1.

For the equilibrium N_(2)+3H_(2)hArr2NH_(3),K_(c) at 1000K is 2.73xx10^(-3) if at equlibrium [N_(2)]=2M,[H_(2)]=3M, the concentraion of NH_(3) is

Answer»

`0.00358` M
`0.0358` M
`0.358` M
`3.58` M

Solution :`K_(c)=([NH_(3)]^(3))/([N_(2)][H_(2)]^(3))`
`2.37xx10^(-3)=(X^(2))/([2][3]^(3))=x^(2)=0.12798impliesx=0.358M.`


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