Chemical Equilibrium : General Characteristics

Whenever we hear the word Equilibrium immediately a picture arises in our mind an object under the influence of two opposing forces. For chemical reactions also this is true. A reaction also can exist in a state of equilibrium balancing forward and backward reactions.

Equilibrium and its dynamic nature

(1) Definition : “Equilibrium is the state at which the concentration of reactants and products do not change with time. i.e. concentrations of reactants and products become constant.”

(2) Characteristics : Following are the important characteristics of equilibrium state,

(i) Equilibrium state can be recognised by the constancy of certain measurable properties such as pressure, density, colour, concentration etc. by changing these conditions of the system, we can control the extent to which a reaction proceeds.

(ii) Equilibrium state can only be achieved in close vessel, but if the process is carried out in an open vessel equilibrium state cannot be attained because in an open vessel, the reverse process will not take place.

(iii) Equilibrium state is reversible in nature.

(iv) Equilibrium state is also dynamic in nature. Dynamic means moving and at a microscopic level, the system is in motion. The dynamic state of equilibrium can be compared to water tank having an inlet and outlet. Water in tank can remain at the same level if the rate of flow of water from inlet (compared to rate of forward reaction) is made equal to the rate of flow of water from outlet (compared to rate of backward reaction). Thus, the water level in the tank remains constant, though both the inlet and outlet of water are working all the time.

(v) At equilibrium state,

Rate of forward reaction = Rate of backward reaction

(vi) At equilibrium state, DG = 0, so that DH = TDS.

(3) Types : Equilibrium in a system implies the existence of the following types of equilibria simultaneously,

(i) Thermal equilibrium : There is no flow of heat from one part to another i.e. T = constant.

(ii) Mechanical equilibrium : There is no flow of matter from one part to another i.e. P = constant.

(iii) Physical equilibrium : There is the substance exist in three states: solid, liquid and gaseous.

(iv) Chemical equilibrium : There is no change in composition of any part of the system with time.

Physical equilibrium

The physical equilibrium is a state of equilibrium between the same chemical species in different phases (solid, liquid and gaseous). The various equilibria which can exist in any physical system are,

Solid	Liquid
Liquid	Vapour
Solid	Gas (vapour)
Solid	Saturated solution of solid in a liquid
Gas (vapour)	Saturated solution of gas in a liquid

(1) Solid-liquid equilibrium

Rate of transfer of molecules from ice to water = Rate of transfer of molecules from water to ice

Rate of melting of ice = Rate of freezing of water

Free energy change and solid-liquid equilibrium in water : For ice-water system free energy change (DG), at 273 K, and one atmosphere pressure is zero i.e., DG = 0; Ice ⇌ Water; H₂O (s) ⇌ H₂O(l)

(i) At temperature higher than 273 K, and 1 atm pressure, DG < 0. Thus, the process in the forward direction would become favourable and ice will melt to give more water.

(ii) At temperature less than 273 K, and 1 atm pressure, DG > 0. Thus, the reverse reaction will become favourable, and more ice will be formed from liquid water.

(2) Liquid-vapour equilibrium : A liquid placed in an open container disappears completely. After some time vapours of the liquid held in an open container can escape out to the atmosphere. Thus, when vapour of liquid exists in equilibrium with the liquid, then

Rate of vaporisation = Rate of condensation,

H₂O(l) ⇌ H₂O

Conditions necessary for a liquid-vapour equilibrium

(i) The system must be a closed system i.e., the amount of matter in the system must remain constant.

(ii) The system must be at a constant temperature.

(iii) The visible properties of the system should not change with time.

(3) Solid-vapour equilibrium : Certain solid substances on heating get converted directly into vapour without passing through the liquid phase. This process is called sublimation. The vapour when cooled, gives back the solid, it is called disposition.

Solid ⇌ Vapour

The substances which undergo sublimation are camphor, iodine, ammonium chloride etc.

For exampie, Ammonium chloride when heated sublimes.

NH₄Cl (s) $\xrightleftharpoons[\text{Cool}]{\text{Heat}}$ NH₄Cl (v)

(4) Equilibrium between a solid and its solution :

When a saturated solution is in contact with the solid solute, there exists a dynamic equilibrium between the solid and the solution phase.

