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ACTIVE SITE EDUTECH - 9844532971, , CHEMICAL EQUILIBRIUM, Total Sessions – 06, SESSION – 1, AIM - To introduce the concept of equilibrium, TYPES OF REACTIONS, 1. Irreversible Reactions:, The chemical reactions in which the products formed do not combine to give, back the reactants are known as irreversible reaction., Properties of irreversible reactions, a. Reactions in which reactant react to form product only., b. Reactions proceed in one single direction., c. Always proceed to completion., d. In this type of reactions, if product is gaseous in state, then they can, escape from reacting site and if they are solid in state they will, precipitate., Examples:, a. Thermal decomposition in open vessel. 2KClO3 ⟶ 2KCl + 3O2, b. All the neutralization reaction of strong acid and strong base., NaOH + HCl ⟶ NaCl + H2 O, , c. Ionic precipitation reaction., d. Oxidation-reduction reaction, , AgNO3 + NaCl ⟶ AgCl + NaNO3, SnCl2 + 2FeCl2 ⟶ SnCl4 + 2FeCl2, , 2. Reversible Reactions:, The reactions in which the products can react with one another to give back, the reactants again under suitable conditions called as reversible reaction., Properties of reversible reactions, a. In these types of reactions, reactant react to form product and product, further react to form reactant., b. Reactions proceed in both directions.
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ACTIVE SITE EDUTECH - 9844532971, , State Of Equilibrium, In general, reactions (physical and chemical) do not proceed to completion, when they are carried out in a closed container. Consider vaporization of, water, in closed vessel, Liquid water ⇌ Water vapour, , • At any temperature, vaporization of water takes place, initially the, concentration of water is much greater than the concentration of vapour, but, with the progress of time, concentration of vapour increases whereas that of, water remains constant and after a certain interval of time, there is no change, in concentration of vapour this state is known as state of physical equilibrium., • Similarly, in chemical reactions, for exp, when PCl5(g) is heated in a closed, container, its dissociation starts with the formation of PCl3(g)and Cl2(g). Initially,, only PCl5(g) was taken, but with the progress of reaction, PCl3(g) and Cl2(g) are, formed due to dissociation of PCl5(g) . After a certain interval of time, the, concentration ofPCl5(g), PCl3(g) and Cl2(g)each becomes constant., It does not mean that at this point of time, dissociation of PCl5(g), and its formation from PCl3(g) and Cl2(g) has been stopped. Actually, the rate, of dissociation of PCl5 and the rate of formation of PCl5(g)becomes equal. This state, is called the state of chemical equilibrium. So, the state of chemical equilibrium is, dynamic., ex:, , PCl5 (g) ⇌ PCl3 (g) + Cl2 (g), , This can be shown graphically., Forward Rate, , Rate, , Concentration, Mole/litre, , Reactant, , Product, , Backward Rate, , Equilibrium, , Equilibrium, Time, , Time, , So, state of chemical equilibrium in a reversible reaction at which both forward and, can be defined as the state backward reactions occur of the same speed.
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ACTIVE SITE EDUTECH - 9844532971, , Classification of equilibrium:, A. Equilibrium in Physical Process:, 1. Solid - Liquid equilibrium:- Certain amount of ice & water are taken in, a thermo flask., Exp- H2 O(s) ⇌ H2 O(L), When temperature remains constant, the mass of ice & water remains, constant. Some liquid molecules adhere to ice and simultaneously some, molecules of ice enter into liquid. The no. of molecules of water forming ice &, no. of molecules of ice forming water are same i.e., eq. is attained ∴ no, change in mass of ice & water., The temperature at which the solid & the liquid phase are at equilibrium at, atmospheric pressure is called freezing point., 2. Liquid – Vapour equilibrium:- Liquid water is taken in closed vessel at, room temp, it starts evaporating., Exp- H2 O(l) ⇌ H2 O(g), As process continuous, more water molecules escape & pressure increases, later, condensation takes place. Finally, rate of evaporation becomes equal, to rate of condensation., The pressure exerted by the vapour over the liquid when it is in equilibrium, with it is called vapor pressure of the liquid., The temperature at which the Vapour Pressure of liquid is equal to, atmospheric Pressure, is called Boiling Point., Different liquids have different vapour pressure. At the same temperature,, the liquid which has a higher Vapour Pressure is more volatile or boils at a, lower temperature., 3. Solid – Vapour equilibrium:- This exists when solid sublimes to vapour., I2(s) ⇌ I2(g), , When we heat some I2 in a closed vessel, it sublimes & vessel filled with, violet vapours. After sometime, intensity of violet colour remains constant. At
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ACTIVE SITE EDUTECH - 9844532971, , this state equilibrium is attained i.e., Rate of sublimation of solid I2 to form, vapour is equal to rate of condensation of I2 vapour to give solid I2 ., 4. Solid – Solution equilibrium:, Sugar (in solution) ⇌ Sugar(Solid), , When more and more amount of sugar is added into a fixed volume of water, at room temperature, after sometime no more sugar dissolves and settles at, the bottom. The solution is now said to be saturated and the concentration, of sugar in the solution remains constant., This indicates that a state of equilibrium has been reached, between the undissolved sugar & dissolved sugar., 5. Gas – Solution equilibrium: In carbonated drinks,, CO2(g) ⇌ CO2 (in solution), , At a given temperature, a liquid can dissolve only a, certain definite mass of the gas. This suggests that a state of equilibrium, exists between the molecules in the gaseous state and the molecules dissolved, in the liquid., The solubility of a gas in the liquid depends on the pressure., It is explained by Henry’s law., i.e., the mass of a gas dissolved in a given mass of solvent at a particular, temp is directly proportional to the pressure of the gas above the solvent., m ∝ p ⟹ m = kp, , B. Equilibrium in Chemical Process:, Ex 1. Decomposition of 𝐂𝐚𝐂𝐎𝟑:, When CaCO3 is heated in a closed vessel at 8000 C, it decomposes into, CaO & CO2 . Due to production of CO2 , pressure will increases in the vessel., After sometime, it is observed that pressure becomes constant at constant, temperature, even though some CaCO3 is still present. This constancy confirms, that equilibrium has been reached., CaCO3 (s) ⇌ CaO(s) + CO2 (g)
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ACTIVE SITE EDUTECH - 9844532971, , Ex 2. Reaction between 𝐇𝟐 & 𝐈𝟐:, When H2 & I2 are heated in a closed vessel, HI is formed. At the beginning, the reaction mixture is deep violet in colour due to presence of I2 . As the, reaction proceeds the intensity of the colour decreases due to decrease in, concentration of I2 and increase in concentration of HI . After some time,, the intensity of colour remains constant. This confirms that equilibrium has, been attained., H2 + I2 ⇌ 2HI, , Characteristics of Chemical Equilibrium:, 1. Equilibrium is attained for reversible reactions only in a closed vessel., 2. At equilibrium, the rate of forward reaction is equal to rate of backward, reaction., 3. For a reversible reaction, the equilibrium constant for the forward reaction is, inverse of the equilibrium constant for the backward reaction., In general, K forward reaction = K′ 1, backward reaction, , 4. At Equilibrium, all observable or measurable properties such as pressure,, concentration, colour and density etc remains constant., 5. It is dynamic in nature i.e. reaction takes place in both the directions at, same speed although appears to be stopped., 6. At equilibrium, concentration of reactants and products remains constant., 7. Catalyst has no effect on the position of chemical equilibrium, but it helps to, attain the Equilibrium state rapidly., A catalyst has the same effect on both forward and backward reactions., 8. Change in pressure, temperature or concentration favours either forward or, backward reaction and thus shifts the equilibrium point in one direction., 9. Change in free energy is equal to zero at equilibrium., 10. Equilibrium can be homogeneous or heterogeneous.
