
Solution:
We can imagine that the reaction takes place in two steps, each of which involves only a standard enthalpy of formation. In the first step CO (carbon monoxide) is decomposed to its elements:
`2CO(g) -> 2C(s) + O_2(g)` `DeltaH_m = DeltaH_1`
Since this is the reverse of formation of 2 mol CO from its elements, the enthalpy change is
`DeltaH_1 = 2* -DeltaH_(f) [CO(g)] = 2*[-(-110.5kJmol^(-1))]=+221.0 kJmol^(-1)`
In the second step the elements are combined to give 2 mol CO2(carbon dioxide):
`2C(s) + 2O_2(g) -> 2CO_2(g)` `DeltaH_m = DeltaH_2`
In this case
`DeltaH_2 = 2*DeltaH_(f) [CO_2(g)] = 2*(-393.5kJ mol^(-1))= -787.0kJmol^(-1)`
You can easily verify that the sum of Equations 3.10.23.10.2 and 3.10.33.10.3 is
`2CO(g) + 2O_2(g) -> 2CO_2(g)` `DeltaH_m = DeltaH_"net"`
Therefore
`DeltaH_"net" = DeltaH_1 + DeltaH_2 = 221.0kJmol^(-1) -787.0mol^(-1) = -566.0 mol^(-1)`
Note carefully how Example 3.10.13.10.1 was solved. In step 1 the reactant compound `CO(g)` was hypothetically decomposed to its elements. This equation was the reverse of formation of the compound, and so `DeltaH_1` was opposite in sign from `DeltaH_f`. Step 1 also involved 2 mol `CO(g)` and so the enthalpy change had to be doubled. In step 2 we had the hypothetical formation of the product `CO_2(g)` from its elements. Since 2 mol were obtained, the enthalpy change was doubled but its sign remained the same.
Any chemical reaction can be approached similarly. To calculate `DeltaH_m` we add all the `DeltaH_f` values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of `DeltaH_f` for the reactants had to be reversed in step 1, we subtract them, again multiplying by appropriate coefficients. This can he summarized by the equation
`DeltaH_m = sum DeltaH_(f) " products" - sum DeltaH_(f) " reactants"` (4)
The symbol `sum` means “the sum of.” Since `DeltaH_f` values are given per mole of compound, you must be sure to multiply each `DeltaH_f` by an appropriate coefficient derived from the equation for which `DeltaH_m` is being calculated.
EXAMPLE 9 Use the table of standard enthalpies of formation at 25°C to calculate `DeltaH_m` for the reaction
`4NH_2(g) + 5O_2(g) -> 6H_2O(g) + 4NO(g) `
Using equation 4, we have
`DeltaH_m = sum DeltaH_(f) " products" - sum DeltaH_(f) " reactants"`
`= [6DeltaH_(f) (H_2O) + 4DeltaH_(f) (NO)] - [4DeltaH_(f) (NH_3) + 5DeltaH_(f) (O_2)]`
`= 6(-241.8 kJ mol^(-1) + 4(90.3) kJ mol^(-1) - 4(-46.1 kJ mol^(-1))-5*0`
`=-1450.8 kJmol^(-1) + 361.2 kJ mol^(-1) + 184.4 kJ mol^(-1)`
`= -905.2 kJ mol^(-1)`
Note that we were careful to use `DeltaH_f [H_2O(g)]` not `DeltaH_f [H_2O(l)]`. Even though water vapor is not the most stable form of water at 25°C, we can still use its `DeltaH_f` value. Also the standard enthalpy of formation of the element `O_2(g)` is zero by definition. Obviously it would be a waste of space to include it in the table above.
