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

2 Answers
Aug 22, 2017

An equilibrium constant #K_(eq)# 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

Answer:

Well for #A+BrightleftharpoonsC+D#, #K_"eq"=([C][D])/([A][B])#

Explanation:

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

#2H_2O(g)rightleftharpoons2H_2(g) + O_2(g)#

#K_"eq"=([H_2(g)]^2[O_2(g)])/([H_2O(g)]^2)#.....

....which looks like #B# to me.