initial (PCl3) = 0.60 mol/L
initial (Cl2) = 0.70 mol/L
equilibrium (PCl5) = 0.040 mols/L
.....................PCl3(g) + Cl2(g)⇄PCl5(g) Kc=0.11
I....................0.60 M......0.70 M........0
C......................-x M.........-x .....M...+x M
E................0.60-x M...0.70-x M........+x M
The problem tells you that 0.60-x = 0.040 M. Calculate x from that and evaluate the three terms of the E line. That will give you the concentrations in M of the three constituents.
Post your work if you get stuck., I worked this for someone yesterday but couldn't find it today. Perhaps you're posting the same question,
PCl3(g)+Cl2(g)⇄PCl5(g) Kc=0.11
A 0.60mol sample of PCl3(g) and a 0.70mol sample of Cl2(g) are placed in a previously evacuated 1.0L rigid container, and the reaction represented above takes place. At equilibrium, the concentration of PCl5(g) in the container is 0.040M . Find the concentration of each reactant at equilibrium
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