1.

The decomposition of N_(2)O_(4) to NO_(2) is carried out at 280^(@)C in chloroform. When equilibrium is reached, 0.2 mol of N_(2)O_(4) and 2xx10^(-3) mol of NO_(2) are present in 2 litre solution. The equilibrium constant for the reaction, N_(2)O_(4)iff2NO_(2) is

Answer»

`1xx10^(-3)`
`2xx10^(-3)`
`1xx10^(-5)`
`2xx10^(-5)`

Solution :`K=([NO_(2)]^(2))/([N_(2)O_(4)])=((2xx10^(-3)//2)^(2))/((0.2//2))=(10^(-6))/(10^(-1))=10^(-5)`


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