Solid substance ⇌ Solution of the substance

Example : Sugar and sugar solution. In a saturated solution, a dynamic equilibrium is established between dissolved sugar and solid sugar.

Sugar (s) ⇌ Sugar (aq)

At the equilibrium state, the number of sugar molecules going into the solution from the solid sugar is equal to the number of molecules precipitating out from the solution, i.e., at equilibrium,

Rate of dissolution of solid sugar = Rate of precipitation of sugar from the solution.

(5) Equilibrium between a gas and its solution in a liquid : Gases dissolve in liquids. The solubility of a gas in any liquid depends upon the,

(i) Nature of the gas and liquid.

(ii) Temperature of the liquid.

(iii) Pressure of the gas over the surface of the solution.

Henry’s law : “At a certain temperature, the mass of a gas which dissolves in a definite volume of a liquid is proportional to the pressure of the gas in equilibrium with the solution.”

m ∞ P or m = KP ; (where K is the proportionality constant)

Thus, at a constant temperature, the ratio of the molar concentration of the gas in the solution and into the atmosphere is constant.

Limitations of Henry’s law

Henry’s law is applicable to ideal gases only. Henry’s law should be applied only at low pressures because real gases behave like ideal gases at low pressures.
Henry’s law is not applicable to gases which react chemically with the solvent.
Henry’s law will not apply to the solution of gases like ammonia (NH₃) and hydrogen chloride (HCl) in water because these gases react chemically with water.

Note : A chilled soda water bottle fizzes out when opened because, soda water is a solution of carbon dioxide gas, in water at high pressure. As soon as the bottle is opened under normal atmospheric conditions, the dissolved gas escapes out to reach a new equilibrium state, so that the pressure of the gas inside the bottle becomes equal to the atmospheric pressure. At low pressure, the solubility of the gas in water decreases.

Chemical equilibrium

The equilibrium between different chemical species present in the same or different phases is called chemical equilibrium. There are two types of chemical equilibrium.

(1) Homogeneous equilibrium : The equilibrium reactions in which all the reactants and the products are in the same phase are called homogeneous equilibrium reactions.

Example:

(i) C₂H₅OH(l) + CH₃COOH(l) ⇌ CH₃COOC₂H₅(l) + $\underset { liquid\quad phase }{ { H }_{ 2 }O(l) }$

(ii) N₂(g) + 3H₂(g) ⇌ 2 $\underset { gas\quad phase }{ N{ H }_{ 3 }(g) }$

(iii) 2SO₂(g) + O₂(g) ⇌ 2 $\underset { gas\quad phase }{ S{ O }_{ 3 }(g) }$

(2) Heterogeneous equilibrium : The equilibrium reactions in which the reactants and the products are present in different phases are called heterogeneous equilibrium reactions.

Example : (i) CaCO₃(s) ⇌ CaO(s) + CO₂(g)

(ii) 2NaHCO₃(s) ⇌ Na₂CO₃(s) + CO₂(g) + H₂O(g)

(iii) H₂O(l) ⇌ H₂O(g)

(iv) Ca(OH)₂(s) + H₂O(l) ⇌ Ca²⁺(aq) + 2OH^–(aq)

Note : The equilibrium expression for heterogeneous reactions does not include the concentrations of pure solids because their concentrations remain constant.

Reversible and irreversible reactions

A chemical reaction is said to have taken place when the concentration of reactants decreases, and the concentration of the products increases with time. The chemical reactions are classified on the basis of the extent to which they proceed, into the following two classes;

(1) Reversible reactions : Reactions in which only a part of the total amount of reactants is converted into products are termed as reversible reactions.

