A mixture of `H_(2), N_(2)` and `NH_(3)` with molar concentration `5.0xx10^(-3) "mol" L^(-1), 4.0xx10^(-3) "mol" L^(-1)` and `2.0xx10^(-3) "mol" L^(-1)` respectively was prepared and heated to `500K`. The value of `K_(c)` for the reaction:
`3H_(2)(g)+N_(2)(g)hArr2NH_(3)(g)` at this temperature is 60. Predict whether ammonia tends to form or decompose at this stage of concenration.

Explanation

`underset(4xx10^(-3))(N_(2))+underset(5xx10^(-3))(3H_(2))hArr underset (2xx10^(-3) "initial concentrations")(2NH_(3))`
`Q=((2xx10^(-3))^(2))/((4xx10^(-3))(5xx10^(-3))^(3))=8xx10^(3)`
`K_(c)=60`
`QgtK_(c)`
Since Q is greater than `K_(c)`, backward reaction takes place. So decomposition of `NH_(3)` takes place.