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For a reversible reaction: X+2Y rarr 2Z,...

For a reversible reaction: `X+2Y rarr 2Z`, the equilibrium concentrations of `X, Y` and `Z` are `0.32, 0.40` and `0.35` moles `L^(-1)` respectively at `25^(@)C`.
a. If unitially the system contained only `X` and `Y` and then reached the state of equilibrium, what were the initial concentrations of `X` and `Y`.
b. If at the start only `X` and `Z` were present, what were the initial concentrations?

Text Solution

Verified by Experts

`{:(X,+,2Y,hArr,2Z),(0.32,,0.4,,0.35 M):}`
`rArr K=([Z]^(2))/([X][Y]^(2))=2.392`
a. `{:(X,+,2Y,hArr,2Z),(a,,b,,-),(a-x,,b-2x,,2x):}`
`rArr2x=0.35rArrx=0.175`
`rArr a=[X]_(0)=x+0.32=0.495 M`
`b=[Y]_(0)=2x+0.4=0.75 M`
Since we start `X` and `Z`
`rArr 'Z'` will decomposes to produce `X+Y`
`{:(2Z,hArr,X,+,2Y,,),(p,,q,,,,),(P-y,,q+y//2,,y,,rArr y=0.4 M):}`
`P=[Z]=0.35+y=075 M`
`q=[X]=0.33-y//2=0.12 M`
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