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

Problem 14.54.

In a sealed 1.75–L vessel at 250 ^{o}C, equilibrium is established between PCl_{5}(*g*) and its dissociation products, PCl_{3}(*g*) and Cl_{2}(*g*). The quantities found at equilibrium are 0.562 g PCl_{5}, 1.950 g PCl_{3}, and 1.007 g Cl_{2}. What is the value of K_{c} for the reaction PCl_{5}(*g*)PCl_{3}(*g*) + Cl_{2}(*g*)? What is the value of K_{p}?

From the balanced reaction and the definition of K_{c} we can write:

PCl_{5}(*g*)PCl_{3}(*g*)
+ Cl_{2}(*g*)

To find K_{c}, then, we need to find the equilibrium concentration
of each component in the reaction from the given masses, the volume of
the container, and the molar mass of each species.

PCl_{5}: [30.97 + 5(35.45)] = 208.22 g/mol.

PCl_{3}: [30.97 + 3(35.45)] = 137.32 g/mol.

Cl_{2}: [2(35.45)] = 70.90 g/mol.

Plugging each of these concentrations into the expression for K_{c} gives:

K_{p} = K_{c}(RT)^{n},
n = (1+1) – 1 = 1; R = 0.0821 L•atm/mol•K;
T = 250 + 273 = 523 K

K_{p} = 4.28×10^{–2}[(0.0821)(523)]^{1} = 1.84