Equilibrium constant:

Chemical reaction for the dissociation of $H_{2}O$ is given by:

$H_{2}O{H}_2+\frac{1}{2}{O}_2$

If the mole fractions were measured to be ${}_{H_{2}O}=0\mathrm{.}9,{}_{H_2}=0\mathrm{.}03$ and ${}_{O_2}=0\mathrm{.}07.$ And the total pressure is $0\mathrm{.}8$atm, find the equilibrium constant $K_p$ for this reaction.

Hints and formulae:

$K_P=prod{p}_i^{v_i}=p_{\text{t}\text{o}\text{t}\text{a}\text{l}}^{{\displaystyle \underset{?}{\overset{?}{}}}{v}_i}$prod${}_i^{v_i}$