Chemical Equilibrium

Equilibrium

Content

• Chemical equilibrium
• Henry's law
• Law of chemical equilibrium
• Le- Chatelier's principle
• Acids, Bases and salts
• p^H scale
• Buffer solution
• Hydrolysis of salts
• Common ion effect
• A chemical equilibrium is a state when the two opposing forces are exactly balanced. It is a state -
  At which reactants concentration and products concentration do not change with time. Chemical equilibrium is of two types.
• Dynamic When the reaction does not stop but continue in both directions with same speed.
• Static When the reaction completely stops.
• Equilibrium in physical processes:-
  The characteristics of system at equilibrium are better understood if we examine some physical processes. The most familiar example are phase transformation processes e.g.
  
  Solid Liquid
  Liquid Gas
  Solid Gas
• Henry's law:
  Henry's law states that "The mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the pressure of the gas above the solvent" i.e.,
  Partial pressure of the gas in solution = K_H x mole fraction of the gas in solution
  Where K_H = Henry's law constant
• General characteristics of physical equilibrium are as follows-
• For solid-liquid equilibrium, there is only one temperature at which the two phases can co-exist at a given pressure. This is called melting point or freezing point.
• For liquid-gas equilibrium, there is only one temperature at which the two phases can co-exist at a given pressure. This is called boiling point
• For dissolution of solid in liquids, the solubility is constant at a given temperature.
• Physical equilibrium is stable and dynamic in nature. At equilibrium two opposite processes occurs at the same rate.
• The measurable properties of the physical system in equilibrium remain constant.
• Physical equilibrium can only be attained in a closed system.
• Law of chemical equilibrium and equilibrium constant-
  Let us consider a general reversible reaction
  A+B C+D
  
  where A & B are the reactants, C & D are the products in the balanced chemical equation. On the basis of experimental studies of many reversible reactions, the Norwegian chemists Cato Maximillion Guldberg and Peter waage proposed in 1864 that the concentrations in an equilibrium mixture are related by the following equilibrium equation
  
  Where K_C is the equilibrium constant and this expression is also known as law of chemical equilibrium.
• This equilibrium equation is also known as law of mass action. According to this law the rate of reaction is directly proportional to the product of the molar concentrations (active masses) of the reactants with each concentration term raised to the power equal to the number of times that reactant appear in the balanced chemical equation
  Let xA + yB mC+nD
  
  According to law of mass action.
  Rate of forward reaction r_f [A]^x [B]^y
  Rate of backward reaction rb [C]^m [D]ⁿ
  
  Hence, Equilibrium constant K_C = r_f / r_b = [C]^m [D]ⁿ / [A]^x [B]^y as at equilibrium the rate of forward reaction equals to the rate of backward reaction
  
  
  Note: The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.
• In a homogenous system, all the reactants and products are in the same phase. For example, in a gaseous reaction, N₂ (g) +3H₂ (g) 2NH₃ (g) reactants and products are in the homogenous phase. For reactions involving gases, however it is usually more convenient to express the equilibrium constant in terms of partial pressure. The ideal gas equation is written as-
  PV = nRT
  or P = n/v RT
  or P = CRT i.e., at constant temperature, the pressure of the gas is directly proportional to its concentration means P [gas] here R = 0.0831 bar lit/mol K
• For gaseous reactions the equilibrium constant can be calculated from the respective partial pressure of the gaseous species. The equilibrium constant determined from the partial pressure is represented by K_p. e.g., for the equilibrium, N₂(g) + 3H₂(g) 2NH₃ (g)
  
  
  
  The relationship between Kp and Kc is expressed by the equations
  Kp = Kc(RT)ⁿ
  Here n = change in the number of gaseous moles
• when n = 0, K_p = K_c
• when n > 0, K_p > K_c
• when n < 0, K_p < K_c
• Equilibrium in a system having more than one phase is called heterogeneous equilibrium.
  
