LECTURE 9: Chemical Equilibrium

Transcription

1 LECTURE 9: Chemical Equilibrium

2 REACTION GIBBS ENERGY Consider the simple reaction A B» Examples d-alanine to l- alanine» Suppose an infintesimal amount of n A converts to n B, then dn A =-dn B» If we define (ksi) as the extent of reaction then d = -dn A =dn B Dimensions of are moles For finite reaction the change is and n A goes to n A - and n B goes to n B +

3 REACTIONS GIBBS ENERGY Reaction Gibbs Energy is the slope of the Gibbs energy vs. or r G= (G/) p,t» Remember, dg = µ A dn A + µ B dn B so dg = -µ A d + µ B d or r G = (dg/d ) p,t = µ B -µ A» Reaction Gibbs energy is the difference of the product chemical potential and the reactant chemical potential

4 GIBBS ENERGY AND EQUILIBRIUM Initially µ B <µ A so r G<0 and the reaction is spontaneous» Called an exergonic reaction (work producing) Gibbs Energy, G G < 0 r G = 0 r G > 0 r 0 Extent of Reaction,

5 GIBBS ENERGY AND EQUILIBRIUM If µ B >µ A so r G>0 and the reverse reaction is spontaneous» Called an endergonic reaction (work consuming) When µ B =µ A so r G = 0 or system is at equilibrium» Thus a minimum in the reaction Gibbs energy plot represents equilibrium» At a minimum, the slope = 0

6 IDEAL GAS EQUILIBRIA Consider homogenous gas phase reactions» Reactants and products are gases» Gases to 1st approximation behave as ideal gases Since r G = (dg/d ) p,t = µ B -µ A and µ = µ + RT ln(p), r G= µ B + RT ln(p B ) - (µ A + RT ln(p A )) = µ B - µ A + RT ln(p B /p A ) r G= r G + RT ln(q) where

7 IDEAL GAS EQUILIBRIA r G = µ B -µ A, the standard reaction Gibbs energy Like std reaction enthalpy it is the difference in standard free energies of formation of the products - that of reactants» Q = p B /p A, the reaction quotient which ranges from 0 (pure A) to infinity (pure B) At equilibrium, r G = 0, so r G = - RT ln(k) or K = exp(- r G /RT)» K is the equilibrium value of Q or K = (p B /p A ) equilibrium

8 IDEAL GAS EQUILIBRIA The minimum in Gibbs energy arises from the mixing of reactant and products» If there were no mixing, G would change linearly in proportion to the amount of B formed Slope of G vs. would be r G xb

9 IDEAL GAS EQUILIBRIA» In a previous chapter, we learned mix G = nrt(x A ln(x A ) + x B ln(x B ) which is a U shaped function (minimum at 50% B) If r G <0 then ln(k) >0 so product is favored If r G >0 then ln(k) <0 so reactant is favored

10 IDEAL GAS EQUILIBRIA Consider now a complicated case, aa(g) bb(g) r G o where K RT ln K p p is the equilibrium constant, given by K p (P (P B A /P /P o o ) ) b a P P b B a A (P o ) ab

11 IDEAL GAS EQUILIBRIA: SAMPLE PROBLEM From the given thermodynamic data below, calculate the equilibrium constant for the reaction at 298 K: N 2 (g) + 3H 2 (g) 2NH 3 (g) f G o (NH 3 ) = kj/mol

12 IDEAL GAS EQUILIBRIA At 298 K, the partial pressures of gases in the reaction mixture described in the previous example are P N2 = 190 torr. P H2 = 418 torr, and P NH3 = 722 torr. Calculate the value of r G for the reaction.

