How would you write the equilibrium constant expression for the following #H_2O_((g))# #rightleftharpoons# #H_2##(g)# #+# #1/2# #O_2##(g)# ?

Your balanced chemical equation looks like this

#"H"_2"O"_text((g]) rightleftharpoons "H"_text(2(g]) + color(red)(1/2)"O"_text(2(g])#

As you know, the equilibrium constant is defined as the ratio between the equilibrium concentrations of the products and the equilibrium concentrations of the reactants, all raised to the power of their stoichiometric coefficients.

In this case, you have

#K_c = (["H"_2]^1 * ["O"_2]^color(red)(1/2))/(["H"_2"O"]^1) = (["H"_2] * ["O"_2]^color(red)("1/2"))/(["H"_2"O"])#

To determine the units for #K_c#, simply use the fact that molarity is measured in moler per liter, or molar, #"M"#.

Using just the units, you would have

#K_c = (color(red)(cancel(color(black)("M"))) * "M"^color(red)("1/2"))/(color(red)(cancel(color(black)("M")))) = "M"^(1/2)#

Alternatively, you can write this as

#sqrt("M") = sqrt("mol L"^(-1)) = "mol"^(1/2) "L"^(-1/2)#

Notice what would happen to the equilibrium constant if you were to rewrite the equation as

#color(blue)(2)"H"_2"O"_text((g]) rightleftharpoons color(blue)(2)"H"_text(2(g]) + "O"_text(2(g])#

This time ,you would have

#K_c^' = (["H"_2]^color(blue)(2) * ["O"_2])/(["H"_2"O"]^color(blue)(2))#

This is actually equal to

#K_c^' = ((["H"_2] * ["O"+2]^("1/2"))/(["H"_2"O"]))^2 = K_c^2#

The units here would be

#K_c = (color(red)(cancel(color(black)("M"^2))) * "M")/color(red)(cancel(color(black)("M"^2))) = "M" = "mol L"^(-1)#