Chapter-5 CHEMICAL EQUILIBRIUM

Transcription

1 hapter-5 HEMIAL EQUILIBRIUM On the basis of etent of reactions, chemical reactions can be classified in two categories reversible and irreversible reactions. Irreversible reactions goes for completion whereas in reversible reactions, we are left with some quantity of reactants even after along period of time. Eamples of Reversible and irreversible reactions Reversible Irreversible l 5 (g) l 3 (g) + l (g) Nal + AgNO 3 Agl(s) + NaNO 3 HI(g) H (g) + I (g) lo 3 l + 3O ROOH + ROH ROOR + H O b(no 3 ) bo + NO + O N + 3H NH 3 ertain reactions are reversible in closed vessel and irreversible under open conditions. eg. ao 3(s) ao (s) + O (g) (closed vessel) ao 3(s) ao (s) + O (g) (open vessel) hemical equilibrium is study of reversible reactions, there etent of completion and factors affecting it. LAW OF MASS ATION The law of mass action states that under a given set of conditions, rate of chemical reactions is directly proportional to the active masses of the reacting substances. e.g. onsider the reaction A + B roducts Rate [A] [B] [B] Thus Rate [A] [B] or Rate k [A] [B] here k is termed as rate constant or velocity constant Active Mass Active mass is represented by [ ] and epressed as Active Mass activity coefficient oncentration. For dilute liquid solution and gases behaving ideally activity coefficient is taken as unity. Thus active mass is equal to concentration for dilute liquid solutions and ideal gases. Active mass of liquids and solids is taken as unity. [08]

2 hemical equilibrium As reaction proceeds concentration of reactants decrease and concentration of products increase, thus rate of forward reaction decreases and rate of backward reaction increases with time and after some time rate of forward reaction equals to rate of backward reaction. Rate forward Rate backward that means the amount of products formed by forward reaction is equal to amount of products consumed in backward reaction, similarly amount of reactants consumed in forward reaction is equal to amount of it formed by backward reaction, thus net concentration of reaction species remain constant and rate of two reactions also get constant. This is termed as state of chemical equilibrium. haracteristics of chemical equilibrium. It is a dynamic equilibrium state. It can be approached from both sides 3. At equilibrium the concentrations of the reactants and products remain constant. 4. atalyst does not affect the equilibrium state but hastens the approach of equilibrium state. 5. At equilibrium the change in free energy, G 0 6. Equilibrium conditions can be affected by change in eternal variables such as change in concentration, temperature, pressure etc. Equilibrium constant Let us consider a general reversible reactions aa + bb m + nd. a b Rate (forward) f A B m n Rate (backward) b c D At equilibrium Rate (f) Rate (b) a b m n f A B b c D f b m a A n D n B Where A B D are concentrations under equilibrium conditions. The ratio f b Equilibrium constant c and When Active masses are taken in terms of concentration (moles per litre), equilibrium constant is termed as concentration equilibrium constant. c. In case of gas phase reactions, the terms active masses may be replaced by partial pressure of each gas at equilibrium, and corresponding equilibrium constant is termed as pressure equilibrium constant p. [09]

3 Hence p p m a A p p n D b B p Where p A, p B, p and p D are partial pressure of each component at equilibrium. Relationship between c and V nrt, where V volume in litre V n RT ; RT [ V n moles/litre] p A A RT and thus p ( m c RT) ( RT) a A ( ( D B RT) RT) n b m a A n D b B (RT)n g p c (RT)n g where n g (m + n) (a + b) n g () n g (R) haracterstics of Equilibrium constant () Its value is independent of (a) Original concentrations of reactants (b) Volume (c) resence of catalyst (d) resence of inert gas (e) Nature and number of steps in the reactions as long as the stoichiometry is not changed. () Its value depends upon temperature : Effect of temperature is represented as log H T T.303R T T at constant pressure (a) (b) log E T T.303R T T at constant volume For endothermic reactions, increases with increase in temperature For eothermic reactions, decreases with increase in temperature. (3) Larger the value of equilibrium constant, higher will be the yield of the reaction. (4) The value of depends on stoichiometry of reactants and products. e.g. A + B AB [ AB] [ A ][ B ] [0]

