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The relationship between \(K_c\) and \(K_p\)

What is the relationship between the equilibrium constant in terms of concentration \(K_c\) and the equilibrium constant in terms of partial pressure \(K_p\)?

For an ideal gas,

p=\frac{n}{V}RT\; \; \Rightarrow\; \; p=[concentration]RT\; \; \; \; \; \; \; \; 1

Substituting eq1 into the general equilibrium constant expression in terms of concentration gives:

K_c=\frac{[C]^p[D]^q...}{[A]^m[B]^n...}=\frac{\left ( \frac{p_C}{RT} \right )^p\left ( \frac{p_D}{RT} \right )^q...}{\left ( \frac{p_A}{RT} \right )^m\left ( \frac{p_B}{RT} \right )^n...}

which rearranges to:

(RT)^{(p+q+...)-(m+n+...)}K_c=\frac{p_C^{\; \; \; p}p_D^{\; \; \; q}...}{p_A^{\; \; \; m}p_B^{\; \; \; n}...}\; \; \; \; \; \; \; \; 2

Substituting the general equilibrium constant expression in terms of partial pressures, Kp, into eq2 yields:

K_p=K_c(RT)^{\Delta n}

where Δn = sum of stoichiometric coefficients of products – sum of stoichiometric coefficients of reactants.

Note that Kp = Kc if Δn = 0.

 

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Why is water sometimes omitted from the equilibrium constant?

Why is water sometimes omitted from the equilibrium constant? To answer this equation, we begin by noting that the equilibrium constants

K_c=\frac{[C]^p[D]^q...}{[A]^m[B]^n...}\; \; and\; \; K_p=\frac{p_C^{\; \;\; p}p_D^{\; \; \; q}...}{p_A^{\; \;\; m}p_B^{\; \;\; n}...}\; \; \; \; \; \; \; \; 3

are approximations of the thermodynamic definition of the equilibrium constant, which is:

K=\frac{a_C^{\; \;\; p}a_D^{\; \; \; q}...}{a_A^{\; \;\; m}a_B^{\; \;\; n}...}\; \; \; \; \; \; \; \; 4

where ai is the activity of species i.

For a dilute solution,

a_i=\frac{[i]}{[i]^o}\; \; \; \; \; \; \; \; 5

where [i]o is the concentration of the pure species at standard conditions of 1 bar and 298.15K.

For an ideal gas,

a_i=\frac{p_i}{p_i^{\: o}}\; \; \; \; \; \; \; \; 6

where pio is the pressure of the pure species at standard conditions of 1 bar and 298.15K.

Combining eq3 through eq6,

K_c=\frac{ \left (\frac{[C]}{[C] ^o} \right )^p\left (\frac{[D]}{[D] ^o} \right )^q...}{\left ( \frac{[A]}{[A]^o} \right )^m\left ( \frac{[B]}{[B]^o} \right )^n...}\; \; and\; \; K_p=\frac{ \left (\frac{P_C}{P_C^{\: o}} \right )^p\left (\frac{P_D}{P_D^{\: o}} \right )^q...}{\left ( \frac{P_A}{P_A^{\: o} }\right )^m\left ( \frac{P_B}{P_B^{\: o}} \right )^n...}\; \; \; \; \; \; \; \; 7

Consider the following reversible reaction:

CH_3COOH(l)+H_2O(l)\rightleftharpoons CH_3COO^-(aq)+H_3O^+(aq)

with

K_c=\frac{ \left (\frac{[CH_3COO^-]}{[CH_3COO^-] ^o} \right )\left (\frac{[H_3O^+]}{[H_3O^+] ^o} \right )}{\left ( \frac{[CH_3COOH]}{[CH_3COOH]^o} \right )\left ( \frac{[H_2O]}{[H_2O]^o} \right )}\; \; \; \; \; \; \; \; 8

Water, being in excess, is assumed to have a constant concentration throughout the reaction, approximately equal to that of its pure state. Hence \frac{\left [ H_2O \right ]}{\left [ H_2O \right ]^o}\approx1 and eq8 becomes

