5 Steps to a 5 AP Chemistry

If the reaction above for the formation of water were reversed, the sign of ΔHwould 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.

ΔHis dependent upon the state of matter. The enthalpy change would be different for

the formation of liquid water instead of gaseous water.

ΔHcan also indicate whether a reaction will be spontaneous. A negative (exothermic)

value of ΔHis 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 lawstates 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 ΔHmust also be reversed. Also, partic-

ular attention must be used if the reaction stoichiometry has to be adjusted. The value of

an individual ΔHmay need to be adjusted.

It doesn’t matter whether the steps used are the actual steps in the mechanism of the

reaction because ΔHreaction(ΔHrxn) 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) ΔH=−393.5 kJ

H 2 (g) +(1/2)O 2 (g) →H 2 O(1) ΔH=−285.8 kJ

C 2 H 2 (g) +(5/2)O 2 (g) →2 CO 2 (g) +H 2 O(1) ΔH=−1299.8 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(1) −285.8 kJ

2 CO 2 (g) +H 2 O(1) →C 2 H 2 (g) +(5/2) O 2 (g) −(−1299.8 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), ΔHf, 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 Hindicates a standard state,

ΔH°. The standard enthalpy of formation of a substance (ΔHf°) 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

ΔH°rxn.

ΔHf°of an element in its standard state is zero.

ΔHf°rxncan be determined from the tabulated ΔHf°of the individual reactants and prod-

ucts. It is the sum of the ΔHf°of the products minus the sum of the ΔHf°of the reactants:

ΔH°rxn=ΣΔHf°products −ΣΔHf°reactants

Thermodynamics  127

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