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Problem 14.65

To establish equilibrium in the following reaction at 250 ^{o}C,

PCl

_{3}(g) + Cl_{2}(g) PCl_{5}(g) K_{c}= 26 at 250^{o}C

0.100 mol each of PCl_{3} and Cl_{2} and 0.0100 mol PCl_{5} 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.

PCl |
+ |
Cl |
PCl |
||

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 K_{c})

Since x must be less than 0.0156 (so that the PCl_{3} and Cl_{2} concentrations are positive at equilibrium), the correct value for x is 0.0027.

Thus, [PCl_{3}]_{e} = 0.0156 – 0.0027 = 0.0129 M, [Cl_{2}]_{e} = 0.0156 – 0.0027 = 0.0129 M, and [PCl_{5}]_{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:

PCl_{3}: 0.0129×6.40 = 0.0826 mol

Cl_{2}: 0.0129×6.40 = 0.0826 mol

PCl_{5}: 0.0043×6.40 = 0.028 mol