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I understand that the equilibrium constant does not take into account the concentrations/ pressures of pure solids and liquids. How about the reaction quotient?

In the following reversible reaction, am I right to say that since water is a pure liquid, it is not reflected in the reaction quotient, hence removing water (without changing the temperature) will not shift the equation left/right after applying Le Chatelier's Principle?

$$\ce{H3PO4 (aq) + H2O <=> H3O+ (aq) + H2PO4- (aq)}$$

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  • $\begingroup$ You're wrong, it will. $\endgroup$ – Mithoron May 17 '16 at 19:14
  • $\begingroup$ @Mithoron: Thank you so much for your reply. Could you help me understand why? Do we actually have to include pure solids/liquids in the reaction quotient Q? Or do do we not use reaction quotient in applying Le Chatelier's Principle? Grateful for your help! $\endgroup$ – Psteo2891988 May 19 '16 at 3:26
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    $\begingroup$ They shouldn't appear in the reaction quotient, indeed. However, by removing water, you concentrate the solution, which leads to a shift of the equilibrium. $\endgroup$ – Thomas Jungers May 29 '17 at 11:03
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It isn't that solids don't appear in equilibrium constants and/or reaction quotients. They are simply assumed to have an activity equal to the value of 1! Hence people don't show them, but they should. I think it is important to understand and remember this assumption.

Also, equilibrium constants and reaction quotients are fundamentally a ratio of activities! The use of concentration and/or pressures in equilibrium constants/ reaction quotients is the results of approximations in themselves and this simplicity permeates chemistry and should be stopped or at least respected.

for the reaction you show, the equilibrium constant is

$$K_\mathrm{eq} = \frac{a_\ce{H2PO4-}a_\ce{H3O+}}{a_\ce{H3PO4}a_\ce{H2O}}$$

See the Chemistry LibreTexts for more details about solid and liquid activities and equilibrium constants.

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