##### 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?

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