Given the reaction between 2 gases represented by `A_(2)` and `B_(2)` to given the compound AB(g). `A_(2)(g) + B_(2)(g)hArr 2AB(g)`
At equilibrium, the concentrtation
of `A_(2) = 3.0xx10^(-3) M`
of `B_(2) = 4.2xx10^(-3) M`
of `AB = 2.8xx10^(-3) M`
If the reaction takes place in a sealed vessel at `527^(@)C` . then the value of `K_(c)` will be

Options

A. `2.0`
B. `1.9`
C. `0.62`
D. `4.5`

Answer

C

Explanation

`A_(2)(g) +B_(2)(g) hArr 2AB(g)`
`K_(c)=[[AB]^(2)]/[[A_(2)][B_(2)]]`
`=((2.8 xx 10^(-3) M^(2)))/((3.0 xx 10^(-3)M)xx (4.2 xx 10^(-3) M))`
`=(2.8 xx 2.8)/(3.0 xx 4.2) =0.62`