4 If we multiply it with integer m. then equation become : ma + mb mab [ AB] m [ A ] m [ B ] m () m Hence ( ) m (5) If the final equation at equilibrium is written as the sum of two different equations then the final equilibrium constant is the product of the equilibrium constants of each individual steps. e.g. N + O NO [NO] [N ][O ] Suppose this equation takes place in two steps Step - I N + O NO [NO] [N ][O ] Step - II NO + O NO [NO ] [NO] [O ] Then Also when equations are subtracted then the equilibrium constant obtained will be the equilibrium constant of one reaction divided by the equilibrium constant of the other. Le hatelier s rinciple:- When a system at equilibrium is subjected to stress or strain in the form of temp, pressure, volume, concentration etc then system moves in that direction in which the stress or strain is nullified. () Effect of change in concentration (a) (b) When concentration of products is raised system moves in backward direction to attain equilibrium. When product concentration decreases system moves in forward direction to attain equilibrium () Effect of pressure: The change in pressure depends upon the sign. of n g. []

5 A. When n g ve (a) (b) Increasing pressure system will move in forward direction to attain equilibrium Decreasing pressure system moves in backward direction to attain equilibrium B. When n g +ve (a) (b) Increasing pressure system moves in backward direction to attain equilibrium. Decreasing pressure system moves in the forward direction. (3) Effect of temperature c changes with temperature according to the equation k log k H.303R T TT T (at constant pressure) k log k E.303R T TT T (at constant volume) (a) For Eothermic reation (i) When temperature is increases c value decreases so system moves in backward direction. (ii) When temperature is lowered c increases and hence system moves in forward direction. (b) For endothermic reaction (i) Increasing the temperature will increase c value so that the system moves in forward direction. (ii) Decreasing the temperature will decrease c value so that the system moves in backward direction. (4) Adding Inert gas at constant volume Addition of inert gas to a system at equilibrium at constant volume does not change equilibrium state. (5) Adding Inert gas at constant pressure (a) (b) Adding inert gas to a system at equilibrium at constant pressure will increase the volume and hence if n g ve the system moves in backward direction attain equilibrium. Adding inert gas to a system at equilibrium at constant pressure will increase the volume and if n g +ve then system moves in forward direction attain the equilibrium []

6 (6) Adding catalyst Addition of catalyst to a system at equilibrium does not shift equilibrium because rate of forward and rate of backward direction both are increased. Degree of dissociation by vapour density measurements for l 5 l 3 + l or N O 4 NO D d where D Density of pure l 5 Density of ure N O 4 d Density of miture of l 5 + l 3 + l or N O 4 + NO [3]

7 SOLVED EXAMLES Eample :If concentration are epressed in moles L and pressure in atmospheres, what is the ratio of p to c for the reaction. SO (g) + O (g) SO 3 (g) at 5º? Solution : n g Hence p c (RT) or p / c /RT Eample :alculate the percent dissociation of H S(g) if 0. mole of H S is kept in 0.4 litre vessel at,000. For the reaction H S(g) H (g) + S (g) the value of c is Solution : H S(g) H (g) + S (g) At equb. 0. / Molar conc. at equb. 0. V V V ( / V ) ( /V ) c 0. V V (0.) % dissociation %. 0. Eample 3 :What happens to the equilibrium l 5 (g) (i) at constant volume (ii) at constant pressure? Solution : (i) The state of equilibrium remains unaffected. (ii) Dissociation increases (i.e., equilibrium shifts forward). l 3 (g) + l (g), if Nitrogen gas is added to it Eample 4 :Ammonia under a pressure of 5 atm at 7º is heated to 347º in a closed vessel in the presence of a catalyst. Under the conditions, NH 3 is partially decomposed according to the Solution : equation, NH 3 N + 3H. The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. alculate the percentage of NH 3 actually decomposed. Let the degree of decomposition NH 3 N (g) + 3 H [Total number of moles] Initial No. of moles 0 0 At equilibrium / 3/ + Applying gas equation V nrt [4]