  Reaction Quotient (Qc) :
• It is an expression that has the same form as the equilibrium constant expression, but all concentration values are not necessarily those at equilibrium.
  a A + b B c C + d D
  Qc =
  i) If Q_c > K_c, backward reaction takes place
  ii) If Q_c < K_c, forward reaction takes place.
  iii) If Q_c = K_c, the reaction mixture is at equilibrium
• Factors Affecting Equilibrium Constant:
  
  1. Methods of representing the equation.
  If, for the reaction N₂ + 3H₂ 2NH₃, the equilibrium is K_c
  
  Hence, now for the reaction NH₃, the equilibrium constant K_c ' will be equal to
  
  
  2. Temperature: this can be seen as
  log
  
  Le-Chatelier principle-
• If any kind of stress (such as change in concentration, temperature or pressure) is applied on equilibrium, it shifts in a direction that tends to undo the effect of the stress This is applicable to all physical and chemical equilibrium. e.g., Increase in pressure favours melting of ice into water. At higher pressure, melting point of ice is lowered whereas boiling point of water is increased.
• With the help of this principle it is possible to predict favourable conditions for the reactions
  i) If the temperature is raised, reaction will proceed in a direction in which heat is absorbed and if the temperature is lowered, reaction will proceed in a direction in which heat is evolved so that temperature remains constant.
  ii) If the pressure is increased, reaction will take place in a direction in which volume decrease so that the product of pressure and volume remains constant.
  iii) If volume is decreased, reaction should be carried out at high pressure and vice-versa.
  iv) If the concentration of any one of the species of the reaction is increased, then reaction takes place in that direction in which concentration gets decreased.
• Addition of an Inert Gas at Constant Volume or Constant Pressure i)
  
  At Constant Volume: The addition of an inert gas at constant volume has no effect. It only increase the total pressure but does not alter the partial pressure of various species.
  ii) At Constant Pressure : The addition of an inert gas at constant pressure will favour the direction of reaction where total no. of moles at equilibrium show an increase.
• Acids, Bases and salts
• According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions H⁺ (aq) and bases are substances that produce hydroxyl (OH) ions. The ionization of an acid HX can be represented by the following equations-
  
  HX (aq) H⁺(aq) + x^-(aq)
  
  
  Note: H⁺ is very reactive and can't exist freely in aqueous solutions. Thus it bonds to the oxygen atom of a solvent water molecule to give triagonal pyramidal hydronium ion H₃O⁺
• Similarly, a base molecule like MOH ionizes in aqueous solution according to the equation MOH (aq) M⁺ (aq) + OH^-(aq)
• Arrhenius theory fails to explain the behaviour of acids and bases in non-aqueous solutions. It fails to explain the acidic character of certain salts like AlCl₃, BF₃, and basic character of NH₃ PH₃ also etc. It fails to explain the neutralization reaction giving salt in the absence of solvent e.g. NH₃ + Cl NH₄Cl
• According to Bronsted-Lowry theory, acid is a substance that is capable of donating a hydrogen ion H+ and bases are substances capable of accepting a hydrogen ion, H⁺
  Consider the example of dissolution of NH₃ in H₂O represented by the following equation.
  
• The acid-base pair that differs only by one proton is called a conjugate acid-base pair. Therefore, OH- is called the conjugate base of an acid H₂O and NH4+ is called conjugate acid of the base NH₃
  Note: If Bronsted acid is a strong acid then its conjugate base is a weak base and vice-versa. The conjugate acid has one extra proton and each conjugate base has one less proton.
• G.N. Lewis in 1923 defined an acid as a species which accepts electron pair and base which donates and electron pair. BF₃ does not have a proton but still acts as an acid and reacts with NH₃ by accepting its lone pair of electrons. The reaction can be represented by BF₃ +: NH₃ BF₃: NH₃
• Electron deficient species like AlCl₃, CO₃+, Mg₂+ etc. can act as Lewis acids while species like H₂O, NH₃, OH- which can donate a pair of electrons can act as Lewis bases.
  pH = - log [H⁺]
• Calculation of pH
  1. Strong acid - [H+] = Normality
  2. Strong Base - [OH-] = Normality
  3. pH of mixture of strong acids / strong bases
  We calculate the normality of final solution.
  4. pH of mixture of strong acids and strong bases
  we calculate normality of final solution.
  i) If equivalents of acids > eq. of base
  Final solution will be acidic and normality = [H⁺]
  ii) If eq. of base > eq. of acid
  Final solution will be alkaline and normality= [OH^-]
  iii) If eq. of acid = eq. of base,
  final solution will be neutral and pH = 7 at 25°C.
  