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ACTIVE SITE EDUTECH - 9844532971, , 11. Equilibrium can be attained in either direction. The value of an equilibrium, constant tells the extent to which a reaction proceeds in the forward or, reverse direction., CLASS EXERCISE, 1. An example of reversible reaction is:, a) Pb(NO3 )2 + 2 NaI = PbI2 + 2NaNo3, b) AgNO3 + HCl = AgCl + HNO3, c) 2Na + 2H2 O = 2 NaOH + H2, d)KNO3 + NaCl = KCl + NaNO3, 2. Which of the following is not a reversible reaction?, a) 2HI(g) = H2 (g) + I2 (g), b)PCl5(g) = PCl3 (g) + Cl2 (g), c) 2KClO3 (s) = 2KCl(s) + 3O2 (g) d)CaCO3 (s) = CaO(s) + CO2 (g), 3. Which one is not correct for a reversible reaction?, a) The reaction is never completed, b) The reactants are present in the initial stage but after that the, reactants and products are always present in the mixture., c) At equilibrium only products are present, d)When the reaction is carried out in a closed vessel, it attains equilibrium, state after suitabletime., 4. The state of equilibrium refers to, a) State of rest, b) Dynamic state, c) Stationary state, d) State of inertness, 5. In time kiln, the reversible reaction;CaCO3(g) ⇌ CaO(s) + CO2(g) proceeds, to, completion because:, a) Of high temperature, b) CO2 escapes out, c) CaO is removed, d) Of low temperature
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ACTIVE SITE EDUTECH - 9844532971, , HOME EXERCISE, 1. A reversible reaction is one which, a) proceeds in both directions, b) proceeds in one direction, c) proceeds spontaneously, d) All the above statements are wrong, 2. Which of the following is a characteristic of a reversible reaction?, a) It can never proceed to completion, b) It can be influenced by a catalyst, c) Number of moles of reactants and products are equal, d) None of the above, 3. All reactions which have chemical disintegration are, a) exothermic, b) reversible, c) reversible and exothermic, d) reversible or irreversible and endothermic or exothermic, 4. In any chemical reaction, equilibrium is supposed to be establish when, a) mutual opposite reactions undergo, b) velocity of mutual reactions become equal, c) concentration of reactants and resulting products are equal, d) the temperature of mutual opposite reactions become equal, 5. In chemical reaction,A ⇌ B,the system will be known in equilibrium when, a) 50% of A changes to B, b) A completely changes to B, c) only10% of A changes to B, d) the rate of change of A to B and B to A on both the sides are same, NCERT Text Book Questions :7. 1, 7. 32.
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ACTIVE SITE EDUTECH - 9844532971, , SESSION – 2AND 3, AIM - To introduce rate or velocity, active mass, equilibrium constant through Law, of Mass Action., Rate or Velocity of a reaction – is the change in molar concentration of the, reactants or products in unit time., It is denoted by, , 𝐝𝐱, 𝐝𝐭, , , where dx is the change in concentration of the reactants or, , products in a given time dt., For a reaction, Rate =, , −𝐝𝐑, 𝐝𝐭, , R ⟶ P, , ,where -dR indicates the decrease in the concentration id reactants., , For product, Rate =, , 𝐝𝐏, 𝐝𝐭, , , +dp indicates the increase in concentration of products., , Unit of the rate of reaction is mol L-1 s-1, Active Mass- is also known as molar concentration.It is shown by square brackets [ ]., The number of g molecules present in unit volume (say 1 litre) is called active, mass., g molecule =, Active mass =, , Weight of subs tance in g, Molecular weight, gram molecule, Volume (in liter), Weight of subs tance in g, Molecular weight Volume, , or Active mass =, The unit of active mass is g molecule/Litre, The active mass or partial pressure of solids is regarded as unity because molecules, are closely packed in solids, Example: What should be the respective active masses in g mole/litre when 4g of, H2 and 128g of HI are present in a two-litre container, 4g, 2 2, , Solution : (a) Active mass of H2 =, = 1 gm mol/lit, (Molecular weight of H2 = 2)
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ACTIVE SITE EDUTECH - 9844532971, 128, 128 2, , (b) Active mass of HI =, = 0.5g mole/litre, (Molecular weight of HI = 1 + 127 = 128), Remember 1 g molecule H2 = 2g, 1 g molecule O2 = 32g, 1 g molecule N2 = 28g, 1 g molecule NH3 = 17g, 1 g molecule I2, , =, , 254g, , Example: If 4g molecule of H2 is present in a two-litre container, the active mass, will be, Solution: Active mass =, , g molecule, Volume, , =, , 4, 2, , = 2 g mole/litre, , LAW OF MASS ACTION, Guldberg and Waage in 1807 gave this law and according to this law, “At, constant temperature, the rate of a reaction is directly proportional to the, product of active masses(molar concentrations) of the reactants”., For a reactation, A ⟶ Product, Rate,, , For a reaction,, , 𝐝𝐱, = K[A], 𝐝𝐭, , where K is Rate constant., , A + B ⟶ Product, Rate = K [A] [ B], , For a reaction,, , 2A + 3B ⟶ Product, Rate = K [A] [A] [ B] [B] [B], Rate = K [A]2 [ B]3, aA + bB ⟶ Product, Rate = K [ A]a [ B]b, , EX.1: 8.5grams of ammonia are dissolved to form 4L aqueous solution. Calculate, the active mass.
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ACTIVE SITE EDUTECH - 9844532971, , n=, , Sol., , weight, 8.5, =, = 0.5;, Grammolecular weight 17, n 0.5, =, V, 4, , Active mass =, = 0.125 mol L-1, EX.2: What is the active mass of one litre of Nitrogen gas at NTP?, Sol. At STP, active mass of Nitrogen = 1 atm, At STP,22.4 litres of Nitrogen= 1 mole, Active mass, , =, , n, 1, =, = 0.0446, V 22.4, , mol L-1, , EX.3: Determine the active mass of NaOH in a solution containing 4gm of NaOH in, 500ml., Solution, , Active mass =, , no.of moles, Vol.of solution inL, , =, , 4, 40, , ×, , 1, 0.5, , = 0.2mL−1, , EX.4: Calculate the active masses of ethanol and carbon tetrachloride taking, their densities to be 0.8 and 1.58g per ml respectively., Solution - Active mass = Molar conc. = No. of Moles Litres–1., (i) Active mass of ethanol, 1litre of ethanol = 1000ml of ethanol, = 1000 × 0.8m of ethanol., [Q density of C2H5OH = 0.8gl–1] = 800gm of ethanol., No. of moles/ litre of ethanol = 800, = 17.39 [Q Mol wt. of C2H5OH=46], 46, Hence active mass of ethanol = 17.39moles/litre., (ii) Calculation of active mass of CCl4, 1 litre of CCl4 = 1000ml of CCl4, [Q density of CCl4 =1.58 gl–1], = 1580g of CCl4, No. of moles/litre = 1580, = 10.26 [Q Mol wt. of CCl4 =, 154, , = 1000 × 1.58g, , Hence active mass of CCl4 =, , 10.26moles/litre, , 154]
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ACTIVE SITE EDUTECH - 9844532971, , Equilibrium Constant Kc in terms of concentration:, Consider the following reversible reaction, aA + bB ⇌ cC + dD, , Rate of forward reaction,, , R f ∝ [A]a [B]b, , R f = k f [A]a [B]b --------(i), , Where k f =rate constant of forward reaction, Rate of reverse reaction, R b ∝ [C]c [D]d, R b = k b [C]c [D]d --------(ii), Where k b =rate constant of reverse reaction, At equilibrium,, Rate of forward reaction = Rate of reverse reaction, i.e .R f = R b, So, from equations (i) and (ii) we get, k f [A]a [B]b = k b [C]c [D]d, , or,, , kf, kb, , [C]c [D]d, , = [A]a [B]b, [C]c [D]d, kf, = Kc =, [A]a [B]b, kb, , Where, 𝐊 𝐜 is the equilibrium constant in terms of molar concentration., The equilibrium constant, at a given temperature, is the ratio of rate constant of, forward and backward reactions., Or, 𝐊 𝐜 is defined as the ratio of the product of molar concentration of, the products and to the product of the molar concentration of the reactants in, which concentration terms raised to the power of respective stoichiometric, coefficient in a balanced chemical equation at constant T.
