phase equilibrium acid-base equilibrium Fe 3+ (aq) + e - Fe 2+ (aq) redox equilibrium becomes faster as more products are formed

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1 CHEMICAL EQUILIBRIUM Most of the chemistry you have studied so far has been concerned with reactions which are one-way or irreversible (represented by the sign). Any chemical reaction of this type should go to completion, with at least one of the reactants being used up completely. There are many processes which can go either way or reversible (represented by the sign). Chemical reactions of this particular type never go to completion, with both reactants and products co-exist at any time. The following examples are some of the chemical equilibria that we would soon come across in detail : H 2 O (l) H 2 O (g) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) phase equilibrium acid-base equilibrium Fe 3+ (aq) + e - Fe 2+ (aq) redox equilibrium Any reversible reaction can be thought of as comprising a forward and a backward reaction taking place simultaneously in opposite directions : rate Forward reaction At the beginning rapid rate zero rate As time passes (before equilibrium is reached) At equilibrium becomes slower as more reactants are used up Backward reaction becomes faster as more products are formed rate of forward reaction = rate of backward reaction time At the stage when forward and backward rates become equal, there is no further change in the amount of reactants and products. The system is said to reach a balanced state or in equilibrium. Dynamic Equilibrium There are two types of equilibria : Static equilibrium : children on a see-saw. At the balance point (i.e. the equilibrium position) no movement of the children or the see-saw occurs. Dynamic equilibrium : boy ascending escalator at the same rate as the escalator descends. At the balance point (i.e. the equilibrium position) the boy and the escalator are moving at the same rate in opposite directions. Thus, the equilibrium position is maintained. All equilibria in chemistry are dynamic in nature, with both forward and backward reactions continue to proceed at the same rate so that there would be no NET change in amount after equilibrium has been reached. Characteristics of Chemical Equilibrium 1 Dynamic equilibrium can exist only in a closed system (one that does not allow exchange of matter with surroundings). 2 At equilibrium, all reactants and products are present, with amounts (or concentrations) unchanged with time. 3 A state of equilibrium is reached when rate of forward change is equal to rate of backward change. 4 Dynamic equilibrium can be reached by starting from either side of the reversible reaction. 5 Any change in physical condition (e.g. conc., pressure, temp.) would result in a shift of equilibrium position.

2 Factors Affecting Equilibrium Position Eqm 2 The direction of shifting in equilibrium position brought by a change in physical condition can be easily predicted by applying the Le Chatelier s Principle, which states that : If a system in equilibrium is subjected to a change which disturbs the equilibrium, the system will respond in such a way as to counteract the effect of the change. Consider the following equilibrium : N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H = -ve Determine the change (if any) on the equilibrium position if it is subjected to each of the following changes : (a) adding more hydrogen gas : (b) removing some nitrogen gas : (c) carrying out at a higher pressure : (d) carrying out at a higher temperature : (e) adding more iron powder as catalyst : Consider the following equilibrium : N 2 O 4 (g) 2 NO 2 (g) H = kj mol -1 colorless dark brown Describe and explain any observable change if the above equilibrium is subjected to each of the following changes : (a) adding NO 2 (g) : (b) decrease the temperature : (c) increase the pressure : Consider the following equilibrium in Contact Process : 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) H = - ve Suggest THREE optimum conditions by which the yield of the product could be raised. conc. time

3 Eqm 3 The following table summarizes the effect of concentration, pressure (for gaseous systems), temperature and catalyst on equilibrium position : Change in physical condition The system responds by Equilibrium position increase in concentration of increase in pressure for gas reactions where (1) reactant (2) product (1) total no. of moles of gases is the same on both sides (2) total no. of moles of gaseous products less than that of gaseous reactants (3) total no. of moles of gaseous products greater than that of gaseous reactants consuming reactant at a faster rate to give more product consuming product at a faster rate to give more reactant reducing the applied pressure, hence shifting to the side with less amount of gaseous particles shifts to right shifts to left no change shifts to right shifts to left increase in temperature for (1) endothermic reaction (2) exothermic reaction absorbing the heat to lower the temperature, hence shifting to the side with a higher enthalpy (energy content) shifts to right shifts to left Presence of catalyst increasing the rates of forward and backward reactions to the same extent no change The Equilibrium Law If a reaction in equilibrium is represented by the equation : a A + b B c C + d D it is found experimentally that c [C] [D] a [A] [B] d b = K C where K C, the equilibrium constant, is a constant value at a given temperature. A gaseous mixture of 2.0 mole H 2 (g) and 1.0 mole I 2 (g) is allowed to react at a particular temperature and pressure. Upon equilibrium, the amount of HI(g) is found to be 1.8 mole. (a) Calculate the equilibrium constant, K c, for the reaction H 2 (g) + I 2 (g) 2 HI(g) (b) What would be the amount of HI(g) formed if 1.0 mole of H 2 (g) is allowed to react with 1.0 mole of I 2 (g) at the same temperature and pressure?