  
  5. pH of weak monobasic acid or weak monoacidic base
  [H+] = [OH^-] =
  Here =
  Note: i) In above formula for a, we have assumed a is very small compared to one and hence neglected compared to one.
  ii) In case when we use above formula and a > 0.1, we do not apply above approximation and if a 0.1, approximation is valid
  
  
  6. pH of mixture of two weak acids
  we must consider ionisation of two acids separately in which
  Total [H⁺]= [H⁺] produced from acid (1) and those from acid (II) = C₁₁ + C₂₂
  Where C₁,C₂ are concentration of two acids and a1 and a2 are degree of dissociation of acids in presence of each other.
  
  Salt Hydrolysis :
  Reaction in which cation or anion of the salt react with water to convert water acidic or basic in nature, is known as salt hydrolysis.
  
  1. Salt of a Strong Acid and Weak Base
  Consider the salt BA, on hydrolysis it will give strong acid (HA) and weak base (BOH) B+ + A- + H₂O BOH + H+ + A- or
  B⁺ + H₂O BOH + H+
  i) K_H = K_W / K_b
  ii)
  iii)
  iv) pH = 1/2 [pK_w - log c - pK_b].
  
  
  2. Salt of a Strong Base and Weak Acid:-
  When this salt is added to water, we have the following equilibrium.
  B+ + A- + H₂O B⁺ + OH^- + HA or A- + H₂O HA + OH^-
  i) K_H = K_w / K_a
  ii)
  iii)
  iv) pH = 1/2 (pK_w + log c + pK_b)
  
  
  3. Salt of Weak Acid and Weak Base:
  When this salt is added to water, we have the following equilibrium
  B⁺ + A^- + H₂O HA + BOH
  i) K_H = K_w / K_a . Kb
  ii) h =
  ii) [H⁺] =
  iv) pH = 1/2 [pK_w + pK_a - pK_b].
  
  Note 1: All the formulae for salt hydrolysis are for univalent salts. The term 'c' in the above equations however represents the concentration of ion that undergoes hydrolysis.
  2: In all the formulae mentioned above we have neglected a compared to one
  Buffer Solution:
  
  Acidic buffer: Weak acid & its salt with strong base. e.g. CH₃COOH and CH₃COONa
  Basic buffer: Weak base and its salt with strong acid e.g. NH₄OH and NH₄Cl
  
  pH for Buffer Solution
  Acidic buffer: pH = pK_a + log [anion]/[acid] or pH = pK_a + log [salt]/[acid]
  Basic buffer: pOH = pK_a + log [cation]/[base] or pOH = pK_a + log [salt]/[base]
  
  
  Solubility Product (K_sp):
  Insoluble substances like AgCl, BaSO₄, PbCl₂, etc., are infact not completely insoluble when present in an aqueous medium. A very small amount of these dissolves and is present as ions. Further, there exists an equilibrium between the un dissolved and the dissolved salt. For AgCl, the equilibrium equation may be written as, AgCl(s) Ag+(aq) + Cl^- (aq)
  Applying the law of mass action,
  [AgCl] is assumed to be constant because of the fact that very little of this solid dissolved in aqueous solution (by definition) K_sp = [Ag⁺] [Cl^-]
  
  Note:
  1. Let the solubility of salt of weak acid and strong base is s₁ in pure water, s₂ in basic buffer and s₃ in acidic buffer then s₃ > s₁ > s₂
  2. For Preciptiation, K_sp < K_ip (Ionic product)
  3. Mutual solubility of two sparingly soluble salts
  Let Ksp(AgCl) = x x and y are close
  K_sp(AgBr) = y
  
  [Ag⁺] =
  Knowing [Ag⁺], we can calculate [Cl^-] and [Br^-] in the solution which will be the mutual solubility of AgCl and AgBr respectively.