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ACTIVE SITE EDUTECH - 9844532971, , From ideal gas equation, , n, , or, P = V RT, , PV = nRT, 𝐧, , At constant temperature, 𝐏 ∝ 𝐕, In a mixture of gases, Partial pressure of any component (say A), PA ∝ [A], , Similarly,, , PB ∝ [B], Pc ∝ [C], PD ∝ [D], , So, equation, , (1), , can be rewritten as, Kp =, , Pcc × PDd, PAa × PBb, , Relationship between KP & KC, aA(g) + bB(g) ⇌ cC(g) + dD(g), [C]c [D]d, , K c = [A]a [B]b, , and, K p =, , PcC ×Pd, D, PaA ×Pb, B, , From the ideal gas equation,, PV = nRT, n, P = RT, V, So, PA = [A]RT ; PB = [B]RT;, , PC = [C]RT ;, , PD = [D]RT, , Substituting the values of PA, PB, PC and PD, we get, [C]c (RT)c × [D]d (RT)d, Kp =, [A]a (RT)a × [B]b (RT)b, [C]c [D]d, , (RT)(c+d), , Or K p = [A]a [B]b × (RT)(a+b), Or K p = K c (RT)Δn, Where, Δn = (c + d) − (a + b), =no of moles of gaseous products - no. of moles of gaseous reactants., , i.e., Δn, Case (i): If Δn = O; K c = K p, Ex: H2(g) + I2 (g) ⇌ 2HI(g), Case (ii): If Δn is positive; K p > K c, Ex: PCl5(g) ⇌ PCl3(g) + Cl2 (g), , Δn = 2 − 1 = 1
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ACTIVE SITE EDUTECH - 9844532971, , ∴ K p = K c (RT), , Case (iii): If Δn is negative; K p < K c, Ex: N2(g) + 3H2(g) ⇌ 2NH3(g), Δn = 2 − 4 = −2, Kp = Kc, , (RT)−2, , Example 1. The value of KP for the reaction 2H2O(g) + 2Cl2(g) ⇌ 4HCl(g) + O2(g) is, 0.035atm at 400oC, when the partial pressures are expressed in, atmosphere. Calculate KC for the reaction 12 O2(g) + 2HCl(g) ⇌ Cl2(g) + H2O(g), Solution:, , K P = K C (RT)Δn, Δn =, , Moles of product – moles of reactance, , =5−4=1, , R = 0.082Latm/molK, T = 400 + 273 = 673K, ∴ 0.035 = K C (0.082 × 673), K C = 6.342 × 10−4 moll−1, ∴ K′C, , for the reverse reaction would be, ∴ K′C =, , 1, KC, , ., , 1, = 1576.8(moll−1 )−1 ., 6.342 × 10−4, , When a reaction is multiplied by any number n (integer or a fraction), the K′C or K′P becomes(KC )n or (KP )n of the original reaction., 1, ∴ K C for 2 O2 (g) + 2HCl(g) ⇌ Cl2 (g) + H2 O(g) is √1576.8 = 39.7(mol. l−1 )−½, Problem 2. K P for the equilibrium, FeO(s) + CO(g) ⇌ Fe(s) + CO2(g) at 1000°C is 0.4. If, CO(g) at a pressure of 1atm and excess FeO(s)are placed in container, at 1000oC, what are pressures of CO(g) and CO2 (g) when equilibrium is, attained?, Solution: Acc, to ideal gas equn, partial pressures are proportional to the no. of, moles present. Since moles of CO2 formed equals moles of CO consumed,, the drop in partial of CO will equal the partial pressure of CO2, produced. Let the partial pressure of CO2 at equilibrium be ′x′ atm., Then, partial pressure of CO will be (1 − x)atm., P, x, Since K p = PCO2 = 1−x, = 0.4 ⇒ x = 0.286, CO, Hence PCO = 1 − x = 𝟎. 𝟕𝟏𝟒𝐚𝐭𝐦
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ACTIVE SITE EDUTECH - 9844532971, , Problem 3. At temperature T, a compound AB2 (g) dissociates according to the, reaction: 2AB2(g) ⇌ 2AB(g) + B2 (g) with a degree of dissociation ′x′ which is, small compared to unity. Deduce the expression for ′x′ in terms of the, equilibrium constant K Pand the total pressure P., Solution, 2AB2 (g) ⇌ 2AB(g) +, B2 (g), Mole before dissociation, 1, 0, 0, 𝑥, (1 − x), Mole after dissociation, x, 2, x, x, Total mole at equilibrium (∑n) = 1 − x + x + 2 = 1 + 2, Now,, , Kp =, , KP =, , or, , nB2 ×(nAB )2, , x, ,(x)2, 2, (1−x)2, 3, , x= √, , (nAB2 ), , ×[, , P, , P Δn, , ×[ ], ∑n, , 1, x] =, , 1+2, , 𝑥3, 2, , [Q, , x is, , small,, , 1 − x 1, , and1 + x 1 ], 2, , 2KP, P, , Problem 4. In which case does the reaction go farthest to completion:, K = 1; K = 1010 ; K = 10−1, , Solution, , [Product], , The ratio [Reactant] is maximum when, farthest to completion when K = 1010 ., , and why?, K = 1010, , and thus reaction goes, , Problem 5. The K C for A2 (g) + B2 (g) ⇌ 2AB(g) at 100°C is 50. If one litre flask, containing one mole of A2 is connected with a two litre flask containing, 2mole of B2 . how many mole of AB will be formed at 100°C?, Solution, A2 (g), +, B2 (g) ⇌, AB2, Initial mole, 1, 2, 0, (2 − x), Final mole at equilibrium (1 − x), 2x, Total volume of both containers on joining becomes 3 litre., At, , 2−x, 2x, equilibrium, [A2 ] = 1−x, ; [B2 ] =, ; [AB] =, 3, 3, 3, , KC =, , [AB]2, [A2 ][B2, , =, ], , 4x2, 1−x 2−x, 9( 3 )( 3 ), , 4x2, , = (2−3x+x2 ) = 50, , , , 4𝑥 2 = 100 − 150𝑥 + 50𝑥 2, , or, , 46x 2 − 150x + 100 = 0
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ACTIVE SITE EDUTECH - 9844532971, , x = 0.93, , Mole of, , mol and, , 2.326, , which is not valid since, , x | 2, , AB = 2x = 2(0.93) = 𝟏. 𝟖𝟔, , Problem 6. NH3 is heated at 15atm from 27°C to 247°C assuming volume constant., The new pressure becomes 50atm at equilibrium of the reaction, 2NH3 ⇌ N2 + 3NH2 ., , Calculate, , %, , of mole of, , Solution, , actually decomposed., , NH3, , 2NH3, , ⇌, , Initial mole, a, (a − 2x), Mole at equilibrium, Initial pressure of a mole of NH3 of, , , , 15, , 0, , 0, , x, , 3x, , atm at, , 347°C, , 620, , P = 31atm, , At constant volume and at, a + 2x ∝ 50, 𝑎+2𝑥, 𝑎, , , , x=, , , , %, , mole, , pressure, , (After equilibrium), , 50, , = 31, , 19, 62, , of, , 347°C, , (Before equilibrium), , a ∝ 31, , , , NH3 = P, , 3H2, , P, , =, , 300, , +, , N2, , a, , NH 3 decomposed, , =, , 2x, a, , × 100 =, , 2×19a, 62×a, , × 100, , = 𝟔𝟏. 𝟑%, , Characteristics of Equilibrium Constant:, The value of equilibrium constant is independent of the following factors:, 1. Initial concentrations of the reactants involved, 2. The presence of a catalyst, 3. The direction from which the equilibrium has been attained, 4. The presence of inert materials
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ACTIVE SITE EDUTECH - 9844532971, , Factors determining the value of equilibrium constant:, The value of equilibrium constant depends in the following factors:, 1. The mode of representation of the reaction:, A+B⇌ C+D, , The equilibrium constant for the reaction,, [C][D], , …(i), , K c = [A][B], , Now, for backward reaction,, C+D⇌A+B, , The equilibrium constant for the reaction is, K ′c =, , [A][B], … … … … . (ii), [C][D], 1, , The equilibrium constant, K ′c is the reciprocal of K c i.e., K ′c = K, , c, , 2. Stoichiometry of the chemical equation:, Example 1 :, 2NO2 ⇌ N2 + 2O2 … (i), , Or, , 1, , NO2 ⇌ N2 + O2 ….(ii), 2, , For equation (i), the value of, , [N2 ][O2 ]2, , Kc =, , [NO2 ]2, 1, , For equation (ii), the value of K ′c =, , [N2 ]2 [O2 ], [NO2 ], , Thus, the two constants are related as, 𝐊 ′𝐜 = √𝐊 𝐜, , In general, when a given reversible system is divided by n , the numerical, value of equilibrium constant is obtained by taking n√k of the original system., Example 2:, 1, 2, , 1, , H2 + I2 ⇌ HI, 2, , H2 + I2 ⇌ 2HI, , … (i), …(ii), 1, , For equation (i), the value of, , Kc =, , 1, , [H2 ]2 [I2 ]2, [HI]