4 Eqm 4 (c) Calculate the amount of H 2 (g) required to react with 1.0 mole of I 2 (g) so that half the amount of I 2 (g) would be converted into HI(g) at the same temperature and pressure. Notice that : 1 the law is only valid after equilibrium has been reached, i.e. [A], [B], [C] and [D] refer to the respective concentrations AT equilibrium 2 it is essential to relate any numerical value for K c to the particular equation concerned : 2 SO 2 + O 2 2 SO 3 K c = SO 2 + ½ O 2 SO 3 K c = the unit for K c are different, depending on the relative amount of reactants and products 4 the convention to express equilibrium law is to write product terms in the numerator while putting reactant terms in the denominator, in doing so providing a parallel relationship between the magnitude of K c and the yield : large K c means that at equilibrium the reaction almost goes to completion to the product side (i.e. higher yield) 5 the magnitude of K c tells us nothing about the RATE of a reaction; i.e. an equilibrium reaction with a large K c value may not even occur if it has a reaction rate too slow Homogeneous and Heterogeneous Equilibria An equilibrium can be classified, according to the phase of the reacting species involved, into : 1 homogeneous - reacting species are in the same phase Fe 3+ (aq) + SCN - (aq) Fe(SCN) 2+ (aq) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) aqueous state gaseous state 2 heterogeneous - at least one reacting species (reactant or product) are in a different phase (usually solid state) from the rest Fe 2+ (aq) + Ag + (aq) Fe 3+ (aq) + Ag(s) For homogeneous equilibria in gaseous phase, equilibrium constants in terms of partial pressure, K p, is a more convenient quantity instead of those in terms of concentration, K c K p = P P C A P P D B where P X represents the partial pressure of X. For the same reaction in the class work on p.3, give the expression for the equilibrium constant, K p and calculate its value.

5 Eqm 5 At 1100 K, K p = 0.13 atm -1 for the system 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) If 2.0 mol of SO 2 and 2.0 mol of O 2 are mixed and allowed to react, what must the total pressure be to give a 90% yield of SO 3? For heterogeneous equilibria, the equilibrium law can usually be simplified by considering the fact that concentration of solids or liquids are effectively constant, which can then be incorporated into the equilibrium constant to yield another constant : e.g. CaCO 3 (s) CaO(s) + CO 2 (g) K = [ CaO( s)] [CO2 ( g )] [ CaCO ( s)] K [ CaCO3( s)] = [CO 2 (g)] [ CaO( s)] K = [CO 2 (g)] Solid ammonium hydrogen sulphide dissociates into hydrogen sulphide and ammonia as represented by the following equation : NH 4 HS(s) H 2 S(g) + NH 3 (g) At 295 K, when solid ammonium hydrogen sulphide is introduced into an evacuated vessel, the total pressure at equilibrium of the gas mixture is found to be 0.54 atm. (i) Calculate the equilibrium constant, K p, for the reaction. 3 (ii) If the vessel initially contains hydrogen sulphide gas at a pressure of atm, what is the partial pressure of ammonia when the system is allowed to reach equilibrium at the same temperature? Experimental Determination of Equilibrium Constant General procedure : 1 Mix known volumes of known concentrations of the reactants for the reaction 2 Allow the reaction to equilibrate in a closed system and at a constant temperature 3 Freeze the equilibrium so that the equilibrium position would not shift significantly during the subsequent titration 4 Determine the equilibrium concentration of one of the reacting species by employing a chemical (via titration) or physical method (e.g. colorimetry) 5 The equilibrium concentrations of the rest of reacting species can be obtained by calculations

