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

`N_(2)+3_(2_((g))) hArr 2NH_(3_((g)))`
`Q_(C)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
`=((2xx10^(-3))^(2))/(5xx10^(-3)xx(4xx10^(-3))^(3))`
`=(2^(2)xx10^(-6))/(5xx4^(3)xx10^(-3)xx10^(-9))`
`=(2^(3)xx10^(6))/(64xx5)`
`=(2^(3)xx10^(6))/(64xx5)`
`=(8)/(64xx5)xx10^(6)`
`=(10^(6))/(40)=(1)/(4)xx10^(5)=25xx10^(3)`
Given `K_(c)=60`
`K_(c) lt Q_(c)`
`:.` Backward reaction i.e., decomposition of `NH_(3)` take place