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

Problem 14.79.

An analysis of the gaseous phase [S2(g) and CS2(g)] present at equilibrium at 1009 oC in the reaction C(s) + S2(g)CS2(g) shows it to be 13.71% C and 86.29% S, by mass. What is Kc for this reaction?




Answer:

For the reaction C(s) + S2(g)CS2(g)

, i.e., the volumes cancel for this reaction.

Since volume is unimportant for finding the equilibrium constant, suppose the volume of gas chosen for analysis contained 100.0 g of sample. This means that of the 100.0 g, 13.71 g of the sample was C and 86.29 g of the sample was S, based on the analysis. The number of moles of C in the gas phase sample = 13.71 g / 12.011 g/mol = 1.141 mol. All of the carbon in the gas phase is in the carbon disulfide, so the 100.0 g sample must contain 1.141 mol CS2. The total number of moles of sulfur in the 100.0 g sample = 86.29 g / 32.066 g/mol = 2.691 mol. Of this total, 2(1.141) = 2.282 mol of sulfur is used in CS2, leaving 2.691 – 2.282 = 0.409 mol sulfur in S2, or 0.409/2 = 0.205 mol S2. Thus, at equilibrium, in the 100.0 g sample there is 1.141 mol CS2 and 0.205 mol S2. Now Kc can be found:
Kc = (1.141)/(0.205) = 5.57.