K_c=\frac{ \left (\frac{[CH_3COO^-]}{[CH_3COO^-] ^o} \right )\left (\frac{[H_3O^+]}{[H_3O^+] ^o} \right )}{\left ( \frac{[CH_3COOH]}{[CH_3COOH]^o} \right )}\; \; \; \; \; \; \; \; 9

Since the standard state of a solute is defined as 1 mol dm-3, eq9 approximates to

K_c=\frac{\left [ CH_3COO^- \right ]\left [ H_3O^+ \right ]}{\left [ CH_3COOH \right ]}\; \; \; \; \; \; \; \; 10

Therefore, water is excluded from the equilibrium constant if it acts as a solvent. However, for the reactions

CH_3COOH(l)+C_2H_5OH(l )\rightleftharpoons CH_3COOC_2H_5(l)+H_2O(l)

4HCl(g)+O_2(g)\rightleftharpoons 2H_2O(g)+2Cl_2(g)

water is not a solvent but a reactant, and the respective equilibrium constants are

K_c=\frac{\left [ CH_3COOC_2H_5 \right ]\left [ H_2O \right ]}{\left [ CH_3COOH \right ]\left [ C_2H_5OH \right ]}

K_c=\frac{\left [ H_2O \right ]^2\left [Cl_2 \right ]^2}{\left [ HCl \right ]^4\left [O_2 \right ]}

In the case of a reversible reaction containing one or more solid species, e.g.,

HCl(g)+LiH(s)\rightleftharpoons LICl(s)+H_2(g)

the concentrations of the solid compounds are assumed to be the same as that of their respective pure states. Hence,

K_c=\frac{\left [ H_2 \right ]}{\left [HCl\right ]}\; \; or\; \; K_p=\frac{p_{H_2}}{p_{HCl}}

 

Question

Write the equilibrium constants for the following reactions:

a) CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)

b) CH_3COOC_2H_5(l)+H_2O(l)\rightleftharpoons CH_3COOH(l)+C_2H_5OH(l)

c) Co(H_2O)_6^{\; 2+}(aq)+4Cl^-(aq)\rightleftharpoons CoCl_4^{\; 2-}(aq)+6H_2O(l)

Answer

a) K_p=p_{CO_2}

b) K_c=\frac{\left [ CH_3COOH \right ]\left [ C_2H_5OH \right ]}{\left [ CH_3COOC_2H_5 \right ]\left [ H_2O \right ]}

H2O is not a solvent in the hydrolysis reaction but a reactant. A typical ester hydrolysis reaction involves adding, for example, 0.10 M of CH3COOC2H5, 0.10 M of H2O and a catalyst in an inert organic solvent.

c) K_c=\frac{\left [ CoCl_4^{\; \; 2-} \right ]}{\left [ Co(H_2O)_6^{\; \; 2+} \right ]\left [ Cl^- \right ]^4}

The reaction occurs in an aqueous solution, i.e., the solvent is water.

 

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Ion-product constant for water

The ion-product constant for water describes the relationship between molecular water and its dissociated ionic components at dynamic equilibrium.

Pure water dissociates partially to give the hydroxonium and hydroxide ions.

2H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)

Since water is the solvent, its activity approximately equals to one and the equilibrium constant is

K_w=\left [ H_3O^+ \right ]\left [ OH^- \right ]\; \; \; \; \; \; \; \; 11

Kw is called the ion-product constant for water. Conductivity measurements of pure water reveals that [H3O+] = 10-7 M at 25oC. Since the concentration of hydroxonium ions and hydroxide ions are formed only from the dissociated of water, [H3O+] = [OH–], giving Kw = 10-14 M at 25oC . It is important to note that Kw remains equal to 10-14 M at 25oC, regardless of whether the H3O+ and OH– ions originate from the dissociation of water or from added acids or bases.

 

Question

What is the concentration of OH– when HCl is added to water at 25oC to give a pH of 2?

Answer

According to Le Chatelier’s principle, the increased [H3O+] shifts the position of the equilibrium of eq11 to the left to attain a new equilibrium where

[OH^-]=\frac{10^{-14}}{10^{-2}}=10^{-12}\: M

 

 

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Distribution coefficient (partition coefficient)

The distribution coefficient describes the relative concentrations of a chemical compound in two immiscible solvents.

