Physical Chemistry Third Edition

2.8 Calculation of Energy Changes of Chemical Reactions 95

The productPVis given by

PVP(V(s)+V(l)+V(g)) (2.8-2)

whereV(s) is the volume of all of the solid phases,V(l) is the volume of all of the

liquid phases, andV(g) is the volume of the gas phase. Under ordinary conditions the

molar volume of a gas is several hundred to a thousand times larger than the molar

volume of a solid or liquid. If there is at least one gaseous product or reactant we can

ignore the volume of the solid and liquid phases to an adequate approximation,

PV≈PV(g)

so that

∆(PV)≈∆PV(g) (2.8-3)

If the products and reactants are at the same temperature and if we use the ideal gas

equation as an approximation,

∆U∆H−∆PV(g)≈∆H−∆n(g)RT (2.8-4)

where∆n(g) is the change in the number of moles of gaseous substances in the reaction

equation. If 1 mol of reaction occurs, then

∆n(g)∆ν(g)

∑s

i 1

νi

(gases only)

(2.8-5)

which defines the quantity∆ν(g), equal to the number of moles of gas in the product

side of the balanced chemical equation minus the numbers of moles of gas in the

reactant side of the balanced equation.

EXAMPLE2.33

a.Find∆(PV) and∆U◦for the reaction of Eq. (2.7-1) at 298.15 K.

b.Using the fact that the molar volume of liquid water is 18 cm^3 mol−^1 at 298.15 K, make

a more accurate calculation of∆(PV) for the reaction of the previous example.

Solution

a.The reaction is

2H 2 (g)+O 2 (g)−→2H 2 O(l)

Since the single product, H 2 O, is liquid,∆ν(g)−3 and∆n(g)− 3 .000 mol

∆(PV)∆ν(g)RT(−3)(8.3145 J K−^1 mol−^1 )(298.15 K)

−7437 J mol−^1 − 7 .437 kJ mol−^1

∆H◦ 2 ∆fH◦(H 2 O)− 0 − 0 − 571 .660 kJ mol−^1

∆U◦− 571 .660 kJ mol−^1 −(− 7 .437 kJ mol−^1 )− 564 .223 kJ mol−^1