6 Colorimetric method in determining K c of the reaction : Fe 3+ (aq) + SCN - (aq) Fe(SCN) 2+ (aq) 1 The concentration of Fe(SCN) 2+, which is blood red in colour, can be determined either by visual comparison or by using a colorimeter. Eqm 6 2 Calibrate a colorimeter by measuring the absorbance (i.e. colour intensity) of several solutions which contain known concentrations of Fe(SCN) 2+ (aq). 3 By mixing comparable volumes of standard Fe(NO 3 ) 3 (aq) and KSCN(aq) solutions and measure the [Fe(SCN) 2+ (aq)] formed with the calibrated colorimeter, [Fe 3+ (aq)], [SCN - (aq)] and [Fe(SCN) 2+ (aq)] at equilibrium can be found. Factors Affecting Equilibrium Constant The numerical value of the equilibrium constant remains unchanged at one particular temperature. This means that K c and K p are unaffected by catalysts or by changes in pressure and concentration. The equilibrium constant does, however, vary with temperature (i.e. temperature-dependent), as illustrated by the following K p values at different temperatures for two important reactions : T / K N 2 (g) + 3H 2 (g) 2NH 3 (g) p K p = ( NH ) 3 p ( p ) N H atm -2 N 2 O 4 (g) 2 NO 2 (g) p K p = ( NO ) 2 p Explain whether these two reactions are exothermic or endothermic. N O atm The relationship between K and T is best illustrated by the following equation : ln K = constant - H RT The close resemblance of this and Arrhenius equation reveals a parallel relationship between K and k : while the values of equilibrium constants tell us about the yield of a reaction from an energetic point of view ( H); the values of rate constants tell us about the rate of a reaction from a kinetic point of view (E a ). Notice that : 1 changes in concentration and pressure result in a shift in equilibrium position in order to preserve the K c value 2 changes in temperature result in a shift in equilibrium position because K c has been adjusted to a new value Applying the Principles of Reaction Rates and Equilibria to Industrial Processes Haber Process optimum condition to increase the rate optimum condition to increase the yield real condition in industry N 2 (g) + 3H 2 (g) 2NH 3 (g) Contact Process 2 SO 2 (g) + O 2 (g) 2 SO 3 (g)

7 Predicting Equilibrium Shift by Reaction Quotient Eqm 7 In order to predict the direction of shifts in a chemical equilibrium, we can compare the reaction quotient, Q, of the reaction with the equilibrium constant. The reaction quotient is defined in the same way as the equilibrium constant except that the concentrations of reactants and products can be taken at any moment of the reaction (i.e. not necessarily at the eqm). For the hypothetical reaction: a A + b B c C + d D the expression of reaction quotient in terms of concentrations is: c d [C] [D] Q c = a b [A] [B] There are three possible cases when the reaction quotient is compared with the equilibrium constant of the reaction: 1 Q = K system is at equilibrium 2 Q > K system is not at eqm as concentrations of products are greater than their eqm values eqm shift to LHS (reactant side) until eqm is reached 3 Q < K system is not at eqm as concentrations of reactants are greater than their eqm values eqm shift to RHS (product side) until eqm is reached A gaseous mixture of 10-6 M NH 3 (g), 0.001M N 2 (g) and 0.002M H 2 (g) is found inside a container at 500 C. Given the equilibrium constant of the following reaction is dm 6 mol -2 at 500 C. (a) Predict the equilibrium shift based on reaction quotient. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) (b) Calculate the equilibrium concentrations of all species, K c, for the reaction Distribution Law when a solute is added to a mixture of two immiscible liquids, it will dissolve to some extent in both of them solute particles can move from one liquid to another as soon as the rate of flow of solute in both directions is equal, a dynamic equilibrium is established : solute in solvent 1 solute in solvent 2 and can be described by the Equilibrium Law : [solute] in solvent 1 = K d [solute] in solvent 2 solvent 1 solvent 2 the distribution law states that when a non-volatile solute is added to two immiscible solvents, the ratio of concentration of solute in two solvents at equilibrium is a constant at the same temperature, irrespective of the amount of solute added