(i) Characteristics of reversible reactions

(a) These reactions can be started from either side,

(b) These reactions are never complete,

(d) This sign (⇌) represents the reversibility of the reaction,

(e) Free energy change in a reversible reaction is zero (DG = 0),

(ii) Examples of reversible reactions

(a) Neutralisation between an acid and a base either of which or both are weak e.g.,

CH₃COOH + NaOH ⇌ CH₃COONa + H₂O

(b) Salt hydrolysis, e.g., FeCl₃+ 3H₂O ⇌ Fe(OH)₃ + 6HCl

CaCO_3(s) ⇌ CaO_(s) + CO_2(g) ; 2HI_(g) ⇌H_2(g) + I_2(g)

(d) Esterification, e.g., CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

(e) Evaporation of water in a closed vessel, e.g., H₂O_(l) ⇌ H₂O_(g) – Q

(f) Other reactions, e.g., N_2(g) + 3H_2(g) ⇌ 2NH_3(g) + Q;

2SO_2(g) + O_2(g) ⇌ 2SO_3(g) + Q

(2) Irreversible reactions : Reactions in which the entire amounts of the reactants are converted into products are termed as irreversible reactions.

(i) Characteristics of irreversible reactions

(a) These reactions proceed only in one direction (forward direction),

(b) These reactions can proceed to completion,

(d) In an irreversible reaction, DG < 0,

(ii) Examples of irreversible reactions

(a) Neutralisation between strong acid and strong base e.g.

NaOH + HCl → NaCl + H₂O + 13.7 kcal

(b) Double decomposition reactions or precipitation reactions e.g.

BaCl_2(aq) + H₂SO_4(aq) → BaSO_4(s) ↓ + 2HCl_(aq) ;

AgNO_3(aq) + NaCl_(aq) → AgCl_(g) ↓ + NaNO_3(aq)

2Pb(NO₃)₂ $\underrightarrow { \quad Heat\quad }$ 2PbO + 4NO₂ + O₂ ↑ ;

NH₄NO₂ $\underrightarrow { \quad Heat\quad }$ N₂ ↑ + 2H₂O ↑

(d) Redox reactions, e.g., SnCl_2(aq) + 2FeCl_3(aq) → SnCl_4(aq) + 2FeCl_2(aq)

(e) Other reactions, e.g., NaCl + H₂SO₄ → NaHSO₄ + HCl ↑

Law of mass action

On the basis of observations of many equilibrium reactions, two Norwegian chemists Guldberg and Waage suggested (1864) a quantitative relationship between the rates of reactions and the concentration of the reacting substances. This relationship is known as law of mass action. It states that

“The rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants at a constant temperature at any given time.”

The molar concentration i.e. number of moles per litre is also called active mass. It is expressed by enclosing the symbols of formulae of the substance in square brackets. For example, molar concentration of A is expressed as [A].

Let us consider a simple reaction between the species A and B : A + B → Products

According to law of mass action, rate of reaction, r µ [A] [B] = k [A] [B]

Where [A] and [B] are the molar concentrations of the reactants A and B respectively, k is a constant of proportionality for the forward reaction and is known as rate constant. The rate constant is also called velocity constant. Now, if the concentration of each of the reactants involved in the reaction is unity, i.e., [A] = [B] = 1, then,

rate of reaction, r = k × 1 × 1 or r = k

Thus, the rate constant of a reaction at a given temperature may be defined as “the rate of the reaction when the concentration of each of the reactants is unity.”

For a general reaction, aA + bB + cC → Products

The law of mass action may be written as : Rate of reaction,

r = k[A]^a[B]^b[C]^c

Thus, the law of mass action may be restated as, “The rate of a chemical reaction at any particular temperature is proportional to the product of the molar concentrations of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactants taking part in the reaction.”

The number of molecules of a reactant taking part in a reaction is also called its stoichiometric coefficient. For example, a, b and c…. in the above equation are called stoichiometric coefficients of A, B and C…. respectively.

Equilibrium constant

(1) Equilibrium constant in terms of law of mass action : The law of mass action may be applied to a reversible reaction to derive a mathematical expression for equilibrium constant known as law of chemical equilibrium.

Let us consider a simple reversible reaction, A + B ⇌ X + Y in which an equilibrium exists between the reactants (A and B) and the products (X and Y). The forward reaction is,

A + B → X + Y

According to law of mass action,

Rate of forward reaction ∞ [A][B[ = k_f[A][B]

Where k_f is the rate constant for the forward reaction and [A] and [B] are molar concentrations of reactants A and B respectively.

Similarly, the backward reaction is ; X + Y → A + B

Rate of backward reaction ∞ [X][Y] = k_b[X][Y]

Where k_b is the rate constant for the backward reaction and [X] and [Y] are molar concentrations of products X and Y respectively.

