126 ❯ STEP 4. Review the Knowledge You Need to Score High
If the previous reaction for the formation of water were reversed, the sign of DH would
be reversed. That would indicate that it would take 483.6 kJ of energy to decompose 2 mol
of water. This would then become an endothermic process.
DH is dependent upon the state of matter. The enthalpy change would be different for
the formation of liquid water instead of gaseous water.
DH can also indicate whether a reaction will be spontaneous. A negative (exothermic)
value of DH is associated with a spontaneous reaction. However, in many reactions this is
not the case. There is another factor to consider in predicting a reaction’s spontaneity. We
will cover this other factor a little later in this chapter.
Enthalpies of reaction can be measured using a calorimeter. However, they can also
be calculated in other ways. Hess’s law states that if a reaction occurs in a series of steps,
then the enthalpy change for the overall reaction is simply the sum of the enthalpy changes
of the individual steps. If, in adding the equations of the steps together, it is necessary to
reverse one of the given reactions, then the sign of DH must also be reversed. Also, particu-
lar attention must be used if the reaction stoichiometry has to be adjusted. The value of an
individual DH may need to be adjusted.
It doesn’t matter whether the steps used are the actual steps in the mechanism of the
reaction, because DHreaction (DHrxn) is a state function, a function that doesn’t depend on
the pathway, but only on the initial and final states.
Let’s see how Hess’s law can be applied, given the following information:
C(s) + O 2 (g) → CO 2 (g) DH = - 393.5 kJ
H 2 (g) + (1/2)O 2 (g) → H 2 O(l) DH = - 285.8 kJ
C 2 H 2 (g) + (5/2)O 2 (g) → 2 CO 2 (g) + H 2 O(l) DH = - 1,299.8 kJ
Find the enthalpy change for:
2C(s) + H 2 (g) → C 2 H 2 (g)
Answer:
2[C(s) + O 2 (g)→ CO 2 (g)] 2 (-393.5 kJ)
H 2 (g) + (1/2) O 2 (g) → H 2 O(l) -285.8 kJ
2 CO 2 (g) + H 2 O(l) → C 2 H 2 (g) + (5/2) O 2 (g) -(-1,299.8 kJ)
2C(s) + H 2 (g) → C 2 H 2 (g) 227.0 kJ
Enthalpies of reaction can also be calculated from individual enthalpies of formation
(or heats of formation), DHf, for the reactants and products. Because the temperature,
pressure, and state of the substance will cause these enthalpies to vary, it is common to
use a standard state convention. For gases, the standard state is 1 atm pressure. For a
substance in an aqueous solution, the standard state is 1 molar concentration. And, for a
pure substance (compound or element), the standard state is the most stable form at 1 atm
pressure and 25°C. A degree symbol to the right of the H indicates a standard state, DH°.
The standard enthalpy of formation of a substance (DHf°) is the change in enthalpy when
1 mol of the substance is formed from its elements when all substances are in their standard
states. These values are then tabulated and can be used in determining DH°rxn.
DHf° of an element in its standard state is zero.
DHf°rxn can be determined from the tabulated DHf° of the individual reactants and prod-
ucts. It is the sum of the DHf° of the products minus the sum of the DHf° of the reactants:
DH°rxn = S DHf° products - S DHf° reactantsKEY IDEA