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

Problem 14.54.

In a sealed 1.75–L vessel at 250 oC, equilibrium is established between PCl5(g) and its dissociation products, PCl3(g) and Cl2(g). The quantities found at equilibrium are 0.562 g PCl5, 1.950 g PCl3, and 1.007 g Cl2. What is the value of Kc for the reaction PCl5(g)PCl3(g) + Cl2(g)? What is the value of Kp?




Answer:

From the balanced reaction and the definition of Kc we can write:

PCl5(g)PCl3(g) + Cl2(g)

To find Kc, 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.

PCl5: [30.97 + 5(35.45)] = 208.22 g/mol.

PCl3: [30.97 + 3(35.45)] = 137.32 g/mol.

Cl2: [2(35.45)] = 70.90 g/mol.

Plugging each of these concentrations into the expression for Kc gives:

Kp = Kc(RT)n, n = (1+1) – 1 = 1; R = 0.0821 L•atm/mol•K; T = 250 + 273 = 523 K

Kp = 4.28×10–2[(0.0821)(523)]1 = 1.84