8 the constant K d, known as partition coefficient, has the following properties : (1) it has no unit ; (2) its value depends on the order by which the 2 solvents are specified ; e.g. K d for Br 2 in H 2 O and CCl 4 = 4 but K d for Br 2 in CCl 4 and H 2 O = 0.25 (3) being an equilibrium constant itself, K d is temperature dependent Eqm 8 distribution law is no longer applicable when the solute does not have the same molecular form in two solvents e.g. partition of CH 3 COOH between water and CCl 4 ( forms dimer in non-polar solvent ) solute does not associate / dissociate in either solvent partition coefficient for a three-component system can be measured experimentally by knowing : -- the conc. of solute in solvent 1 (usually aqueous layer), which can be easily found by titrations -- the conc. of solute in solvent 2, which can be found if the initial amount of solute is known, ie. amount of solute in solvent 2 = initial amount of solute - amount of solute in solvent 1 Suppose 1g of iodine was shaken with 50 cm 3 of water and 50 cm 3 of CCl 4. If 25 cm 3 of the aqueous layer requires 4.5 cm 3 of 0.01M sodium thiosulphate solution to reach the end-point, calculate the partition coefficient of iodine between CCl 4 and water. (relative atomic mass of I = 127) Do you think that iodine is more soluble in water or in CCl 4? Application of Partition Coefficient (1) Solvent Extraction a purification method to extract solute from one solvent into another, which is immiscible with the first one ether is particularly useful in extracting organic compounds, partly owes to (1) its ability to dissolve a wide range of organic compounds, and mostly because of (2) its high volatility, which can be easily removed by distillation Q.1 The partition coefficient of butanoic acid in ether and water is 3.5 and butanoic acid happens to be more soluble in organic solvent than in aqueous. Calculate the amount of butanoic acid extracted into the ether layer by shaking 100 cm 3 of water containing 1 mole of butanoic acid with 100 cm 3 of ether.

9 Eqm 9 Q.2 What would be the total amount of butanoic acid extracted if two more extractions, using 100 cm 3 of ether each time, is performed? Q.3 Compare the result in Q.2 with the amount of butanoic acid being removed in a single extraction using 300 cm 3 of ether. Comment on the difference. It is always more efficient if the same amount of extracting solvent is used in SEVERAL SMALL portions rather than ONE LARGE portion. In general, the more extractions the better. The same principle applies to cleaning things (ie. many small rinses are more efficient than one large rinse using the same amount of solvent) (2) Paper Chromatography Chromatography is a special type of solvent extractions in that many repeated extractions are performed in one single operation. Chromatography describes a whole family of analytical techniques which depends on the relative adherence of a sample between a stationary phase and a mobile phase (the eluent). One of the most familiar technique is paper chromatography. In paper chromatography, the cellulose of which the paper is made holds a thin film of water, and this forms the stationary phase. Paper chromatography operates mainly by the partition of the sample between the stationary water film and a mobile solvent. A very small spot of sample solution is placed near the bottom of a strip of chromatography paper which is then dipped into a suitable solvent. As the solvent moves up the paper by capillary action, the substances in the sample distribute between the mobile and stationary phases. distance travelled by solvent distance travelled by spot Due to the fact that various substances have different partition coefficients between the mobile and stationary phases, they would be carried forward to different extents. Substances which are more soluble in the eluent than in the water film are carried along more rapidly, while those more soluble in water tend to move slowly. solvent R f = distance travelled by spot distance travelled by solvent Fresh eluent is constantly moving past the sample spot and the situation is rather similar to an enormous number of successive solvent extractions. When the solvent front has moved a suitable distance, the process is stopped by removing the paper from the solvent. For each spot, a R f value (retardation factor) can be calculated. Under the same condition (temp, solvent, etc.), substances with the same R f values are likely to be identical. When the R f values of various substances under a particular set of conditions are known, it may be possible to identify a certain substance by paper chromatographic analysis. Extensive application of such technique is found in the identification of constituent amino acids from proteins.