Physical Chemistry , 1st ed.

Example 5.2

The following reaction is set up with the initial amounts of each substance

listed below.

6H 2  P 4 →4PH 3

18.0 mol 2.0 mol 1.0 mol

In each of the following scenarios, show that no matter which species is used

to determine , the value for is the same.

a.All the P 4 reacts to form products.

b.All the PH 3 reacts to form reactants.

Solution

a.If 2.0 mol P 4 reacts, no P 4 will be left, so nP 4 0.0 mol. Of the H 2 ,

12.0 mol will have reacted, leaving 6.0 mol H 2 (nH 2 6.0). This produces

8.0 mol PH 3 , which in addition to the 1.0 mol initially will give nPH 3 

9.0 mol. Using the definition of and the appropriate values for each chem-

ical species:

     2.0 mol using H 2



0.0 mol





1

2.0 mol

 2.0 mol using P 4



9.0 mol





4

1.0 mol

 2.0 mol using PH 3

Note that we have used positive or negative values of (^) i, as appropriate, and
that extent has units of mol.
b.If all of the PH 3 reacts,nPH 3 would be zero and H 2 and P 4 would have
gained 1.5 mol and 0.25 mol, respectively. Therefore,
 0.25 mol using H 2
 0.25 mol using P 4
0.0 mol




4

1.0 mol 0.25 mol using PH

3

These examples should convince you that has the same value no matter

which species is used and therefore is a consistent way to follow the course of

a chemical reaction. In addition, we also see that is positive when a chemical

process moves to the right side of the reaction, and negative when it moves to

the left side of a reaction.

When a reaction proceeds, the amounts nichange. The infinitesimal change

in each amount,dni, can be written in terms of the extent using the relation-

ship in equation 5.1:

dni

id (5.2)

As the nivalue changes, so does the Gibbs free energy of the system, according

to equation 4.48 from the last chapter:

dGS dTV dp

0

i

idni

2.25 mol 2.0 mol



 1

19.5 mol 18.0 mol



 6

6.0 mol 18.0 mol

 6

122 CHAPTER 5 Introduction to Chemical Equilibrium