1.

For the reaction equlibrium N_(2)O_(4)hArr2NO_(2(g))the concentration of N_(2)O_(4) and NO_(2) at equlibrium are 4.8xx10^(-2)and 1.2xx10^(-2)"mol lite"^(-1) respectively. The value of K_(c) for the reaction is

Answer»

`3.3xx10^(2)"mol litre"^(-1)`
`3xx10^(-1)"mol litre"^(-1)`
`3xx10^(-3)"mol litre"^(-1)`
`3xx10^(3)"mol litre"^(-1)`

Solution :`K=([NO_(2)]^(2))/([N_(2)O_(4)])=([1.2xx10^(-2)])/([4.8xx10^(-2)])=0.3xx10^(-2)=3xx10^(-3)`


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