13 IDEAL GAS EQUILIBRIA If you have a number of products and reactants (all ideal gases), Dalton s Law tells you P i V= v i RT where v i is the number of moles of the ith component (stoichiometric factor in the equation) For each component, the Gibbs energy difference G i -G i = RT ln(p i ) At constant temperature, dg =VdP =nrtd lnp so» R G - R G = RT v i ln(p i ) equilibrium R G =0 Remember R G i = R G i (products ) - R G i (reactants) = RT( v i ln(p i (prod) - v i ln(p i (reactants) )» Thus - R G = RT v i ln(p i ) equilibrium or - R G /RT = v i ln(p i ) equilibrium R G is a constant at t=constant so exp(- R G /RT ) = K Further, v i ln(p i ) = ln( P i vi ) equilibrium So K = P i vi v i is positive for products and negative for reactants In terms of activities (a A = p A /p* A ), K =( a i vi ) equilibrium

14 IDEAL GAS EQUILIBRIA: NOTES For pure solids and liquids a = 1 so they don t contribute to the reaction quotient (or K) For ideal gases (a A = p A /p* A ), but for real gases (a A = f A /p* A ), where f A is the fugacity K is a function of temperature and R G is also a function of temperature K is independent of total pressure and variation in partial pressures K is called the thermodynamic equilibrium constant» Like activities it is dimensionless» You can approximate K by replacing fugacities by partial pressures or molalities or molar concentrations

15 REAL GASES Fugacity (f) replaces partial pressure for real gases. o RTln f fugacity f P o fugacity fugacity would f coeficient : P standard state : state at be1 bar if pressures to 1 bar. the gas behaved which the fugacity ideally from low

16 REAL GASES: For aa bb K f (f (f B A / 1 bar ) / 1 bar) b a K f b B a A (PB / (PA / 1bar 1bar ) ) b a K K p

17 REAL GASES: SAMPLE PROBLEM What is the partial pressure of NO gas produced in the earth s atmosphere from the reaction of atmospheric nitrogen and oxygen? N 2 (g) + O 2 (g) 2NO(g) Data: G f (N 2 ) = 0; G f (O 2 ) = 0; G f (NO ) = kj/mol At atmospheric pressure P N2 = 0.78 and P O2 = 0.21

18 SAMPLE PROBLEM: For the equation, H 2 O(g) H 2 (g) + 1/2O 2 (g), what is the mole fraction of oxygen resulting from passing steam through a tube at 2000 K if the Gibbs energy is kj/mol and P = 200kPa? K = exp(- R G /RT ) = exp (( kj/mol)/(8.3145*2000 J/mol)) =exp(-8.13)=2.945 x 10-4 If x moles of water dissociate then fraction left is 1-x, fraction of hydrogen = x and fraction of oxygen=x/2 Total pressure = {(1-x) + x +x/2}p = {1 + 1/2x} [p=200 kpa] Partial pressures :water = (1-x)p/(1+1/2x); H 2 = xp/(1+1/2x); O 2 = 1/2xp/(1+1/2x) K = {(xp/(1+1/2x)( 1/2xp/(1+1/2x)) 0.5 )/[ (1-x)p/(1+1/2x)]; if 1>>x this simplifies to K = (1/2) 0.5 (xp) 1.5 /p = (p/2) 0.5 (x) 1.5 = x 10-4 Substituting for p (200 kpa/10 5 Pa)and solving for x, (x) 1.5 = x 10-4 ;so x = Mole fraction of oxygen = x/2=

19 REACTIONS IN SOLUTION aa(aq) bb(aq) r G o RT ln K m K m or if K c (m (m ( ( B A B A / / m m o o / 1M ) / 1M ) ) ) b a b a we express in Molarities m are molalities

20 HETEROGENOUS EQUILIBRIA Solids and liquids, a = 1; do not include in equilibrium constant expressions. Example: CaCO 3 (s) CaO(s) + CO 2 (g)

21 SAMPLE PROBLEM: Calculate the equilibrium constant for the following reaction at 298 K, using the following thermodynamic data: 2H 2 (g) + O 2 (g) 2H 2 O(l) G o (H 2 O) = kj/mol

