Scienuk Posted July 10, 2016 Report Share Posted July 10, 2016 (edited) Can someone please explain the rules for writing equilibrium constant expressions as shown in my book: 1. For an aqueous reaction the concentration of the solvent water does not appear in the equilibrium constant expression as its concentration does not change during the reaction. 2. If the reaction takes place in a non-aqueous solution, water must be included in the Kc expression as any other reactant or product. Please provide examples Edited July 10, 2016 by Scienuk Reply Link to post Share on other sites More sharing options...
kw0573 Posted July 10, 2016 Report Share Posted July 10, 2016 If water is the solvent and is involved in the reaction, it's implied that that its concentration only changes very marginally, and hence does not affect rate, hence does not affect equilibrium. Kw = [H3O+][OH-]. Note that high rate constant does not mean high equilibrium constant, but rather if the forward reaction rate become really high, then the products become more favoured in equilibrium. In addition, only aqueous, gaseous, and miscible liquids appear in the equilibrium expression, because solids and immiscible liquids do not change (to any great extent) the number of molecules exposed to the solvent and other reactants (remain the same surface area). When water is not present at a really high amount (AND is miscible or gaseous), any changes in water levels will be significant. Equilibrium of combustion of methane: Kc = [CO2][H2O]2/([CH4][O2]2) In this case, water is a gaseous product. 1 Reply Link to post Share on other sites More sharing options...
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