1.

One mole of N_(2)O_(4) is heated in a flask with a volume of 10dm^(3). At equilibrium 1.708 mole of NO_(2) and 0.146 mole of N_(2)O_(4) were found at 134^(@)C. The equilibrium constnt will be

Answer»

250 mol `dm^(-3)`
300 mol `dm^(-3)`
2 mol `dm^(-3)`
230 mol `dm^(-3)`

SOLUTION :`N_(2)O_(4)hArr2NO_(2)`
`{:("Moles at "eqb^(m),0.146,1.708),(Conc.,,),((mol//dm^(3)),[(0.146)/(10)],[(1.708)/(10)]):}`
`thereforeK_(c)=([NO_(2)]^(2))/([N_(2)O_(4)])=([(1.708)/(10)])/([(0.146)/(10)])`
`impliesK_(c)=(2.917264)/(100)xx(10)/(0.146)`
`IMPLIES K_(c)=(2.917264)/(1.46)=2"mol dm"^(-3)`


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