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 December 7th, 2010, 07:45 PM #1 Member   Joined: May 2010 Posts: 45 Thanks: 0 Equilibrium 2SO3 + heat <-> 2SO2 + O2 A 4 liter container originally contains 0.6 moles of $SO_3$. At Equilibrium 0.12 moles of $O_2$ had been produced. Calculate $K_{eq}$ I was arguing with someone else over how this problem is done. Here is how I did it. [SO3]=0.6-0.12=0.48 [O2]=0.12 So since 2SO2: SO2=2[O2]=2(0.12)=0.24 [SO3]=0.6-0.12=0.48-0.24=0.24 [SO2]=0.24 [O2]=0.12 $K=\frac{[O_2][SO_2]^2}{[SO_3]^2}$ $K=\frac{(0.12)(0.24)^2}{(0.24)^2}$ $K=0.12$ $K=12x10^{-2}$ Am I right?
 December 8th, 2010, 06:11 AM #2 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 Re: Equilibrium i think u r right except for one important thing : in the formula of computing keq u should use concentrations not quantities so u should actually have had C(O2)=.12/4=0.03 C(so3)=0.24/4=0.06 C(SO2)=0.24/4=0.06 (C means concentration which is quantity / volume) then keq=0.03=3*10^(-2)
 December 8th, 2010, 02:35 PM #3 Member   Joined: May 2010 Posts: 45 Thanks: 0 Re: Equilibrium I thought of that during school today. I guess we were both wrong, he forgot about the ratio. I also wrote K=12x10^-2, not scientific notation. K=1.2x10^-1 Thank you for your help.

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