When a solute, X, with different solubilities in two immiscible solvents, e.g., water and hexane, is shaken in a separating funnel containing both solvents and left to settle, a dynamic equilibrium is established such that the rate of the solute moving from the aqueous layer to the organic layer is the same as the rate of the solute moving from the organic layer to the aqueous layer.

X(aq)\rightleftharpoons X(org)

The equilibrium constant, called the distribution coefficient (or partition coefficient) is given by:

K_d=\frac{[X(org)]}{[X(aq)]}

If the compound is more soluble in the organic phase, the distribution coefficient will be greater than 1. If it is more soluble in the aqueous phase, the coefficient will be less than 1. This coefficient is important in various fields, like pharmacology, where it helps predict how a drug will behave in the body (its absorption, distribution, and excretion), and in environmental science, where it can indicate how a pollutant might move between water and soil.

 

Question

10.00 g of benzoic acid that is dissolved in 100 ml of water is shaken with 50 ml of chloroform and left to settle. 20 ml is then extracted from the aqueous layer and titrated with 6.40 ml of 0.1000 M of NaOH. Calculate the distribution coefficient for benzoic acid in the two solvents.

Answer

C_6H_5COOH(l)+NaOH(aq)\rightarrow C_6H_5COO^-Na^+(aq)+H_2O(l)

K_d=\frac{\frac{\frac{10.00}{122.05}-(0.1000\times 0.00064\times 5)}{0.05}}{\frac{0.1000\times 0.0064\times 5}{0.1}}=49.2

 

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Le Chatelier’s principle

Henry Louis Le Chatelier, a French chemist, developed a principle of chemical equilibrium in the late 1800s, which states:

A change in concentration, pressure or temperature to a system at dynamic equilibrium causes the position of the equilibrium to shift in the direction that minimises the change.

The principle is based on the thermodynamic properties of the equilibrium constant, which is dependent only on temperature for a particular reaction. Let’s see how the factors (concentration, pressure and temperature) affect the equilibria of chemical reactions in the next few articles.

 

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Solubility product

The solubility product of a chemical compound is the equilibrium constant for the dissolution of the compound in its solid state in an aqueous solution.

The solubility of a solute is the maximum amount of the solute in grammes that can dissolve in a 100 ml (or sometimes 1 dm3 or 1 kg) of a solvent at a particular temperature. For example, the solubility of AgCl in 100 ml of water at 25oC is about 1.92×10-4 g or 1.34×10-5 moles.

To illustrate the concept of solubility product, let’s begin by adding 1.0×10-5 moles of solid AgCl in 100 ml of water at 25oC. The solid completely dissolves, producing 1.0×10-5 moles each of Ag+ and Cl– ions. As we increase the amount of solid AgCl to 1.34×10-5 moles, it still dissolves completely, yielding the maximum concentration of Ag+ and Cl– ions possible in 100ml of water. At this point, the solution is considered saturated. If we add more than 1.34×10-5 moles of AgCl, any excess solid will remain undissolved. The concentrations of Ag+ and Cl– ions stay constant at 1.34×10-5 mol per 100 ml, and the system reaches equilibrium. At equilibrium, the rate at which solid AgCl dissolves into ions equals the rate at which the ions recombine to form the solid, establishing a dynamic balance:

AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)

Since the concentration of solid silver chloride is assumed to be the same as that of its pure state, the equilibrium constant is:

K_{sp}=[Ag^+][Cl^-]

with

Ksp being the solubility product of AgCl, which is 1.8×10-10 at 25oC.
[Ag+] is the maximum amount of Ag+ in 100 ml of water.
[Cl–] is the maximum amount of Cl– in 100 ml of water.

The solubility product of AgCl is therefore the mathematical product of the solubility of Ag+ (with respect to Cl–) and the solubility of Cl– (with respect to Ag+) raised to the power of their respective stoichiometric coefficients in 100 ml of solvent at a particular temperature. In general, the solubility product of AxBy is

K_{sp,A_xB_y}=[A^{y+}]^x[B^{x-}]^y

 

Question

Calculate the Ksp for PbCl2, given that its solubility is 0.0108 g/ml at 20oC.

Answer

PbCl_2(s)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)

Solubility of Pb2+ with respect to Cl– = \frac{10.8}{\left [ 207.2+\left ( 2\times 35.45 \right ) \right ]}\; mol\, dm^{-3}

Solubility of Cl– with respect to Pb2+ = \frac{2\times 10.8}{\left [ 207.2+\left ( 2\times 35.45 \right ) \right ]}\; mol\, dm^{-3}

K_{sp}=\left [ Pb^{2+} \right ]\left [ Cl^- \right ]^2=\left ( \frac{10.8}{278.1} \right )\left ( \frac{2\times 10.8}{278.1} \right )^2=2.3\times 10^{-4}\; mol^{\: 3}\, dm^{-9}

 

Just as Ka and Kb are only useful for comparing weak acids and weak bases respectively, Ksp is only useful for comparing sparing soluble salts, as highly soluble salts have a higher probability of forming ion pairs. An ion pair consists of a cation and an anion that are electrostatically attracted to each other, rather than being individually surrounded by solvent molecules. This interaction alters their physical properties—for example, their mobility—and can distort the measurement of ion concentrations in solution when using ionic or conductivity methods. As a result, the accuracy of the Ksp value may be compromised. In general, the higher the solubility of a solid, the greater the concentration of ions in the solvent, which increases the likelihood of ion pair formation.

Furthermore, just as Kw is constant at a particular temperature regardless of the source of H3O+ and OH–, Ksp for a solid remains constant at a particular temperature regardless of the source of the dissolved ions. For example, the presence of NaCl in a saturated solution of AgCl causes the latter to precipitate, as Cl– is common to both species. The decrease in the solubility of a dissolved compound in the presence of an ion in common with the dissolved compound is called the common ion effect.

 

Question

Is Ksp dependent on the volume of the solution?

Answer

No. is a thermodynamic equilibrium constant that is governed by the formula

\Delta_rG^{\: o}=-RTlnK

Hence, Ksp is only dependent on temperature. If we dilute a solution of PbCl2 that is in equilibrium with solid PbCl2, the increase in volume of the solution shifts the position of the equilibrium according to Le Chatelier’s principle to produce more aqueous Pb2+ and Cl– such that the saturation concentrations (mole per volume) of Pb2+ and Cl– remains unchanged when the new equilibrium is attained.

Another way to look at it is that Ksp is the mathematical product of the solubility of the ions of a compound, raised to the power of their respective stoichiometric coefficients in a particular volume of solvent at a particular temperature. Since solubility is an intensive property, Ksp is independent of the volume of solvent.

 

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Effect of concentration on equilibrium

What is the effect of concentration on equilibrium?

As mention in an earlier article, the equilibrium constant K of a reaction is:

K=e^{-\frac{\Delta_rG^o}{RT}}\; \; \; \; \; \; \; \; 17

Eq17 shows that the value of K for a particular reaction only varies with temperature and is independent of the concentration of reaction species. With that in mind, how does concentration affect the position of chemical equilibrium of a reaction? Consider the hydrolysis of ethyl ethanoate,

CH_3COOC_2H_5(l)+H_2O(l)\rightleftharpoons CH_3COOH(l)+C_2H_5OH(l)

The equilibrium constant for the reaction is

K=\frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5][H_2O]}

When the reaction reaches dynamic equilibrium at a particular temperature, the ratio of the products and the reactants is a constant. If we increase the concentration of ethyl ethanoate at this stage, the excess ethyl ethanoate reacts with water to give more products to maintain the constant value of K, thereby reducing the concentration of ethyl ethanoate and shifting the position of the equilibrium to the right to attain a new dynamic equilibrium. This is consistent with Le Chatelier’s principle.

If we now increase the concentration of ethanoic acid or decrease the concentration of ethyl ethanoate, the position of the equilibrium will shift to the left to minimise the change, maintaining the same value of K.

We can also explain the above observation using chemical kinetics. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. If we increase the concentration of ethyl ethanoate, the greater number of ethyl ethanoate molecules leads to a higher frequency of collisions between ethyl ethanoate and water molecules, which increases the rate of the forward reaction relative to the reverse reaction. As a result, the position of equilibrium shifts to the right to maintain the value of K.