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ACTIVE SITE EDUTECH - 9844532971, , For equation (ii), the value of K ′c =, , [H2 ][I2 ], [HI]2, , Thus, the two constants are related to each other, 𝐊 ′𝐜 = 𝐊 𝟐𝐜, , In general, when a balanced equation having equilibrium constant K c , is, multiplied by a certain value n, the equilibrium constant for the new equation, will be equal to (K c )n ., 3. Use of partial pressures instead of concentrations: When the reactants and, products are in gaseous state, the partial pressures can be used instead of, concentrations at a definite temperature, as the partial pressure of a, substance is proportional to its concentration in the gas phase., K p = K c (RT)Δn, K p can, , be equal or less than or greater than Kc, depending upon the, , chemical reaction., 4. Temperature: According to Arrhenius equation,, −, , E, RT, , …(i), Where, k = rate constant,, E = activation energy,, R= gas constant., T = absolute temperature and, K = A. e, , log, , k2, k1, , =, , −E, , [, , 1, , 2303R T2, , −, , 1, T1, , ], , e = exponential constant., , …. (ii), , When, T2 > T1, For forward reaction,, kf, Ef, 1, 1, log ( 2 ) = −, [ − ] … . (iii), k f1, 2.303R T2 T1, , For backward reaction,
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ACTIVE SITE EDUTECH - 9844532971, , Or log K1 > log K 2, Or K1 > K 2, The value of equilibrium constant decreases with increase in temperature in the, case of exothermic reactions., Problem 1. At, , 10650C,, , Kp, , = 0.0118atm for the reaction, 2H2S(g), 2H2(g)+ S2(g), , The enthalpy of the reaction is 177.3kJ/mol. Calculate the equilibrium, constant at 12000C. (Given R = 8.314J), Solution: The effect of temperature over the equilibrium constant is given by,, Van’t Hoff equation:, K 11, p, , log, , =, , K 1p, , K 1p, , and, , ΔH, 2.303R, , [, , T2 −T1, T1 T2, , ], , K 11, p, , are the equilibrium constants at temperatures T1 and T2, T1 = 1065 + 273 = 1338K and K = 0.0118, T2 = 1200 + 272 = 1473K and K = ?, 1, p, , 11, p, , log, , K11, p, K1p, , =, , =, =, K11, p, K1p, , 177 .3, 2.303 8.314 10, , −3, , 1473 − 1338 , 1473 1338 , , , , 177 .3 135, 2.303 8.314 1473 1338 10 − 3, , 177 .3 135 10 3, = 0.6342, 2.303 8.314 1473 1338, , = 4.307, , 1, K 11, p =4.307 K p = 4.307 0.0118atm = 0.0508atm, , Problem 2. For the reaction N2(g)+3H2(g), 2NH3(g) at 773K, the equilibrium, constant Kp=1.40410-5Pa-2. Calculate the value of Kc for this equilibrium, with concentration units of mol/L and mol/m3. [R = 8.134J/k/mol], Solution: N2(g) + 3H2(g), 2NH3(g), n= - 2.0, Kp = 1.404 10-5Pa-2 = 1.404 10-5(N/m2)2, Kp = Kc.(RT)-2, Kc = Kp., , (RT)2, , =, , 1.404 , , 10-5 , , N , 2 , m , , −2, , 2, , 8.314 , 2, , (773 K ), K.mol
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ACTIVE SITE EDUTECH - 9844532971, , =, , 1.404 10 − 5 m 4, , =, , 5.798 10 −10 m 6, , N2, , mol 2, , , , (8.314 )2 N 2 m 2 (773 K )2, K 2 .mol 2, , =, , 5.798 10 −8, mol 2 / m 6, , = 5.79810-8C-2, , where C = conc. in mol/m3, Units of Equilibrium Constant- The units of equilibrium constant vary in case of, different reactions., - ‘K’ has no units for a reaction in which total number of moles of, reactants and products are the same., For exp , for dissociation of nitric oxide, Kc has no units., 2NO ⇌ N2 + O2, [N2 ][O2 ], Kc =, [NO]2, , -, , will have units for a reaction in which the total number of moles of, reactants and products are different., For exp, for decomposition of PCl5, the Kc has mol/litre units., 𝐊𝐜, , PCl5 ⇌ PCl3 + Cl2, [PCl3 ][Cl2 ], Kc =, [PCl5 ], , In the formation of ammonia,, N2 + 3H2 ⟶ 2NH3, [NH3 ]2, , K c = [N, , 2 ][H2 ], , 3, , Kc, , has, , 𝐥𝐢𝐭𝐫𝐞𝟐 𝐦𝐨𝐥−𝟐, , units, , In general; unit of K c = [M]Δn, Where, M = mol litr −1, Δn =no, of gaseous moles of products – no, of gaseous moles of reactants., Note: (i) The above relation can be used in homogeneous liquid system., (ii) Similarly, the unit of K p = [atm]Δn
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ACTIVE SITE EDUTECH - 9844532971, , NOTE, • If Kc > 103 , products predominate over reactants. If Kc is very large, the, reaction proceeds almost all the way to completion., • If Kc < 10−3 , reactants predominate over products. If Kc is very small, the, reaction proceeds hardly at all., • If Kc is in the range 10−3 to 103 , appreaciable concentration of both, reactants and products are present., Kc, Reaction proceeds, hardly at all, , 103, , 10–3, , Reaction proceeds to, completion, , Both reactants and products, are present at equilibrium, , Reaction Quotient OR Mass Action Ratio, Let us consider a reaction:, A + B ⇌C + D, [C][D], , Q = [A][B], Q is denoted as Qc or Qp depending upon whether the concentration is taken, in terms of moles per litre or partial pressures respectively., With the help of mass action ratio we can determine whether the reaction is, at equilibrium or not., I. When, 𝐐𝐜 = 𝐊 𝐜 𝐨𝐫 𝐐𝐩 = 𝐊 𝐩 , then the reversible reaction is at equilibrium,, i.e, the rate of forward and backward reaction becomes equal., II. When, 𝐐𝐜 < 𝐊 𝐜 𝐨𝐫 𝐐𝐩 < 𝐊 𝐩 . The reaction will be fast in forward direction,, i.e, reaction has tendency to form product/products., Rate of forward reaction > Rate of backward reaction., III. When, 𝐐𝐜 > 𝐊 𝐜 𝐨𝐫 𝐐𝐩 > 𝐊 𝐩 . The reaction will be fast in backward direction,, i.e, have a tendency to form reactants., Rate of forward reaction < Rate of backward reaction., , Q, , Kc, , Reactants → Products, , Q, , Kc, , Equilibrium, , Q, , Kc, , Reactants Products
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ACTIVE SITE EDUTECH - 9844532971, , Problem 1. For the reaction,, A(g) + B(g), 2C(g) at 25°C, in a 2litre vessel contains, of respectively. Predict the direction of the reaction if, (a) Kc for the reaction is 3, (b) Kc for the reaction is 6, (c) Kc for the reaction is 4.5, Solution:, , A(g) + B(g), , 1, 2, 3, , moles, , 2C(g), , Reaction quotient Q =, , [C]2, [A][B], , 3 2, , =, , (2), , 1 2, ×, 2 2, , 9, , = = 4.5, 2, , (a) ∴ Q > Kc , therefore, backward reaction will be followed, (b) ∵ Q < K c ∵The forward reaction is followed, (c) Q = Kc The reaction is at equilibrium, Problem 2., , For the reaction: A(aq)+ B(aq) ⇌ C(aq) +D(aq) , the net rate of, consumption of B at 25C and at any time 't' is as given below, d [B ], , - dt = {410-4[A] [B] – 1.3310-5 [C] [D]} mol L-1 min-1., Predict whether the reaction will be spontaneous in the direction as, written in reaction mixture in which each A, B, C and D is having a, concentration of 1mol L-1?, Solution:, , K=, =, , Q=, , K1, K2, , 4×10−4, , 1.33×10−5, , = 30, , [C][D], , [A][B], 1×1, , = 1×1= 1 < K, Since Q < K, so the above reaction is spontaneous in the forward direction., CLASS EXERCISE, 1. The concentration of reactants is increased by x, then equilibrium constant, K becomes, K, , a) ln x, , b), , 𝐾, 𝑥, , c) K + x, , d) K
