equilibrium constant question (1 Viewer)

[~GrOoVy~]

COF2 <-> CF4 + CO2 (all of them are in a gaseous state )

carbon oxyfluoride <-> tetrafluoride + Carbon dioxide


okay now to the question:
The reaction is carried out at 200 degrees in a fixed-volume 10L container. Initially, there are 2moles of carbon oxyfluoride gas in the flask. At equilibrium, 80% of the carbon oxyfluoride has decomposed.

Determine the value of the equilibrium constant.

 

[jetfan]

COF2 <-> CF4 + CO2 (all of them are in a gaseous state )

carbon oxyfluoride <-> tetrafluoride + Carbon dioxide


okay now to the question:
The reaction is carried out at 200 degrees in a fixed-volume 10L container. Initially, there are 2moles of carbon oxyfluoride gas in the flask. At equilibrium, 80% of the carbon oxyfluoride has decomposed.

Determine the value of the equilibrium constant.

Well this sin't my strong suit but it lookes like an ICE question (Initial, change, equilibrium )

COF2 <-> CF4 + CO2
I 0.2 0 0
C -0.16 0.08 0.08
E 0.04 0.08 0.08

K = [0.08]^2 / 0.04
= 0.16

If this looks wrong or you can't understand it, feel free to comment

--jetfan--

 

[richz]

hmm.... i dont think i agree with that since it the ratio is 1:1:1 so it would be

K= (0.04)^2/(0.04)

=0.04

i maybe rong, not so sure though

 

[jetfan]

Ok i'll tell you how i figured it, if the reactant loses 80%, it goes from 2mol to 0.4 mol. Therefore there is 1.6 mol used up in the reaction. This means that the products must total 1.6mol, and because they are both in the same mole ratio, they must have used 0.8 mol each. Thus the concentration must be 0.08M for each. This right??

 

[~GrOoVy~]

Thanks Jetfan and xrtzx


the answer is K=4

Jetfan u forgot to balance the equation and to square the bottom number.

 

[jetfan]

yeah, i just figured that it was balanced, apart from that i'd have been fine, ta