# Can someone help me please? I didn't do it right then I got it right. Thanks!

Aug 22, 2017

An equilibrium constant ${K}_{e q}$ is written so that its mass action expression uses the stoichiometric coefficients in the reaction in its exponents (except if they are $1$).

overbrace(color(red)(2)"H"_2"O"(g))^"reactant" rightleftharpoons overbrace(color(red)(2)"H"_2(g) + color(red)(1)"O"_2(g))^"products"

The mass action expression is written with products on top and reactants on the bottom.

color(blue)(K_(eq) = overbrace((["H"_2]^color(red)(2)["O"_2])/(["H"_2"O"]^color(red)(2)))^"mass action expression")

Remember to exclude the pure liquids and solids. Only include aqueous and gaseous substances.

Aug 22, 2017

Well for $A + B r i g h t \le f t h a r p \infty n s C + D$, ${K}_{\text{eq}} = \frac{\left[C\right] \left[D\right]}{\left[A\right] \left[B\right]}$

#### Explanation:

For the given, gas-phase equation, each reagent and product is raised to the power of its stoichiometric coefficient......i.e.

$2 {H}_{2} O \left(g\right) r i g h t \le f t h a r p \infty n s 2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right)$

${K}_{\text{eq}} = \frac{{\left[{H}_{2} \left(g\right)\right]}^{2} \left[{O}_{2} \left(g\right)\right]}{{\left[{H}_{2} O \left(g\right)\right]}^{2}}$.....

....which looks like $B$ to me.