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ACTIVE SITE EDUTECH - 9844532971, , 2. In which of the following reactions ;K p > K c ?, a)N2 (g) + 3H2 (g) ⇌ 2NH3 (g), b) H2 (g) + I2 (g) ⇌ 2HI(g), c)PCl3 (g) ⇌ Cl2 (g) ⇌ Cl5 (g), d)2SO3 (g) ⇌ 2SO2 (g) + O2 (g), 3. The equilibrium constant for a reaction A + 2B ⇌ 2C is 100. The equilibrium, constant for reaction C ⇌ B + 12 A is, a), , 1, , b)(, , 100, , 1, 100, , 1/2, , ), , c)(, , 1, , 2, , ), , 100, , d) 100, , 4. The equilibrium constant for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g) is K and, for the reaction 12 N2 (g) + 32 H2 (g) ⇌ NH3 (g) is K’2 K and ‘K’ will be related, to each other as, a) K = K1, b) K1 = √K, c)K = √K ′, d)K × K1 = 1, 5. If equilibrium constant for the reaction 2HI(g) ⇌ H2 (g) + I2 (g) is 0.25. Then, the equilibrium constant for the reverse reactionwill be, a) 1, b) 2, c) 3, d) 4, 6. In the gas phase reaction C2 H4 + H2 ⇌ C2 H6 , the equilibrium constant can be, expressed in the units of, a) liter −1 mol−1 b)mol2 litre−2 c)litre mol−1, d) mol litre, 7. The reaction quotient (Q) for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g)is given, [𝑁𝐻3 ]2, , by 𝑄𝑐 = [𝑁, , 2 ][𝐻2 ], , The reaction will proceed from right to left if, , a) Q = 0, 8. The ratio of, , b) Q = K c, c)Q < K c, d)Q > K c, K p toK c for the following equilibrium, N2 (g) + 3H2 (g) ⇌ 2NH3 (g) at 400 K is(R=2 Cal/K mole), a)8.21 × 10-2, b)1.56 × 10-6 c)1.56 × 10-4, d)8.21× 10-6, 9. If for the equilibrium; 2SO2(g) + O2(g) ⇌ 2SO3, Kp = 2.0 × 1010 bar at, value of K c at this temperature is, a)7.48 × 109 L mol−1, b) 2.0 × 1010 L mol−1, c) 7.48 × 1011 L mol−1, d)7.389 × 1011 L mol−1, , 450K,, , the
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ACTIVE SITE EDUTECH - 9844532971, , 10. At 3000K, the equilibrium pressure of CO2 , CO and O2 are 0.6, atm respectively. K pfor the reaction,2CO2 ⇌ 2CO + O2 , is, a) 0.089, b) 0.098, c) 0.189, d) 0.198, HOME EXERCISE, 1. The ratio of K p /Kcfor the reaction, , 0.4, , and, , 0.2, , 1, , CO(g) + O2 (g) ⇌ CO2 (g)is, 2, , c) 1, b) RT, c)(RT)1/2, d)(RT)−1/2, 2. For the chemical reaction 2A + B ⇌ C, the units of K pis, a) atm−2, b) atm−3, c)atm−1, d) dimensionless, 3. What will be the effect of increasing temperature on the equilibrium, constant, if the reaction either absorbs heat nor releases heat?, a) Equilibrium constant will remain constant, b) Equilibrium constant will decrease, c) Equilibrium constant will increase, d) Cannot be predicted, 4. At a given temperature, the equilibrium constant for the reactions, 1, NO(g) + O2 (g) ⇌ NO2 (g) and 2NO2 (g) ⇌ 2NO(g)O2 (g) are, 2, K1 and K 2 respectively., , If, , K1 is 4 × 10−3 ,, , then, , will be, a) 8 × 10−3, b)16 × 10−3, c)6.25 × 104, d)6.25 × 106, 5. When two reactants A and B are mixed to give products C and D, the, reaction quotient, Q, at the initial stages of the reaction, a) is zero, b) decreases with time, c) is independent of time, d) increases with time, 6. The active mass of 45g of KCl in a 3L flask would be, a) 0.20, b) 2.0, c) 3, d) 4, K2
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ACTIVE SITE EDUTECH - 9844532971, , 7. The rate constant for forward reaction and backward reaction of hydrolysis, of ester are 1.1 × 10−2 and 1.5 × 10−3 per minute respectively. Equilibrium, constant for the reaction is, 𝐶𝐻3 𝐶𝑂𝑂𝐶2 𝐻5 + 𝐻2 𝑂 ⇌ 𝐶𝐻3 𝐶𝑂𝑂𝐻 + 𝐶2 𝐻5 𝑂𝐻, , a) 33.7, b) 7.33, c) 5.33, d) 33.3, 8. For the reaction, PCl3 (g) + Cl2 (g) ⇌ PCl5 (g) , the value of, 0.67 atm−1 . The value of K c at this temperature will be, a) 15 (, , mol −1, L, , ), , −1, , b)26 (mol, ), L, , −1, , c)35 (mol, ), L, , d), , 52 (, , Kp, , at, , mol −1, L, , NCERT Text Book Questions: 7. 2,7. 4,7. 5,7. 6,7. 7,7. 13, , ), , 250°C is
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ACTIVE SITE EDUTECH - 9844532971, , SESSION – 4 AND 5, AIM - To apply law of Mass Action to different reversible systems, APPLICATIONS OF LAW OF MASS ACTION, 1. Synthesis of Hydrogen Iodide:, Suppose ‘a’ moles of H2 and ‘b’ moles of I2 are heated at 444°𝐶 in a closed, container of volume ‘V’ liter and at equilibrium, 2x moles of HI are formed., H2(g), Initial moles, Initial concentration, (mol L−1 ), Equilibrium moles, Equilibrium, concentration (mol L−1 ), , +, , b, b, V, , o, , a− x, a−x, V, , b− x, b−x, V, , 2x, 2X, v, , H2 , I2 and HI, , in above equation,, , Kc =, , V, , a−x, , (, , Kp =, , V, 2, , b−x, , )(, , V, , ), , =, , 0, , [HI]2, [H2 ][I2 ], , Substituting the equilibrium concentrations of, we get, 2x 2, , ⇄ 2HI(g), , a, a, V, , Kc =, , ( ), , I2(g), , 4x 2, (a − x)(b − x), , P HI, PH2 , PI2, Total no. of moles at equilibrium = (a – x) + ( b – x) + 2x = ( a + b), a−x, , PH2 = P × xH2 = P × a+b, , Similarly, PI2 =, , P×(b−x), (a+b), P×2x, , PHI = (a+b), Kp =, , 2x 2, ), a+b, a−x, b−x, P×, ×P×, a+b, a+b, 2, 4x, , (P×, , ∴ K p = (a−x)(b−x), From, , ----------(iv), , equations (iii) and (iv), that, Kp = Kc, , This is so, because n = 0 for the synthesis of HI from H2 and I2
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ACTIVE SITE EDUTECH - 9844532971, , 2. Thermal dissociation of PhosphorusPentachloride, PCl5(g) dissociates thermally according to the reaction,, PCl5(g) ⇄ PCl3(g) + Cl2 (g), , Let us consider that ‘a’ mole of PCl5 has been taken in a container of volume, V litre and at equilibrium x moles of PCl5(g) dissociates., Thus PCl3(g) + Cl2(g), PCl5(g) ⇄, Initial moles, Initial concentration, (mol L−1 ), Equilibrium moles, Equilibrium, concentration (mol L−1 ), , PCl3(g) +, , Cl2(g), , a, a, v, , 0, , 0, , 0, , 0, , (a – x), a−x, V, , x, x, V, , x, x, V, , According to law of mass action, at constant temperature,, Kc =, , [PCl3 ][Cl2 ], , And K p =, , ..............(i), , [PCl5 ], PpCl3 ×PCl2, , .............(ii), , PPCl5, , Substituting the values of equilibrium concentration, in equation (1), we have, Kc =, , x x, ×, V V, α−x, V, , x2, , or K c = V(a−x), , Now, total number of moles at equilibrium, Mole fraction of, , PCl3, , = a − x + x + x = a + x, a, , = Mole fraction of Cl2 = a+x . P, a−x, , and mole fraction of PCl5 = a+x . P, Suppose total pressure at equilibrium is P, then we have from equation (2),, Kp =, , x, x 2, ×, P, a+x a+x, a−x, P, a+x, , or K p =, , x2 P, (a−x)2, , 3. Thermal dissociation of solid ammonium chloride:, The thermal dissociation of NH4 Cl(s) takes place in a closed container, according to the equation: NH4 Cl(s) ⇌ NH3(g) + HCl(g)