 

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Acid/base dissociation constant

An acid/base dissociation constant is a measure of the strength of an aqueous acid/base.

A weak acid dissociates partially in an aqueous solution to give the hydroxonium ion and a conjugate base.

HA(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+A^-(aq)

Since water is the solvent, its activity approximately equals to one, and the equilibrium constant is given by:

K_a=\frac{\left [ H_3O^+ \right ][A^-]}{[HA]}\; \; \; \; \; \; \; \; 12

Ka is called the acid dissociation constant. The higher the value of Ka of an acid, the greater the extent of dissociation of the acid (i.e., the stronger the acid). It is difficult to determine accurately the Ka of an acid that completely ionises in water (a strong acid), since [HA] → 0. Hence, Ka is a useful measure only for weak acids.

A weak base dissociates partially in an aqueous solution to give a conjugate acid and the hydroxide ion.

B(aq)+H_2O(l)\rightleftharpoons BH^+(aq)+OH^-(aq)

Since water is the solvent, its activity approximately equals to one, and the equilibrium constant is given by:

K_b=\frac{[BH^+][OH^-]}{[B]}\; \; \; \; \; \; \; \; 13

Kb is called the base dissociation constant. The higher the value of Kb of a base, the greater the extent of dissociation of the base (i.e., the stronger the base). Similar to acids, it is difficult to determine accurately the Kb of a base that completely ionises in water (a strong base), since [B] → 0. Hence, Kb is a useful measure only for weak bases.

Note that BH+ is an acid with the following dissociation equilibrium:

BH^+(aq)+H_2O(l)\rightleftharpoons B(aq)+H_3O^+(aq)

K_a=\frac{[B][H_3O^+]}{[BH^+]}\; \; \; \; \; \; \; \; 14

Multiplying eq14 with eq13,

K_aK_b=[H_3O^+][OH^-]\; \; \; \; \; \; \; \; 15

Substituting eq11 from the previous article in eq15

K_aK_b=K_w\; \; \; \; \; \; \; \; 16

Eq16 expresses the relationship for all conjugate acid-base pairs.

 

Question

Given that Ka for HCN is 6.2×10-10 at 25oC, calculate the equilibrium constant for

CN^-(aq)+H_2O(l)\rightleftharpoons HCN(aq)+OH^-(aq)

Answer

K_b=\frac{K_w}{K_a}=\frac{10^{-14}}{6.2\times 10^{-10}}=1.6\times 10^{-5}

 

 

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Effect of pressure on equilibrium

What is the effect of pressure on equilibrium?

Similar to concentration, a change in pressure does not change the value of the equilibrium constant K. So how does pressure affect the position of chemical equilibrium of a reaction? Consider the following reaction:

2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)

The equilibrium constant for the reaction is

K=\frac{p_{NO}^{\; \; \; \; \; \; 2}p_{O_2}}{p_{NO_2}^{\; \; \; \; \; \; \; 2}}

where pi is the partial pressure of gas i.

We can also write the above equation as:

K=\frac{\left ( x_{NO}p \right )^2\left ( x_{O_2}p \right )}{\left ( x_{NO_2}p \right )^2}=\frac{x_{NO}^{\; \; \; \; \; \; 2}\, x_{O_2}\, p}{x_{NO_2}^{\; \; \; \; \; \; \; 2}}=\frac{\left (\frac{n_{NO}}{n} \right )^2\left ( \frac{n_{O_2}}{n} \right )p}{\left ( \frac{n_{NO_2}}{n}\right )^2} =\left ( \frac{n_{NO}^{\; \; \; \; \; \; 2}\, n_{O_2}}{n_{NO_2}^{\; \; \; \; \;\; \;\, 2}} \right )\frac{p}{n}\; \; \; \; \; \; \; \; 18

where xi is the mole fraction of gas i, p is the total pressure of the system, ni  is the number of moles of gas i, and n is the total number of moles of the gas mixture.