8 At constant V, nrt or nt constant or n T n T ( ) ercentage decomposition %. Eample 5 :What two changes on the equilibrium N (g) + 3H (g) Solution : its state undisturbed? NH 3 (g), H 9.5 kj can keep Increase of temperature alongwith suitable increase of pressure or increase of presure alongwith suitable increase of temperature. Eample 6 :At temperature, T compound AB (g) dissociates according to the reaction AB (g) AB(g) + B (g) with a degree of dissociation,, which is small compared with unity. Deduce the pressure for in terms of the equilibrium constant, p and the total pressure, Solution : AB (g) AB(g) + B (g) Initial moles 0 0 Moles of equb. Total moles at equilibrium + + ( pab ) ( pb p ( pab ) p 3 ) or ( ) ( ) p /3 3 ( )( ) Eample 7 :At 87º, p for the reaction between O (g) and ecess hot graphite is 0 atm. Solution : (a) What are the equilibrium concentration of the gases at 87º, and a total pressure of 5 atm.? (b) At what total pressure, the gas contains 5% O by volume? The concerned chemical reaction is O (g) + (s) O(g) Total No. of moles of start 0 No. of moles of equb. + Mole fraction at equb. [5]

9 artial pressure 5 5 p or 0 ( po ) po Hence at equilibrium moles of O moles of O (b) Given % of O 5% by volume % of O 95% by volume 5 Then p O and p O p or (0.095 ) atm. Eample 8 :The pressure of Iodine gas at 73 is found to be 0. atm. whereas the epected presure is atm. The increased pressure is due to dissociation I Solution : I I Initial conc. 0 Eqbm. conc. Total moles of equilibrium + + Since the pressure is proportional to the number of moles. Eperimental value of pressure Epected value of pressure I. alculate p or I Now p I [6]

10 p 4 utting 0.5, 0. atm We get p Eample 9 :For the reaction, O(g) + H (g) Solution : H 3 OH(g), Hydrogen gas is introduced into a five litre flask at 37º, containing 0. mole of O(g) and a catalyst, until the pressure is 4.9 atm. At this point 0. mole of H 3 OH(g) is formed. alculate the equilibrium constants, p and c. Let the total number of moles of all gases at equilibrium point n 4.9 atm. R 0.08 atm l mol V 5 l By applying the formula V nrt T V n 0.5 moles RT alculation of the number of moles of the individual gases at equilibrium point. No. of moles of H 3 OH 0. (given) No. of moles of O (also) 0. Hence No. of moles of H 0.5 ( ) 0.3 Molar concentration will be [H 3 OH] [O] [H 3OH] a [O][H ] (0.06) ; [H ] mole l alculation of p. We know that p c (RT) n ( ) (n 3 ) atm Eample 0:A saturated solution of Iodine in water contains g I / litre. More than this can dissolve Solution : in a I solution because of the following equilibrium. I (g) + I I 3 A 0.00 M I solution (0.00 MI ) actually dissolves.5 g of Iodine/litre, most of which is converted into I 3. Assuming that the concnetration of I in all saturated solutions is same, calculate the equilibrium constant for the above reaction. No. of mole of I in saturated solution of Iodine m n M 54 Moles of I in 0.00 M I solution I + I I [7]

