CHM112 Home Page Reference Materials Homework Assignments Lectures Exams & Quizzes Grades Study Aids URI Home Page


J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry, 4th edition.

Problem 14.65

To establish equilibrium in the following reaction at 250 oC,

PCl3(g) + Cl2(g) PCl5(g)    Kc = 26 at 250 oC

0.100 mol each of PCl3 and Cl2 and 0.0100 mol PCl5 are introduced into a 6.40-L reaction flask. How many moles of each of the gases will be present when equilibrium is established?






Answer:

First, convert the initial amounts of species from moles to molarity. Then do the usual equilibrium problem. Finally, convert the calculated molarities back to moles to obtain the answers.


 

PCl3(g)

+

Cl2(g)

PCl5(g)

 

Initial

0.100/6.40 = 0.0156

 

0.100/6.40 = 0.0156

 

0.0100/6.40 = 0.00156

Change

– x

 

– x

 

+ x

Equilibrium

0.0156 – x

 

0.0156 – x

 

0.00156 + x

Solving with the quadratic equation gives

x = 0.067 or x = 0.0027 (2 significant figures because of Kc)

Since x must be less than 0.0156 (so that the PCl3 and Cl2 concentrations are positive at equilibrium), the correct value for x is 0.0027.

Thus, [PCl3]e = 0.0156 – 0.0027 = 0.0129 M, [Cl2]e = 0.0156 – 0.0027 = 0.0129 M, and [PCl5]e = 0.00156 + 0.0027 = 0.0043 M.

The number of moles of each component is found by multiplying each molarity by the volume of the reaction flask:

PCl3: 0.0129×6.40 = 0.0826 mol

Cl2: 0.0129×6.40 = 0.0826 mol

PCl5: 0.0043×6.40 = 0.028 mol