Calculating `K_c`: Some nitrogen and hydrogen gas are placed in an empty `2.50L` container at `500^@C`. When equilibrium is established, `1.51` mol of `N_2`, `1.05` mol of `H_2`, and `0.283` mol of `NH_3` are present. Calculate `K_c` for the following reaction at `500^@C` :
`N_2(g)+3H_2(g)hArr NH_3(g)`

Explanation

Strategy: To calculate `K_c`, we need equilibrium concentrations, which are obtained by dividing the number of equilibrium moles of each reacting substance by the volume of the container.
Solution:
`C=n/V`
`:. C_(N_2)=(n_(N_2))/(V)=(1.51 mol)/(2.50 L)=0.604M`
`C_(H_2)=(n_(H_2))/(V)=(1.05 mol)/(2.50 L)=0.420 M`
`C_(NH_3)=(n_(NH_3))/(V)=(0.283 mol)/(2.50 L)=0.113M`
The equilibrium constant, `K_c`, is given by
`K_c=(C_(NH_3)^2)/(C_(N_2)C_(H_2)^3)`
Substituting the equilibrium concentrations, we find that
`K_c=((0.113)^2)/((0.604)(0.420)^3)`
`=0.285`