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

## P.S. I am totally open to the idea of just changing this equation... BUT that would be a problem beacuse were supposed to answer this QUESTION Also, if you could (pretty please) determine the units for the expression. Thanks!

Feb 15, 2016

Here's what I got.

#### Explanation:

Your balanced chemical equation looks like this

${\text{H"_2"O"_text((g]) rightleftharpoons "H"_text(2(g]) + color(red)(1/2)"O}}_{\textrm{2 \left(g\right]}}$

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

${K}_{c} = \left(\left[\text{H"_2]^1 * ["O"_2]^color(red)(1/2))/(["H"_2"O"]^1) = (["H"_2] * ["O"_2]^color(red)("1/2"))/(["H"_2"O}\right]\right)$

To determine the units for ${K}_{c}$, simply use the fact that molarity is measured in moler per liter, or molar, $\text{M}$.

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

$\textcolor{b l u e}{2} {\text{H"_2"O"_text((g]) rightleftharpoons color(blue)(2)"H"_text(2(g]) + "O}}_{\textrm{2 \left(g\right]}}$

This time ,you would have

${K}_{c}^{'} = \left({\left[\text{H"_2]^color(blue)(2) * ["O"_2])/(["H"_2"O}\right]}^{\textcolor{b l u e}{2}}\right)$

This is actually equal to

${K}_{c}^{'} = {\left(\left(\left[\text{H"_2] * ["O"+2]^("1/2"))/(["H"_2"O}\right]\right)\right)}^{2} = {K}_{c}^{2}$

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)