At equilibrium, the rates of two opposing reactions become equal. Therefore, at equilibrium,

Rate of forward reaction = Rate of backward reaction

K_f[A][B] = k_b[X][Y]

$\frac { { k }_{ f } }{ { k }_{ b } } =\frac { [X][Y] }{ [A][B] }$ or $K=\frac { [X][Y] }{ [A][B] }$

The combined constant K, which is equal to k_f/k_b is called equilibrium constant and has a constant value for a reaction at a given temperature. The above equation is known as law of chemical equilibrium.

For a general reaction of the type : aA + bB ⇌ cC + dD

The equilibrium constant may be represented as :

where the exponents a, b, c and d have the same values as those in the balanced chemical equation. Thus, the equilibrium constant may be defined as, $K=\frac { { [C] }^{ a }{ [D] }^{ d } }{ { [A] }^{ a }{ [B] }^{ b } }$

“The ratio between the products of molar concentrations of the products to that of the molar concentrations of the reactants with each concentration term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation at a constant temperature.”

(2) Characteristics of equilibrium constant

(i) The value of equilibrium constant is independent of the original concentration of reactants.

For example, the equilibrium constant for the reaction,

Fe³⁺(aq) + SCN^–(aq) = FeSCN²⁺ (aq)

K = $\frac { [FeSC{ N }^{ 2+ }] }{ [{ Fe }^{ 3+ }][SC{ N }^{ - }] }$ 138.0 L mol^–1 (at 298 K)

Whatever may be the initial concentrations of the reactants, Fe³⁺ and SCN^– ions, the value of K comes out to be 138.0 L mol^–1 at 298 K.

(ii) The equilibrium constant has a definite value for every reaction at a particular temperature. However, it varies with change in temperature.

For example, the equilibrium constant for the reaction between hydrogen and iodine to form hydrogen iodide is 48 at 717 K.

H₂(g) + I₂(g) = 2HI(g) ; K = $\frac { { [HI] }^{ 2 } }{ { [H }]_{ 2 }{ [I] }_{ 2 } }$ = 48

For this reaction, the value of K is fixed as long as the temperature remains constant.

(iii) For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the backward reaction.

For example, if equilibrium constant, K, for the reaction of combination between hydrogen and iodine at 717 K is 48

H₂(g) + I₂(g) ⇌ 2HI (g); K = $\frac { { [HI] }^{ 2 } }{ { [H }]_{ 2 }{ [I] }_{ 2 } }$ = 48

Then, the equilibrium constant for the decomposition of hydrogen iodide is the inverse of the above equilibrium constant.

2HI(g) ⇌ H₂(g) + I₂(g) ; KK = $\frac { { [HI] }^{ 2 } }{ { [H }]_{ 2 }{ [I] }_{ 2 } }$ = $\frac { 1 }{ K } =\frac { 1 }{ 48 }$ = 0.02

In general, K_{forward reaction} = $\frac { 1 }{ { K }_{ backward\quad reaction } }$

(iv) The value of an equilibrium constant tells the extent to which a reaction proceeds in the forward or reverse direction. If value of K is large, the reaction proceeds to a greater extent in the forward direction and if it is small, the reverse reaction proceeds to a large extent and the progress in the forward direction is small.

(v) The equilibrium constant is independent of the presence of catalyst. This is so because the catalyst affects the rates of forward and backward reactions equally.

(vi) The value of equilibrium constant changes with the change of temperature. Thermodynamically, it can be shown that if K₁ and K₂ be the equilibrium constants of a reaction at absolute temperatures T₁ and T₂. If ΔH is the heat of reaction at constant volume, then

log K₂ – log K₁ = $\frac { -\Delta H }{ 2.303\quad R } \left[ \frac { 1 }{ { T }_{ 2 }-{ T }_{ 1 } } \right]$ (Van’t Hoff equation)

The effect of temperature can be studied in the following three cases

(a) When ΔH = 0 i.e., neither heat is evolved nor absorbed

log K₂ – log K₁ = 0 or log K₂ = log K₁ or K₂ = K₁

Thus, equilibrium constant remains the same at all temperatures.