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ACTIVE SITE EDUTECH - 9844532971, , 7. Dissociation of ammonium carbonate:, NH2 COCNH4 (s) ⇌ 2NH3 (g) + CO2 (g), 2, , K c = [NH3 ]2 [CO2 ] ;, , K p = [PNH3 ] × [PCO2 ], , NOTE:- Relation between degree of dissociation (α) density of a gaseous system., α=, , D−d, d(n − 1), , D = initial vapour density, d = vapour density at equilibrium, n = no. of molecules formed on dissociation, Homogeneous equilibria,equations K (Equilibrium pressure in atm in a V L flask), n = 0 ; K p = Kc, H 2 + I2 ⇌, Initial mole, Mole, at, Equilibrium, , 1, , (g), , 1, , (1–x), , (1– x), , Total mole at, equilibrium, , N 2 + 3 H 2 ⇌ 2 NH 3, , 2 HI, , (g), , (g), , n 0; K p Kc, , n 0 ; K p Kc, (g), , 0, , 1, , 2x, , (1–x), , (g), , 2 SO 2 + O 2 ⇌ 2 SO 3, , (g), , (g), , 30, , 2, , (g), , 2, , (3–3x), , 2x, , (2–2x), , (4 – 2x), , PCl 5 ⇌ PCl 3 + Cl 2, , (g), , 1, , 0, , 1, , (1–x), , 2x, , (1–x), , 2x, V, , 1 − x 1 − x 2x , , 3, , , V V V , , 2 − 2x 1 − x , , , , V V , , Mole fraction, , 1 − x 1 − x , , , , 2 2 , , 2x, 2, , x, 1− x 31− x , , , 2 (2 − x ) 2 2 − x (2 − x ), , 2 − 2x , , , 3−x , , Partial, pressure, , 1 − x 1 − x 2x , p, , p, p, 2 2 2 , , 1 − x 3(1 − x ) Px, P , , P , , 2(2 − x ) _ 2(2 − x ) (2 − x ), , 2 − 2x , 2x , P, P 1 − x P , , 3−x , 3− x, 3 − x, , Kp, , 4 x 2V 2, , (1 − x ) 2, 4x2, , (1 − x ), , 2x , , , V , , 1 − x 2x , , , , 3− x 3− x, , x 2V, , (1 − x ) 3, , 16 x 2 (2 − x ) 2, , x 2 (3 − x ), , 4, , P (1 − x ), , 2, , 0, x, , 1 − x , , , V , , x, , V, , x, , V, , 1− x x x , , , , , 1 + x 1 + x 1 + x , 1− x , , 1 + x , , x , , 1 + x , , P, , P, , x , , 1 + x , , P, , x2, (1 − x ) V, , 27 (1 − x ) 4, 27 (1 − x ) P, , 2, , 0, , (1 + x), , 1 − x 1 − x , , , , V V , , 4x2, , (g), , x, , (3 – x), , Active masses, , Kc, , (g), , (g), , Px 2, 1− x2, , (, , 3, , ), , Heterogeneous equilibria and equation for equilibrium constant (Equilibrium pressure is P atm), NH 4 HS (s) ⇌ NH 3 (g) + H2S (g), Initial mole, , 1, , Mole at equilibrium, , (1–x), , Total moles at equilibrium, (solid not included), Mole fraction, , 0, , 0, , x, , 1, x, , 2x, , x, 1, =, 2x 2, , Partial pressure, , P, 2, , Kp, , P2, 4, , C(s) + CO2 (g) ⇌ 2CO (g), 1, , (1–x) (1–x), , 0, , 1, , 0, , 2x, , (1–x), , 2x, , (1+x), , 1, 2, P, 2, , NH2CO2 NH4 (s) ⇌ 2 NH3 (g) + CO2 (g), 0, x, , 3x, , 1 − x , , , 1 + x , , 2x , , , 1 + x , , 1 − x 2x , P, P, , 1 + x 1 + x , , 4P x2, (1 − x 2 ), , 1, 3, , 2, 3, , 2P, 3, , 4 P3, 27, , P, 3
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ACTIVE SITE EDUTECH - 9844532971, , CLASS EXERCISE, 1. An equilibrium mixture for the reaction 2H2S(g) ⇌ 2H2(g) + S2(g) has 1mole of, 0.2mole of H2 and 0.8mole of S2 in a 2litre flask. The value of K c in molL−1is, a) 0.004, b) 0.08, c) 0.016, d) 0.160, 2. The degree of dissociation ofPCl5 (α)obeying the equilibrium:, PCl5 ⇌ PCl3 + Cl2 is related to the pressure at equilibrium by, a)α ∝ P, , 1, , b) α ∝ √P, , 1, , c)α ∝ P2, , H2 S ,, , d)α ∝ 1/P4, , 3. When 3moles of a reactant A and 1mole of the reactant B are mixed invessel of, 1litre, the following reaction takes place A(g) + B(g) ⇌ 2C(g) . If 1.5mole of C is, formed at the equilibrium, then equilibrium constant (K c )for the reaction is, a) 0. 12, b) 0.50, c) 0.25, d) 4.0, 4. For a reaction A + B ⇌ C + D, initially we start with equal concentration of A and B., At equilibrium, we find that moles of C are two times of A. The value of equilibrium, constant for the reaction is, a), , 1, 4, , b), , 1, 2, , c) 4, , d) 2, , 5. 4moles each of SO2 and O2 gases are allowed to react to form SO3in a closed vessel. At, equilibrium 25%of O2 is used up. The total number of moles of all the gases present, at equilibrium is, a) 6.3, b) 7.0, c) 8.0, d) 2.0, 6. For the reaction 2A(g) + B(g) ⇌ 3C(g) + 4D(g) two moles each of A and B were taken, into a 1L flask. The following must always be true when the system attained, equilibrium, a) [A] = [B], b)[A] < [B], c)[B] = [C], d[A] + [B] < [C] + [D], 7., , At 550K, the Kc for the following reaction is 104 L mol−1, 1, 1, X(g) + Y(g) ⇌ Z(g)At equilibrium, it was observed that [X] = 2 (Y) + 2 (Z). What is the, value of [Z] in mol L−1 at equilibrium?, a) 2 × 10−4, b) 10−4, , c)2 × 104, , d)104
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2, , ACTIVE SITE EDUTECH - 9844532971, , 8. A sample of HI(g) is placed in a flask at a pressure of 0.2atm. If the partial, pressure of HI(g) at equilibrium is 0.04atm, the value of K p is, a) 2, b) 4, c) 6, d) 8, 9. A nitrogen-hydrogen mixture initially in the molar ratio of 1:3 reached equilibrium to, form ammonia when 25% of the material had reacted. If the total pressure of the, system was 21atm, the partial pressure of ammonia at the equilibrium was, a) 4. 5atm, b) 3.0atm, c) 2.0atm, d) 1.5atm, 10. The vapour density of PCl5 is 104.25 but when heated to 230°C, its vapour density, is reduced to 62. The degree of dissociation of PCl5 at this temperature will be, a) 6.8%, b) 68%, c)46%, d) 64%, HOME EXERCISE, 1. If, in the reaction N2O4 ⇌ 2NO2 , x is that part of, number of molecules at equilibrium, will be, 2., , N2 O4, , which dissociates, then the, , a) 1, b) 3, c) (1 + x), d) (1 + 𝑥 )2, What is K c for the following equilibrium when the equilibrium concentration of, eachsubstance is [SO2] = 0.60M, [O2 ] = 0.82Mand [SO3] = 1.90M?, 2SO2 (g) + O2 (g) ⇌ 2SO3 (g), , a) 11.229, b) 12.229, c) 3.861, d) 7.33, 3. When 3moles of ethyl alcohol are mixed with 3moles of acetic acid, 2moles of esters are formed, at the equilibrium point. The value of equilibrium constant is, a) 4, b) 2/9, c) 2, d) 4/9, 4. 1.1mole of A are mixed with 2.2moles B and the mixture is kept in a 1L vessel till the, equilibrium is established in the reaction. If the molar conc. of C at the equilibrium, point is 0.2mole, the value of equilibrium constant for the reaction A + 2B ⇌ 2C + D is, a) 0.001, b) 0.002, c) 0.003, d) 0.004, 5. 5moles of SO2 and 5moles of O2 are allowed to react to form SO3 in the closed vessel., At equilibrium state, 60% of SO2 is used. The total number of moles of SO2, O2 and SO3, in thevessel now is
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2, , ACTIVE SITE EDUTECH - 9844532971, , 6., , 7., , 8., , 9., , a) 10.0, b) 8.5, c) 10.5, d) 3.9, 1mole of hydrogen and 2moles of iodine are taken initially in a 2litre vessel. The, number of moles of hydrogen at equilibrium is 0.20 Then the number of moles of iodine, and hydrogen iodide at equilibrium are, a) 1.2, 1.6, b) 1.8, 1.0, c) 0.4, 2.4, d) 0.8, 2.0, A sample of pure PCl5 was introduced into an evacuated vessel at 473K. After, equilibrium was attained, concentration of PCl5 was found to be 0.5 × 10−1 mol L−1 . If, value of K c, is 8.0 × 103 , what are concentrations of PCl3 and Cl2 at, equilibrium?, a) 0.01M, b) 0.02M, c) 0.05M, d) 2.82M, XY2 dissociates as XY2 (g) ⇌ XY(g) + Y(g) when the initial pressure of XY2 is 600mm Hg,, the total equilibrium pressure is 800mm Hg. Calculate K for the reaction assuming, that the volume of the system remains unchanged., a) 50, b) 100, c) 166.6, d) 400, For the reaction N2O4(g) ⇌ 2NO2(g) , the reaction connecting the degree of, dissociation(α) of N2O4(g) with the equilibrium constant K pis, a)𝛼 =, , 𝐾𝑝, 𝑃, 𝐾𝑝, 4+ 𝑃, , c)𝛼 = [, , 𝐾𝑝, 𝑃, , 4+𝐾𝑝 \𝑃, , b)𝛼 =, , 𝐾𝑝, 4+𝐾𝑝, , 1/2, , ], , NCERT Text Book Questions:, , d)𝛼 = [, , 𝐾𝑝, 4+𝐾𝑝, , 1/2, , ], , 7.3, 7.8 , 7. 9 , 7. 10 , 7. 11, 7. 12, 7. 14, 7. 15, 7. 16, 7. 17,, , 7. 18, 7. 19, 7. 20, 7.21, 7.22, 7.23, 7.24, 7.27, 7.33, 7. 34.
