1.

For the reaction:N_(2)O_(4)(g)hArr 2NO_(2)(g)the concentration of the equilibrium mixture at 300 K are [N_(2)O_(4)]=4.8xx10^(-2)mol L^(-1) and [NO_(2)]=1.2xx10^(-2)mol L^(-1).K_(c ) for the reaction is :

Answer»

`0.25`
`3xx10^(-1)MOL L^(-1)`
`3xx10^(3)mol L^(-1)`
`3xx10^(-3)mol L^(-1)`

Solution :`K_(C )=([NO_(2)]^(2))/([N_(2)O_(4)])`
`=((1.2xx10^(-2))^(2))/((4.8xx10^(-2)))`
`=3xx10^(-3)mol L^(-1)`


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