(b) When ΔH = + ve i.e., heat is absorbed, the reaction is endothermic. The temperature T₂ is higher than T₁.

log K₂ – log K₁ = +ve or log K₂ > log K₁ or K₂ > K₁

The value of equilibrium constant is higher at higher temperature in case of endothermic reactions.

log K₂ – logK₁ = – ve or K₁ > log K₂ or K₁ > K

The value of equilibrium constant is lower at higher temperature in the case of exothermic reactions.

(vii) The value of the equilibrium constant depends upon the stoichiometry of the chemical equation.

Examples :

(a) If the equation (having equilibrium constant K) is divided by 2, then the equilibrium constant for the new equation is the square root of K i.e. √K. For example, the thermal dissociation of SO₃ can be represented in two ways as follows,

2SO₃(g) ⇌ 2SO₂(g) + O₂(g) and SO₃(g) ⇌ SO₂(g) + 1/2O₂(g)

K = $\frac { [S{ O }_{ 2 }]^{ 2 }[{ O }_{ 2 }] }{ [{ SO }_{ 3 }]^{ 2 } }$ and K’ = $\frac { [S{ O }_{ 2 }][{ O }_{ 2 }]^{ 1/2 } }{ [{ SO }_{ 3 }] }$ ; K’ = √K or (K)^1/2

(b) Similarly, if a particular equation is multiplied by 2, the equilibrium constant for the new reaction (K¢) will be the square of the equilibrium constant (K) for the original reaction i.e., K’ = K²

(c) If the chemical equation for a particular reaction is written in two steps having equilibrium constants K₁ and K₂, then the equilibrium constants are related as K = K₁ × K₂

For example, the reaction N₂(g) + 2O₂(g) ⇌ 2NO₂(g) with equilibrium constant (K) can be written in two steps :

N₂(g) + O₂(g) ⇌ 2NO (g) ; (Equilibrium constant = K₁)

2NO(g) + O₂(g) ⇌ 2NO₂(g) ; (Equilibrium constant = K₂)

Now, K = $\frac { [N{ O }]^{ 2 } }{ [{ N }_{ 2 }][{ O }_{ 2 }] }$ and K₂ = $\frac { [N{ O }_{ 2 }]^{ 2 } }{ [{ N }O]^{ 2 }[{ O }_{ 2 }] }$

Therefore, K₁ × K₂ = $\frac { [N{ O }]^{ 2 } }{ [{ N }_{ 2 }][{ O }_{ 2 }] }$ × $\frac { [N{ O }_{ 2 }]^{ 2 } }{ [{ N }O]^{ 2 }[{ O }_{ 2 }] }$ = $\frac { [N{ O }_{ 2 }]^{ 2 } }{ [{ N }_{ 2 }][{ O }_{ 2 }]^{ 2 } }$ = K

(3) Types of equilibrium constant : Generally two types of equilibrium constants are used,

(i) K_c → It is used when the various species are generally expressed in terms of moles/litre or in terms of molar concentrations.

(ii) K_p → It is used when in gaseous reactions, the concentration of gases expressed in terms of their partial pressures.

K_p is not always equal to K_c. K_p and K_c are related by the following expression, K_p = K_c (RT)^Δn

Where, R = Gas constant = 0.0831 bar dm³ mol^–1 k^–1; T = Temperature in Kelvin

Δn = number of moles of gaseous products – number of moles of gaseous reactants in chemical equation

(4) Unit of equilibrium constant : Equilibrium constant K has no units i.e., dimensionless if the total number of moles of the products is exactly equal to the total number of moles of reactants. On the other hand if the number of moles of products and reactants are not equal, K has specific units.

Units of K_p and K_c and the value of Δn

Value of Δn	Relation between K_p and K_c	Units of K_p	Units of K_c
0	K_p = K_c	No unit	No unit
> 0	K_p > K_c	(atm)^Δⁿ	(mole l^–1)^Δⁿ
< 0	K_p < K_c	(atm)^Δⁿ	(mole l^–1)^Δⁿ

Chemical Equilibrium : General Characteristics

Chemical Equilibrium : General Characteristics

Limitations of Henry’s law

Rate of forward reaction = Rate of backward reaction

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