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2, , ACTIVE SITE EDUTECH - 9844532971, , SESSION – 6, AIM - To introduce LeChateleir’s Principle, to explain factors which can change, the state of equilibrium in a reversible system., LE CHATELIER’S PRINCIPLE, Three main factors which can change the state of equilibrium in a reversible, system i.e., concentration, pressure and temperature., Le Chatelier explained the effect of change in concentration, pressure and, temperature on any reversible system whether physical or chemical., According to him, if a system at equilibrium is subjected to a change of, concentration, pressure or temperature, the equilibrium shifts in the direction, that tends to undo the effect of the change,, (OR), If a system at equilibrium is subjected to a change in concentration,, pressure or temperature, the system adjusts itself in such a way as to, nullifies the effect of that change., , 1., , 2., , Change in concentration: If an additional amount of reactant or product is, added to the system, the stress is relieved as the reaction that consumes, the added substance occurs more rapidly than its reverse reaction,, i.e., in general, increasing the concentrations of the, reactants results in shifting the equilibrium towards products while increasing, concentrations of the products results in shifting the equilibrium towards, reactants., Change of pressure: At equilibrium, when medium consists of gases, then the, concentrations of all the components can be altered by changing the
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2, , ACTIVE SITE EDUTECH - 9844532971, , pressure. When the pressure on the system is increased, the volume decreases, proportionately., i.e., If pressure is increased, then process will move in that, direction where number of moles of gaseous species are fewer and vice versa., 3. Change in temperature: In a reversible chemical reaction, if one reaction is, endothermic, other will be exothermic in nature., When heat energy is added by raising temperature, the system can, relieve itself from the stress if the reaction which absorbs heat moves faster,, i.e, endothermic reaction is always favoured with increase of temperature., So for exothermic process - increase in temperature favours backward, reaction and vice versa., For endothermic process - increase in temperature favours forward reaction, and vice versa., 4. Addition of an inert gas: When inert gas, added to a system at, equilibrium, inert gas neither reacts with reactants nor with products., Case I: Addition of inert gas at constant pressure process will move in, that direction where number of moles of gaseous species are greater., Addition of inert gas will increase the total volume at equilibrium. Now, • When Δn = 0 e.g.,, For this equilibrium, , 2HI ⇌ H2 + I2, Kc = Kp =, , x2, 4(1−x)2, , ; where x is the degree of, , dissociation, The equation is independent of P and V terms and thus there is no effect, of addition of inert gas to this type of equilibrium., • When Δn > 0; eg., PCl5 ⇌ PCl3 + Cl2, Moles at t = 0, 1, 0, 0, Moles at equilibrium (1–x) x x, Kc =, , x2, V(1 − x)
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2, , ACTIVE SITE EDUTECH - 9844532971, , Where x is the degree of dissociation, Since V increases and thus to keep 𝐾𝑐 constant, ‘x’ must increase i.e.,, degree of dissociation of PCl5 increases or [PCl3 ] and [Cl2] increases at equilibrium and, the [PCl5 ]decreases at equilibrium, Case II: Addition of inert gases at constant volume to an equilibrium will increase, the pressure at equilibrium. Now, •, , Δn = 0, 2HI ⇌ H2 + I2, , For the equilibrium, , Kc = Kp =, , x2, 4(1−x2 ), , Where x is the degree of dissociation., Since the equation is independent of P and V terms and thus no effect of, addition of inert gas to this type of equilibrium., • Δn = 0 PCl5 ⇌ PCl3 + Cl2, , For this equilibrium, , Kc =, , x2, V(1−x), , Where ‘x’ is the degree of dissociation., Since volume remains constant during the change and the addition of inert, gas at constant volume will also have no influence on the equilibrium, concentration for this type of equilibrium., Conclusions:, 1. Increase in concentration of any substance favours the reaction in which it is used up., 2. High pressure is favourable for the reaction in which there is decrease in volume., 3. A rise in temperature favours the endothermic reaction., 4. Effect of inert gas addition:, Condition, ΔV = 0, V = ConstantΔn ≠ 0,, = Constant, ΔV ≠ 0, V ≠ ConstantΔV ≠ 0, V, ≠ Constant, , Δn = 0, +ve or − ve, Δn = 0, Δn > 0, Δn < 0, , Effect, No effect, No effect, Forward shift, Backward shift
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2, , ACTIVE SITE EDUTECH - 9844532971, , 5., , Effect of temperature and pressure:, increased 𝛥𝑛 ; side with fewer Effect of increase in P, mole of gas, , Nature of reaction, , Effect, of, temperature, , 1. Exothermic, 2. Exothermic, , backward shift, backward shift, , 0;, -ve; right, , No shift, forward shift, , 3. Endothermic, , forward shift, , -ve; right, , forward shift, , 4. Endothermic, 5. Exothermic, , forward shift, backward shift, , +ve; left, +ve; left, , backward shift, backward shift, , 6. Exothermic, , backward shift, , -ve tight, , forward shift, , 7. Endothermic, , forward shift, , +ve left, , backward shift, , 8. Endothermic, , forward shift, , +ve left, , backward shift, , Applications of LeChatelier’s Principle:, a. Formation of HI:, H2 (g) + I2 (g) ⇌ 2HI(g) + 3000cal, , • Effect of concentration: When concentration of H2 and I2 is increased at, equilibrium, the system moves in a direction which decreases the, concentration, i.e., the rate of forward reaction increases., • Effect of pressure: As there is no change in the number of moles in the, reaction, the equilibrium state remains unaffected by change of pressure., • Effect of temperature: By increasing temperature, the equilibrium state, shifts towards the reaction which moves with absorption of heat. The, formation of HI is an exothermic reaction. Thus, the backward reaction, moves faster when temperature is increased., In short, we can say that the favorable conditions for greater yield of HI are:, (i) High concentrations of H2 and I2 and, (ii) Low temperature
