Kc Calculations

I despise this subject with a passion but I need to learn it!!!

Bascially I need to work out the Kc for this:

= MEANS THE EQUILIBRIUM ARROW SIGNS
PCL5 = CL2 + PLC3

Analysis of the equilibrium mixture showed:
0.0042moles of PL5
0.040moles of PCL3
0.040moles of CL2

Total volume - 2.0 dm3

Calculate the conc of each species at equilibrium and then Kc

I attempted it and got Kc = 0.2moldm-3

But my answer probably is wrong -

Thanks in advance - Rep to first person who works it out!!

Reply 1

cpchem

OK...
[PCl₅] is 0.0042 mol / 2.0 dm³ = 0.0021 mol dm^-3

[Cl₂] is 0.040 mol / 2.0 dm³ = 0.020 mol dm^-3

[PCl₃] is 0.040 mol / 2.0 dm³ = 0.020 mol dm^-3

Given that K_c = [PCl_{3 eqm}][Cl_{2 eqm}]/[PCl_{5 eqm}]

K_c = (0.020 mol dm^-3)(0.020 mol dm^-3)/(0.0021 mol dm^-3) = 0.19 mol dm^-3

Your answer seems OK to me!

Reply 2

nk_angel89

Thanks - woohoo I actually got something right!!!

Rep given

Reply 3

giggsy

i will make sure youre right tmoz... i would do it now but im busy! hehe

Reply 4

giggsy

firstly its too easy a question. Kc= (pcl3)(pc12)/(pcl5)

Each substance has to be divided by the volume so:

pcl3:0.040/2=0.02
pcl5: 0.0042/2= 2.1*10^-3
cl2= 0.040/2=0.02

plug in the numbers

0.02*0.02/ (2.1*10^-3)= 0.19moldm-3

so yeah youre right love :smile: