The thing to remember about a heterogeneous equilibrium, which as you know involves substances in more than one phase, is that changing the amount of a solid will not affect the position of the equilibrium. The same can be said for a pure liquid, with the mention here that the solutions in question must be rather dilute to begin with. For your given equilibrium reaction
adding more ammonium chloride, #"NH"_4"Cl"#, will not affect the position of the equilibrium. This is why the concentration of the solid is not included in the expressions for the two equilibrium constants #K_c# and #K_p#, which for this equilibrium will take the form
Now, without going into detail about this, the reason for why solids do not affect chemical equilibria can be traced back to their chemical activity. In simple terms, chemical activity is a measure of how the concentration of a substance under some specific conditions compares with the concentration of that substance under standard conditions. So, if a substance has the same concentration (you'll sometimes see this referred to as molar density) at some given conditions and at standard conditions, then its chemical activity is equal to #1#. Excluding extreme changes in pressure and temperature, solids will always have the same concentration. In your example, for a given temperature, the partial pressures and concentrations of the products will not depend on the amount of ammonium chloride present. Once the equilibrium is established, adding or removing ammonium chloride will not change its chemical activity, i.e. its concentration, so the position of the equilibrium remains unchanged. The law of chemical equilibrium states that, at any given temperature a chemical system reaches a state in which a particular ratio of reactant and product activities has a constant value. This constant is known as the equilibrium constant. For the generic reaction $mA+nB\rightleftharpoons xC+yD$ the equilibrium constant denoted by Keq is given by ${ K }_{ eq }=\frac { { [C] }^{ x }{ [D] }^{ y } }{ { [A] }^{ m }{ [B] }^{ n } }$ where [X] is the activity of X. The activity of X is equal to the concentration of X if it is a gas or liquid. The activity is equal to 1 if its a pure liquid or solid. In other words, pure solids and liquids do not affect the equilibrium constant as long as there is enough for the reaction to proceed. Their activity is 1, so they do not need to be written in the equilibrium constant. Since activity is a dimensionless quantity, the equilibrium constant, Keq, is also a dimensionless quantity. By convention, the equilibrium concentrations of the substances appearing on the right hand side of the chemical equation (the products) are always placed in the numerator of the equilibrium constant expression; the concentrations of the substances appearing on the left hand side of the chemical equation (the reactants) are placed in the denominator. The larger the value of the equilibrium constant, the more the reaction proceeds to completion. Irreversible reactions can be thought to have an infinite equilibrium constant since there are no reactants left .
The equilibrium constant,denoted by K, is the ratio of products to reactants at equilibrium. Example: Take the reaction: ${ H }_{ 2 }(g)+{ I }_{ 2 }(g)\rightleftharpoons 2HI(g)$ The forward reaction is: ${ H }_{ 2 }+{ I }_{ 2 }\rightarrow HI$ The reverse reaction is: $2HI{ \rightarrow H }_{ 2 }+{ I }_{ 2 }$ The equilibrium constant expression for this reaction is: ${ K }_{ eq }=\frac { { [HI] }^{ 2 } }{ { [H }_{ 2 }]{ [I }_{ 2 }] }$ The progress of an equilibrium reaction can be visualized. At time = 0, the rate of the forward reaction is high and the rate of the reverse reaction is low. As the reaction proceeds, the rate of the forward reaction decreases and the rate of the reverse reaction increases, until both occur at the same rate, reaching equilibrium.
It very much depends on what definition of the equilibrium constant you are looking. The most common usage of the same has quite a variety of possible setups, see goldbook:
The standard equilibrium constant is always unitless, as it is defined differently (goldbook)
It is worthwhile to note, that the first definition is always an approximation to the standard definition and in the standard definition all compounds regardless of their state are included in the equilibrium. If you are looking at a reaction that takes place in gas phase, solid materials will play a constant role, since their partial pressure will solely depend on their vapour pressure and can therefore be regarded as constant. Therefore it can be seen as a part of the equilibrium constant. (The same applies to liquids while deriving for gas phase.) If you would increase the amount of solid in the system you would still not change it's concentration in gas phase. For the reaction $$\ce{H2 (g) + I2 (s) <=> 2HI (g)}$$ you can form the standard equilibrium constant with activities/ fugacities $$K^\circ = \frac{a^2(\ce{HI})}{a(\ce{H2})\cdot{}a(\ce{I2})},$$ with $$a=\frac{f}{p^\circ}.$$ The activity for a pure solid is normally defined as one ($a(\ce{I2})=1$) and therefore $$K^\circ\approx K = \frac{a^2(\ce{HI})}{a(\ce{H2})}$$ For concentration dependent equilibrium constants, the following assumption is to be considered: $c(\ce{H2O})\approx55.6~\mathrm{mol/L}$ and is usually always much larger than that of any other component of the system in the range where the equilibrium approximation can be used. Also in most solution based reactions, the solute itself does not directly influence the reaction. Its concentration will therefore not change (significantly) and can be included in the equilibrium constant. You are correct in assuming that the kinetics of a surface reaction will only depend of the actual area of the surface (and of course the reactants forming the products). But here again this area is most likely to be considered as constant and will result in a scalar for the reaction rate (in the area where an accurate description of the reaction is possible). As of Greg's comment, He is of course right. In an equilibrium process the surface area will affect both directions of the reaction, hence it will not show up in the reaction rate at all. (The only effect it might have is, that the equilibrium state will be reached faster.) The equilibrium constant, K, expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit.This article explains how to write equilibrium constant expressions, and introduces the calculations involved with both the concentration and the partial pressure equilibrium constant. A homogeneous reaction is one where the states of matter of the products and reactions are all the same (the word "homo" means "same"). In most cases, the solvent determines the state of matter for the overall reaction. For example, the synthesis of methanol from a carbon monoxide-hydrogen mixture is a gaseous homogeneous mixture, which contains two or more substances: \[ CO (g)+ 2H_2 (g) \rightleftharpoons CH_3OH (g)\] At equilibrium, the rate of the forward and reverse reaction are equal, which is demonstrated by the arrows. The equilibrium constant, however, gives the ratio of the units (pressure or concentration) of the products to the reactants when the reaction is at equilibrium. The synthesis of ammonia is another example of a gaseous homogeneous mixture: \[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
A heterogeneous reaction is one in which one or more states within the reaction differ (the Greek word "heteros" means "different"). For example, the formation of an aqueous solution of lead(II) iodide creates a heterogeneous mixture dealing with particles in both the solid and aqueous states: \[PbI_{2 (s)} \rightleftharpoons Pb^{+2}_{(aq)} + 2I^-_{(aq)}\] The decomposition of sodium hydrogen carbonate (baking soda) at high elevations is another example of a heterogeneous mixture, this reaction deals with molecules in both the solid and gaseous states: \[ 2NaHCO_{3 (s)} \rightleftharpoons Na_2CO_{3 (s)} + H_2O_{ (g)} + CO_{2 (g)} \] \[ C_{(s)} + O_{2 (g)} \rightleftharpoons CO_{2 (g)} \] This difference between homogeneous and heterogeneous reactions is emphasized so that students remember that solids, pure liquids, and solvents are treated differently than gases and solutes when approximating the activities of the substances in equilibrium constant expressions.
The numerical value of an equilibrium constant is obtained by letting a single reaction proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. The ratio of the product concentrations to reactant concentrations is calculated. Because the concentrations are measured at equilibrium, the equilibrium constant remains the same for a given reaction independent of initial concentrations. This knowledge allowed scientists to derive a model expression that can serve as a "template" for any reaction. This basic "template" form of an equilibrium constant expression is examined here.
The thermodynamically correct equilibrium constant expression relates the activities of all of the species present in the reaction. Although the concept of activity is too advanced for a typical General Chemistry course, it is essential that the explanation of the derivation of the equilibrium constant expression starts with activities so that no misconceptions occur. For the hypothetical reaction: \[bB + cC \rightleftharpoons dD + eE \] the equilibrium constant expression is written as \[ K = \dfrac{a_D^d ·a_E^e}{a_B^b · a_C^c}\] *The lower case letters in the balanced equation represent the number of moles of each substance, the upper case letters represent the substance itself.
