thermodynamics help!!

[anonymous1]

0.482mol of n2 and 0.933mol of o2 are put in a vessel and allowed to react (volume 10L) Kc= 2.0 * 10^ -13
equation is 2N2 + 02 <------> 2N20

what will be the concentrations of each of the individual gases at equilibrium?

what will be Qc if volume is doubled?

thanks

[John President]

This looks like it will turn into a very ugly equation, unfortunately!

The first step though would be to form a RICE equilibrium concentration table to work out the concentrations of N2 and O2 in terms of the concentration of N2O. It looks like you'll end up with a tremendously ugly cubic, good luck with that!

As for the second part of the question, if the volume of the container is doubled that means that the concentration of each reactant is instantly halved. Therefore the new Qc = [N20/2]^2 / {[N2/2]^2 * [O2/2]^2 }. This will be equal to 2*Kc: ie Qc > Kc, hence the reaction will move to the left.

[anonymous1]

for part a they just used the number of mols and volume to find cocentration, and then to find N20 they used kc and the other two concentrations...i don't really get this though

i understand the second part now though, thanks!

[BlueSky_3]

aA<-->bB+cC

the equilibrium law is written as:

Kc=^b+[C]^c
---------------------
[A]^a

i.e. the concentrations of the substances to the power of the coefficient of the reaction.

Then rearrange and solve.

[anonymous1]

thanks bluesky