11 [I [I3 ] ][I At equilibrium, I M. ] [I 3 ] M and [I ] M Eample :One mole of Nitrogen is mied with three moles of Hydrogen in a 4 litre container. If mole of Nitrogen is converted to Ammonia by the following reaction, N (g) + 3H (g) NH 3 (g) calculate the equilibrium constant ( c ) for the following equilibrium. 3 N (g) + H (g) NH 3 (g) Solution: Initial moles 3 0 N (g) + 3H (g) NH 3 (g) Equb. moles Equb. conc [NH3] Now we know that c 3 [N ][H ] Since and are very small, and may be taken as and 3 respectively. Substituting the various values, a litre mol For the equilibrium. N (g) + 3 H (g) [NH 3] c / 3/ [N ] [H ] Eample: For the reaction : H 6 (g) c NH 3 (g) 5 (.48 0 litre mol ) litre mol H 4 (g) + H (g) Gº is.38 J mol at 900. alculate the mole percent of Hydrogen present at equilibrium if pure H 6 is passed over a suitable dehydrogenation catalyst at 900 and.0 atm pressure. Solution: G.303 RT log p log p G.303 RT [8]

12 log p.98 p We have H 6 (g) H 4 (g) + H (g) Initial conc. 0 0 Equb. conc. Total moles at eqbm p H 6 p H 4 p H p p p H4 p. p H6 H p atm., p Neglecting in comparison to, we get p p / p Mole % H %. Eample 3:In a 0 litre evacuated chamber 0.5 mole H and 0.5 mole I are reacted at 448º, at which temperature 50 for H (g) + I (g) HI(g). (a) What is the value of p? (b) What is total pressure in the chamber? (c) How many moles of the iodine remain unreacted at equilibrium? (d) What is the partial pressure of each component in the equilibrium miture? [9]

13 Solution: (a) Since the same number of gas moles (or gas volumes) occurs on both sides of the equation, the reaction is not affected by a volume change (or a pressure change). Hence here p e, i.e. 50. (b) Note that according to reaction, mole of H and mole of I react to give moles of HI, so there is no change in total number of moles before the reaction ( ) and after the reaction. The total pressure can be determined from the ideal gas law, i.e. V nrt Here n mole, V 0 litre T ,? Substituting the values in the equation, nrt atm V 0 (c) H + I HI Moles at start Moles at equb where is the number of moles of I reacted at equilibrium, Thus [HI] [H ][I ] () 50 (0.5) 0.39 () (0.5) (0.5) Number of moles of I remaining unreacted at equilibrium mole (d) Determination of partial pressure of each component at equilibrium. p I Moles of I Total pressure Total moles p H p I 0.65 atm p HI Total pressure (p H + p I ) ( ) atm atm Eample 4: Solution: The equilibrium constant for the equation O(g) + H O(g) O (g) + H (g) is A miture of mole of water vapour and 3 moles of O is allowed to come to equilibrium at a total pressure of atmospheres. Let the number of initial moles of O and H O be a and b respectively. If mole of O and mole of H O have reacted at equilibrium, then Initial moles a b 0 0 O(g) + H O(g) O (g) + H (g) At equilibrium (a ) (b ) Total No. of moles at equilibrium (a ) + (b ) + + a + b [O][H ] p [O][H O] ( a)( b) [0]

14 According to problem, p 0.63, a 3, b (3)( ) Moles of Hydrogen at equilibrium 0.68 artial pressure Moles of substance Total moles p O p H ( a b) ( a) p O ( ab) Total pressure atm.6 atm p H O ( b) ( ab) atm Eample 5:When l 5 is heated it gasifies and dissociates into l 3 and l. The vapour density of the gas Solution: miture at 00º and 50º is 70. and 57.9 respectively. Find the percentage dissociation of l 5 at 00º and 50º. If no dissociation were to occur, Mol. wt the vapour density (D) of l However, observed vapour density (d) at 00º 70. Dd Hence degree of dissociation, d 70. ercentage dissociation at 00º % Similarly, degree of dissociation () at 50º, Dd d % dissociation % It is important to note that the above relation between degree of dissociation and densities is applicable only when the reaction is accompanied by change in total number of moles. This relation is not applicable to reactions in which number of moles of the reactants and products are equal as in the formation or decomposition of HI. H + I HI []