For the formation of `NH_3` from `H_2` at 500 K, The concentration of `N_2, H_2 and NH_3` at equilibrium are `1.5xx10^(-2) M , 3.0 x 10^(-2) M and 1.2 xx 10^(-2) M`, respectively. The equilibrium constant for the reverse reaction is

Options

A. `3.56 xx 10^2`
B. `2.81 xx 10^(-3)`
C. `3.56 xx 10^(-2)`
D. `2.81 xx 10^3`

Answer

B

Explanation

`3H_2 + N_2 to 2 NH_3`
Equilibrium constant
for reverse reaction `K_e = ([H_2]^3 [N_2])/([NH_3]^2)`
`((3.0 xx 10^(-2))^3 (1.5 xx 10^(-5)))/((1.2 xx 10^(-2)))`
`= 2.81 xx 10^(-3)`