Friday
October 31, 2014

Homework Help: Chemistry

Posted by Lena on Sunday, July 19, 2009 at 6:18pm.

Initially, 1.68 mol of PCl5(g) and 0.36 mol of PCl3(g) are in mixed in a 2.00 l container. It is later found that 1.44 mol of PCl5 are present when the system has reached equillibrium. Calculate the value of the equllibrium.

This is what I did:

Equation:
PCl5(g) <--> PCl3(g) + Cl2(g)

PCl5(g): Intial Concentration = 1.68 mol/2.00l = 0.84 mol/L
Change in concentration =0.84 - (1.44 mol/2.00 L) = 0.84 mol/L -0.72 mol/L= 0.12 mol/L
Equallibrium = 0.12 mol/L

PCl3(g): Initial concentration = 0.36mol/2.00 L = 0.18 mol/L
Change in concentration = 0.12 mol/L + 0.18 mol/L = 0.30 mol/L
Equillibrium = 0.30 mol/L

Cl2(g):
Initial Concentration = 0 mol/L
Change in conc. = 0.12 mol/L
Equillibrium = 0.12 mol/L

K= [0.30][0.12]/[0.12]
= 0.3

Did I do this correctly?

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