To avoid the use of activities, and to simplify experimental measurements, the equilibrium constant of concentration approximates the activities of solutes and gases in dilute solutions with their respective molarities. However, the activities of solids, pure liquids, and solvents are not approximated with their molarities. Instead these activities are defined to have a value equal to 1 (one).The equilibrium constant expression is written as \(K_c\), as in the expression for the reaction: \[HF_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + F^-_{(aq)} \] \[ K_c = \dfrac{a_{H_3O^+}· a_{F^-}}{a_{HF} · a_{H_2O}} ≈ \dfrac{[H_3O^+][F^-]}{[HF](1)} = \dfrac{[H_3O^+][F^-]}{[HF]} \] Here, the letters inside the brackets represent the concentration (in molarity) of each substance. Notice the mathematical product of the chemical products raised to the powers of their respective coefficients is the numerator of the ratio and the mathematical product of the reactants raised to the powers of their respective coefficients is the denominator. This is the case for every equilibrium constant. A ratio of molarities of products over reactants is usually used when most of the species involved are dissolved in water. A ratio of concentrations can also be used for reactions involving gases if the volume of the container is known. .
Gaseous reaction equilibria are often expressed in terms of partial pressures. The equilibrium constant of pressure gives the ratio of pressure of products over reactants for a reaction that is at equilibrium (again, the pressures of all species are raised to the powers of their respective coefficients). The equilibrium constant is written as \(K_p\), as shown for the reaction: \[aA_{(g)} + bB_{(g)} \rightleftharpoons gG_{(g)} + hH_{(g)} \] \[ K_p= \dfrac{p^g_G \, p^h_H}{ p^a_A \,p^b_B} \]
To convert Kc to Kp, the following equation is used: \[K_p = K_c(RT)^{\Delta{n_{gas}}}\] where:
Another quantity of interest is the reaction quotient, \(Q\), which is the numerical value of the ratio of products to reactants at any point in the reaction. The reaction quotient is calculated the same way as is \(K\), but is not necessarily equal to \(K\). It is used to determine which way the reaction will proceed at any given point in time. \[Q = \dfrac{[G]^g[H]^h}{[A]^a[B]^b}\]
The same process is employed whether calculating \(Q_c\) or \(Q_p\).
The most important consideration for a heterogeneous mixture is that solids and pure liquids and solvents have an activity that has a fixed value of 1. From a mathematical perspective, with the activities of solids and liquids and solvents equal one, these substances do not affect the overall K or Q value. This convention is extremely important to remember, especially in dealing with heterogeneous solutions.
Example \(\PageIndex{1}\) In a hypothetical reaction: \[ aA_{(s)} + bB_{(l)} \rightleftharpoons gG_{(aq)} + hH_{(aq)} \] The equilibrium constant expression is written as follows: \[K_c = \dfrac{[G]^g[H]^h}{1 \times 1} = [G]^g[H]^h\] In this case, since solids and liquids have a fixed value of 1, the numerical value of the expression is independent of the amounts of A and B. If the product of the reaction is a solvent, the numerator equals one, which is illustrated in the following reaction: \[ H^+_{(aq)} + OH^–_{(aq)} \rightarrow H_2O_{ (l)}\] The equilibrium constant expression would be: \[ K_c= \dfrac{1}{ [H^+][OH^-]}\] which is the reciprocal of the autoionization constant of water (\(K_w\)) \[ K_c = \dfrac{1}{K_w}=1 \times 10^{14}\]
The equilibrium constant expression must be manipulated if a reaction is reversed or split into elementary steps. When the reaction is reversed, the equilibrium constant expression is inverted. The new expression would be written as: \[K'= \dfrac{1}{\dfrac{[G]^g[H]^h}{[A]^a[B]^b}} = \dfrac{[A]^a[B]^b}{[G]^g[H]^h}\] When there are multiple steps in the reaction, each with its own K (in a scenario similar to Hess's law problems), then the successive K values for each step are multiplied together to calculate the overall K.
Activities Because the concentration of reactants and products are not dimensionless (i.e. they have units) in a reaction, the actual quantities used in an equilibrium constant expression are activities. Activity is expressed by the dimensionless ratio \(\frac{[X]}{c^{\circ}}\) where \([X]\) signifies the molarity of the molecule and c is the chosen reference state: \[ a_b=\dfrac{[B]}{c^{\circ}} \] For gases that do not follow the ideal gas laws, using activities will accurately determine the equilibrium constant that changes when concentration or pressure varies. Thus, the units are canceled and \(K\) becomes unitless. Practice Problems
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What is \(K_c\) for the Reaction 1) Kc: 24.5 Kp: 1.002 Atm 2) Qc= 83.33 > Kc therefore the reaction shifts to the left
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What substances do not appear in the equilibrium constant expression?
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