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April 19, 2014

April 19, 2014

Posted by **Anonymous** on Saturday, February 12, 2011 at 10:55pm.

a) Find the concentration of N2O5 in the equilibrium system.

b) Find the concentration of N2O in the equilibrium system.

c) Find the concentration of N2O3 in the equilibrium system.

- chemistry -
**DrBob222**, Sunday, February 13, 2011 at 6:17pmGet a second opinion on this AND check my thinking..

N2O5 ==> N2O3 + O2

N2O3 ==> N2O + O2

Add the two equations to obtain

N2O5 ==> N2O + 2O2 and k for this reaction is k1*k2 = 7.75*4 = 31.0 = (N2O)(O2)/(N2O5).

The final O2 is 4.5 moles; some came from reaction 1 and some from reaction 2. How much came from each. If we call x the amount from 1, then 4.00/7.75 must have come from rxn 2 or

1+(4.00/7.75) = 4.5 and x = 2.97

The amount from rxn 2 then must be 2.97*(4.00/7.75) = 1.53 (total is still 4.50).

From rxn 2,

N2O3 ==> N2O + O2

If O2 from this rxn is 1.53, that means N2O must be 1/2 that or 0.765 (that is part (c)). From

4.00 = (N2O)(O2)/(N2O3), we solve for N2O3 (substitute 4.5 for O2 because the O2 in the final mixture can't tell where it came from and we know the total is 4.5). That gets part(b).

For part a, concn N2O5 at equilibrium, use K1K2 = 31.0 = (N2O3)(O2)/(N2O5)

You know N2O3 and O2.

Just to check things, if we substitute the values I obtained for N2O3 (0.861) and 4.5 for O2 and 0.500 for N2O5, K1 gives 7.75 which was K in the problem. Again, get a second opinion and check my thinking.

- chemistry -
**Nika**, Tuesday, February 15, 2011 at 2:08amHi, I'm doing the same assignment. I got to the part where you added the reactions and got the K3 = 31. Then, I got stuck

I don't understand why you're doing the 4.00/7.75 and then stating that 1 +(4.00/7.75) = 4.5? isn't that 1.52 instead of 4.5?

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