The following concentrations were obtained for the formation of `NH_(3)` from `N_(2)` and `H_(2)` at equilibrium at `500 K`. `[N_(2)]=1.5xx10^(-2) M, [H_(2)]=3.0xx10^(-2)M,` and `[NH_(3)]=1.2xx10^(-2)M`. Calculate the equilibrium constant.

Options

A. `1.08 xx 10^(4)`
B. `3.98 xx 10^(2)`
C. `1.06 xx 10^(3)`
D. `2.93 xx 10^(4)`

Answer

B

Explanation

The equilibrium constant for the reaction
`N_(2)(g) +3H_(2) hArr 2NH_(3)(g)`can be written as
`K_(C) = ([NH_(3)(g)]^(2))/([N_(2)(g)][H_(2)(g)]^(3))`
`= ((1.2 xx 10^(-2))^(2))/((1.5 xx 10^(-2))(3.0 xx 10^(-2))^(3))`
`= 0.0355 xx 10^(4) = 3.55 xx 10^(2)`
`= 3.98 xx 10^(2)`