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2, , ACTIVE SITE EDUTECH - 9844532971, , b. Formation of Nitric Oxide:, N2 (g) + O2 (g) ⇌ 2NO(g) − 43200cal, , • Effect of concentration: When concentration of N2 or O2 is increased,, the system moves in a forward reaction thereby increasing the, concentration of NO., • Effect of pressure:In the formation of nitric oxide, the number of, moles remains the same, i.e, no change in volume occurs. Consequently,, the equilibrium state is not affected by any change in pressure., • Effect of temperature: The formation of NO is endothermic in nature., If the temperature is raised, the equilibrium shifts in the direction in, which heat is absorbed. The concentration of NO will, therefore, be, higher at higher temperature., Thus, favourable conditions for greater yield of nitric oxide are, (i) High concentrations of N2 or O2 or both., (ii) High temperature, c. Dissociation of PCl5 :, PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) − 15000cal, , • Effect of concentration: When concentration of PCl5 is increased, the, rate of forward reaction increases as to decrease the added, concentration. Thus, more of PCl5 & Cl2are formed., • Effect of pressure: The volume increases in the dissociation of PCl5., When pressure is increased, the system moves in the direction in which, there is decrease in volume. Thus, high pressure does not favors, dissociation of PCl5.
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2, , ACTIVE SITE EDUTECH - 9844532971, , • Effect of temperature: The dissociation of PCl5 is an endothermic, reaction. Thus, increase of temperature favours the dissociation., Thus, favorable conditions for dissociation of PCl5 are, (i) High concentration of PCl5., (ii) Low pressure and, (iii) High temperature, d. Formation of Ammonia:, Fe, , N2 (g) + 3H2 (g) →, , 2NH3 (g); ΔH = −22.4 kcal/mol, , • At high pressure reaction will shift in forward direction to form more product, • When concentration of N2 and H2 is raised or concentration of NH3 is, lowered, then equilibrium shifts in forward direction to from more amonmia., • The reaction shifts in forward direction at low temperature., But at very low temperature the rate of reaction becomes very low;, thus moderate temperature is favourable for this reaction., Example 1. At what conditions will the following reaction go in the forward direction?, 1., 2., 3., 4., 5., 6., 7., , Solution:, , 1., 2., 3., 4., 5., 6., 7., , N2 (g) + 3H2 (g) ⇌ 2NH3 (g) + 23kcal., 2SO2 (g) + O2 (g) ⇌ 2SO3 (g) + 45kcal, N2 (g) + O2 (g) ⇌ 2NO(g) − 43.2kcal, 2NO(g) + O2 (g) ⇌ 2NO2 (g) + 27.8kcal, C(s) + H2 O(g) ⇌ CO2 (g) + H2 (g) + Xkcal, PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) − Xkcal, N2 O4 (g) ⇌ 2NO2 (g) − 14kcal, , Low T, High P, excess of N 2 and H 2, Low T, High P, excess of SO2 and O2, High T, any P, excess of N 2 and O2, Low T, High P, excess of NO and O2, Low T, Low P, excess of C and H 2O, High T, Low P, excess of PCl5, High T, Low P, excess of N 2O4
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2, , ACTIVE SITE EDUTECH - 9844532971, , Applications of Le-Chateleirs Principle to Physical Equilibria:, 1. Vapour pressure of a liquid: Consider the equilibrium, Liquid ⇌ Vapour, , Change of a liquid into its vapour is done by absorption of heat whereas the, conversion of vapour into liquid state is done by evolution of heat., Therefore, addition of heat to such a system will shift the equilibrium towards, the right. On raising the temperature of the system, liquid will evaporate. This, will raise the vapour pressure of the system. Thus, the vapour-pressure of a, liquid increases with rise in temperature., 2. Effect of pressure on the boiling point of a liquid: If pressure on the, system is increased, some of the vapours will change into liquid so as to lower, the pressure. Thus, the application of pressure on the system tends to, condense the vapour into liquid state at a given temperature. In order to, counteract it, a higher temperature is needed. This explains the rise of, boiling point of a liquid on the application of pressure., 3. Effect of temperature on solubility:In most cases, when a solute passes into, solution, heat is absorbed, i.e., cooling results. When heat is applied to a, saturated solution in contact with solute, the change will take place in that, direction which absorbed heat (i.e., which tends to produce cooling)., Therefore, some more of the solute will dissolve., i.e., the solubility of the substance increases with rise in temperature., CLASS EXERCISE, 1. In what manner will increase of pressure affect the following equation?, C(S) + H2 O(g) ⇌ CO(g) + H2 (g), , a) Shift in the forward direction, b) Shift in the reverse direction, c) Increase in the yield of hydrogen d) No effect
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2, , ACTIVE SITE EDUTECH - 9844532971, , 2. For the reactionCO(g) + H2 O(g) ⇌ CO2 (g) + H2 (g) at a given temperature,, the equilibrium amount of CO2 (g)can be increased by, a) adding a suitable catalyst, b) adding an inert gas, b) decreasing the volume of the container d) increasing the amount of CO(g), 3. The reaction N2 + O2 ⇌ 2NO is endothermic. The forward reaction is, a) favoured by decrease in T, b) favoured by increase in pressure, c) unchanged on increasing P, d) equilibrium shifts by adding catalyst, 4. In the chemical reaction 2SO2 (g) + O2 (g) ⇌ 2SO3 (g) increasing the total, pressure leads to, a) increase in amount of SO3, b) increase in partial pressure of O2, c) increase in partial pressure of SO2 d) change in equilibrium constant, 5. On the basis of Lechatelier’s principle, predict which of the following, conditions would be unfavourable for the formation of SO3 in the reaction, 2SO2 (g) + O2 (g) ⇌ 2SO3 (g); ΔH = −176.0 kJ, , a) Low temperature, b) High pressure, c) High temperature, d) High conc. of 𝑆𝑂2, 6. If chemical system is in equilibrium, the addition of catalyst would result in, a) increase in rate of forward reaction, b) increase in rate of reverse reaction, c) a new reaction path, d) increase in amount of heat evolved in the reaction, , HOME EXERCISE, 7. Which of the following is not affected by change in pressure?, a) 2NO2 (g) ⇌ N2 O4 (g), b) CO2 (g) ⇌ +H2 O(l) CO2 (solution), c) 2HI(g) ⇌ H2 (g) + I2 (g), d) 2O3 (g) ⇌ 3O2 (g), 8. In which case, the forward reaction is favoured by high pressure?
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2, , ACTIVE SITE EDUTECH - 9844532971, , IONIC EQUILIBRIUM CONTINUED IN NEXT, PART- 2
