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Chemistry LibreTexts

Special Equilibria

Chemical Equilibrium: Previous page: Mass Action Law

Skills to Develop

  • Write the equilibrium constant expression when the solvent is one of the products or reactants.
  • Write the equilibrium constant expression for heterogeneous equilibria.

We treat chemical equilibria in which the solvent is a reactant or product and heterogeneous equilibria as special chemical equilibria. The expressions for equilibrium constants of these systems are discussed. The mass action law is valid for the case when the solvent is a reactant or product. However, due to the large amount of solvent present, the equilibrium constant expression can be simplified. Special consideration also applies to heterogeneous equilibria, in which solids or liquids are involved. Phases such as liquid and solid are not sensitive to pressure, and their "concentrations" are constant as long as these phases are present. Their "concentrations" are not defined, and that is why we use a " " to mark the special meaning of these phrases. Perhaps "activity" or "property" is a better term to use than "concentration" in these cases.

Equilibrium Constant for Reactions Involving Solvents

As a solvent in the system, the concentration of the solvent is so large that its concentration rarely changes during the course of the reaction. For example, when water is used as a solvent, the concentration of water is almost always a constant. Its concentration can be calculated to be 55.6 M. (1 L of water has 1000 g, and molecular weight 18 g/mol; and thus \(\mathrm{[H_2O] = \dfrac{1000}{18} = 55.6\: mol/L}\))

For example, consider the following reaction,

\[\ce{CH3COOH + C2H5OH \rightleftharpoons CH3COOC2H5 + H2O}\]

According to the Mass Action Law, we can use either

\[K' = \mathrm{\dfrac{[CH_3COOC_2H_5] {\color{Red} [H_2O]}}{[CH_3COOH] [C_2H_5OH]}}\]


\[K = \ce{\dfrac{[CH3COOC2H5]}{[CH3COOH] [C2H5OH]}}\]

Of course,

\[K = \mathrm{\dfrac{\mathit K\,' }{ {\color{Red} [H_2O]}}} = \dfrac{K' }{ {\color{Red} 55.6}}\]

Heterogeneous Equilibria

Heterogeneous reactions involve at least two phases. That is two of gas, liquid, and solid are present as reactants or products. In heterogeneous equilibria, the activities (or concentrations) of solid and liquid but not gas are always a constant. In these cases, their activities or concentrations are omitted in the expression of Q or K. Let us look at these expressions for the equilibria:

\ce{CaCO_{3\large{(s)}} &\rightleftharpoons CaO_{\large{(s)}} + CO_{2\large{(g)}}} &&K_{\ce p} = \ce{P(CO2)}\\
\ce{2 H2O_{\large{(l)}} &\rightleftharpoons 2 H2O_{\large{(g)}}} &&K_{\ce p} = \ce{P^2(H2O)}\\
\ce{2 H2O_{\large{(l)}} &\rightleftharpoons 2 H_{2\large{(g)}} + O_{2\large{(g)}}} &&K_{\ce p} = \ce{P^2(H2) P(O2)}\\
\mathrm{AgCl_{\large{(s)}}} &\mathrm{\rightleftharpoons \sideset{ }{_{\large{(aq)}}^{+}}{Ag} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl}} &&K = \ce{[Ag^+] [Cl^- ]}

Heterogeneous equilibria discusses this type of system in more detail.


  1. What is the concentration of water in pure water?

  2. Give the Kp expression for the reaction

    \(\ce{HgO_{\large{(s)}} \rightleftharpoons Hg_{\large{(s)}} + O_{2\large{(g)}}}\)

  3. For the phase transition,

    \(\ce{H2O_{\large{(l)}} \rightleftharpoons H2O_{\large{(g)}}}\),

    the vapor pressure of water depends on temperature. At 298 K, the saturated vapor pressure is 23.8 torr. What is the equilibrium constant based on pressure Kp in atm? (1 atm = 760 torr)

  4. At some low temperature, the reaction of ethyl acetate with water,

    \(\ce{CH3COOC2H5_{\large{(aq)}} + H2O_{\large{(l)}} \rightleftharpoons CH3COOH_{\large{(aq)}} + C2H5OH_{\large{(aq)}}}\)

    has an equilibrium constant of 0.10. If \(\mathrm{[CH_3COOC_2H_5] = 0.90\: M}\) in a system that is at equilibrium, what is \(\ce{[C2H5OH]}\)?

  5. The value of Kp is 81 Pa at some temperature for the reaction,

    \(\ce{NH4Cl_{\large{(s)}} \rightleftharpoons NH_{3\large{(g)}} + HCl_{\large{(g)}}}\).

    What is the partial pressure (in Pa) of \(\ce{HCl}\) in an enclosed system containing solid \(\ce{NH4Cl}\) and its vapor?


  1. 55.6
    For most solutions, \(\mathrm{[H_2O] = \dfrac{1000}{18} =\: ?}\)

  2. \(\ce{P(O2)}\)
    This reaction is used to produce \(\ce{O2}\) from \(\ce{HgO}\).

  3. 0.031 atm
    \(K_{\ce p} = \ce{P(H2O)}\) in this case. Just want you to consider one of the common equilibrium states. The amount of water present does not affect the vapor pressure.

  4. 0.3

    \ce{&CH3COOC2H5_{\large{(aq)}} + H2O_{\large{(l)}} \rightleftharpoons \; &&CH3COOH_{\large{(aq)}} +\; &&C2H5OH_{\large{(aq)}}}\\
    &\hspace{55px}0.9  &&\hspace{40px}x  &&\hspace{30px}x

    \(\dfrac{x^2}{0.90} = K = 0.1\);       \(x =\: ?\)

    What was the original \(\ce{[CH3COOC2H5]}\) before any hydration reaction takes place? (Answer, 1.2)

  5. 9

    \(\ce{P(HCl)} = \sqrt{K_{\ce p}}\)

    \ce{NH4Cl_{\large{(s)}} \rightleftharpoons \; &NH_{3\large{(g)}} + \; &&HCl_{\large{(g)}}}\\
    &\:\:\:x  &&\:\:\:x

    \(K_{\ce p} = x^2\)

    The partial pressure of \(\ce{NH3}\) is the same as that of \(\ce{HCl}\) .