For ideal gases, we can substitute pV = nRT in eq18 to give:

K=\left ( \frac{n_{NO}^{\; \; \; \; \; \; 2}\, n_{O_2}}{n_{NO_2}^{\; \; \; \; \;\; \;\, 2}} \right )\frac{RT}{V}\; \; \; \; \; \; \; \; 19

If we increase the total pressure by decreasing the volume of the system at constant temperature, the denominator of the term in brackets in eq19 must increase to maintain the value of K (or the numerator of the term in brackets must decrease). The position of the equilibrium of the reaction therefore shifts left to increase the amount of nitrogen dioxide. This is consistent with Le Chatelier’s principle, where an increase in pressure shifts the position of the equilibrium from right (3 moles of gases) to left (2 moles of gases) to minimise the change in pressure.

It is important to note that changes in pressure only affects the position of an equilibrium where species are gases, as solids and liquids are relatively incompressible.

 

Question

Consider a fixed-volume reaction vessel with the following reaction at a state of equilibrium at constant temperature:

2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)

If we increase the pressure in the vessel by adding an inert gas (assuming that all gases in the vessel are ideal gases), will the composition of the gases change?

Answer

No, because the partial pressures of the gases remain the same even though the total pressure of the system has increased.

p_i=x_ip=\frac{n_i}{n}p=\frac{n_iRT}{V}

Since the factor of RT/V is a constant, pi does not change if ni remains the same.

 

 

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Effect of temperature on equilibrium

What is the effect of temperature on equilibrium?

The change in temperature of a chemical reaction at equilibrium results in a shift in position of the equilibrium that opposes the temperature change. For example, the increase in temperature of an endothermic reaction encourages the reaction to occur because the increased energy is absorbed to facilitate the reaction, thereby lowering the temperature.

Quantitatively, the effect of temperature on a chemical reaction system is best explained using the van’t Hoff equation (see below for derivation):

\frac{dlnK}{d\frac{1}{T}}=-\frac{\Delta_rH^o}{R}\; \; \; \; \; \; \; \; 20

where ΔrHo is the enthalpy of a reaction at standard conditions, K is the equilibrium constant, T is the temperature of the system and R is the universal gas constant.

Eq20 can be rewritten as:

ln\left ( \frac{K_f}{K_i} \right )=-\frac{\Delta_rH^o }{R}\left ( \frac{1}{T_f}-\frac{1}{T_i} \right )\: \: \: \: \: \: \: \:\: 20a

where the subscripts f and i represent ‘final’ and ‘initial’ respectively.

Let’s consider an increase in T of an endothermic reaction (ΔrHo > 0). The RHS of eq20a under such conditions is positive, which means that Kf > Ki, i.e. the equilibrium constant increases with an increase in T for an endothermic reaction. Therefore, unlike concentration and pressure, temperature affects the value of K. Nevertheless, Le Chatelier’s principle still applies, as illustrated by the following example:

N_2(g)+O_x(g)\rightleftharpoons 2NO(g)\; \; \; \; \; \; \; \Delta_rH^o=+180\: kJmol^{-1}

K=\frac{p_{NO}^{\; \; \; \; \; \; 2}}{p_{N_2}\, p_{O_2}}\; \; \; \; \; \; \; \; 21

If K increases as a result of an increase in T, the numerator of eq21 must increase relative to the denominator. This means that the position of the equilibrium shifts to the right, which is consistent with the definition of Le Chatelier’s principle.

Using the same logic and eq20a, we can describe the effect of temperature on the position of an exothermic reaction’s equilibrium.

In summary,

T

 K

Equilibrium position

Endothermic

 ↑ ↑ →
  ↓ ↓ ←

Exothermic

 ↑ ↓ ←
  ↓ ↑ →

 

Question

Derive the van’t Hoff equation.

Answer

Substituting the definition of the standard Gibbs energy of reaction ΔrGo = ΔrHo – TΔrSo in the Gibbs energy equilibrium constant relation ΔrGo = –RTlnK, we have

lnK=-\frac{\Delta_r H^o}{RT}+\frac{\Delta_rS^o}{R}

If we assume that ΔrHo and ΔrSo are constant for a reaction over a range of temperature, the change of lnK with respect to the change in temperature is:

\frac{dlnK}{dT}=\frac{\Delta_r H^o}{RT^2}

This is the called the van’t Hoff equation, which can also be written as:

\frac{dlnK}{d\frac{1}{T}}=-\frac{\Delta_r H^o}{R}

 

 

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