22 EFFECT OF TEMPERATURE R S RT H ln K RT G ln K T 1 T 1 R H K K ln o r o r o r 2 1 o r 1 2

23 EFFECT OF TEMPERATURE

24 EFFECT OF TEMPERATURE If reaction is exothermic H <0 slope of ln(k) vs. 1/T is positive (>0)» As T increases 1/T decreases so as T increases thus K gets smaller» Increase in temperature favors reactants. If reaction is endothermic H >0 slope of ln(k) vs. 1/T is negative (>0)» As T increases 1/T decreases so as T increases thus K gets larger» Increase in temperature favors products.

25 LE CHATELIER S PRINCIPLE Henry Le Chatelier (1888) and F. Braun (1887) If a system in equilibrium is perturbed by subjecting it to a small variation in one of the variables that define the equilibrium, it will tend to return to an equilibrium state, which is usually somewhat different than the initial state

26 EFFECT OF PRESSURE for the reaction, A(g) 2B(g) K p K p 4P

27 Response of Equilibria to Pressure» Effect of additional inert gas - none because partial pressures unchanged (assumes ideality» Compression (changing volume): Le Chatelier s Principle says reaction will adjust to minimize pressure increase

28 RESPONSE OF EQUILIBRIA TO PRESSURE Example: A 2B; K = p B2 /p A p On compression [A] will increase to compensate for pressure change» If extent of reaction is then we start with n moles the A = (1- )n and b=2 n pb Mole fraction of A, x A = (1- )n/((1- )n + 2 n) = (1- )/(1+ ) Mole fraction of B, x A = 2 /(1+ ) K = [(2 /(1+ ))p] 2 /(1- )/(1+ )p = p[(4 2 /(1+ )(1- )] or K = 4 2 p/(1-2 ) or 4 2 p + K 2 -K= (4p + K) 2 -K = 0 Note : p = p/p Solving for 2 = K/ (4p + K) = 1/(1 + 4p/K) or =1/(1 + 4p/K) 1/2 This says as p increases then the extent of reaction decreases

29 Example: 3H 2 + N 2 2NH 2 K = p NH32 /p 2 H2 p N2 K = x NH32 p 2 /x 3 H2 p 3 x N2 p where p A =x A p K = [x NH3 2 /x 3 H2 x N2 ] [p 2 /p 4 ] = [x NH3 2 /x 3 H2 x N2 ] [p 2 /p 4 ] K = K X 1/ p 2 where, K X = [x NH3 2 /x 3 H2 x N2 ] or K X p 2» Since K is constant, if p increases must increase as the square» Self test 9.3: If there is a ten-fold increase in pressure, what will be the effect on the constituents K X p 2 so a ten fold increase in pressure will increase K X by 100 This means that the products must increase or the reactants must decrease Le Chatleier s principle tells you that by inspection since reactants have the greatest contribution to the total pressure

30 EFFECT OF CATALYST

31 SAMPLE PROBLEM Consider the following reaction: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(g) H = kj/mol How does the equilibrium position shift when: a) 0.20 atm of Argon is added? b) The gaseous water product is liquified as soon as it is formed? c) 0.2 moles of oxygen is added? d) The temperature of the system is increased from 278K to 700K? e) The total system pressure is increased by 100 atm? f) If gaseous HCl is added? Note that NH 3 reacts with HCl to form NH 4 Cl(s)

32 EQUILIBRIUM APPLICATIONS: LIME PRODUCTION FROM LIMESTONE CaCO 3 (s) CaO(s) + CO 2 (g) H = +178 kj/mole limestone lime K p p CO2 Reaction is endothermic; so formation of lime is favored at high temperature. One mole gas is formed; no gas in the reactant so decreasing the pressure will result in the formation of more CaO. In normal practice the reaction is carried out at 900 o C to 1000 o C at atmospheric pressure in lime kilns that are well ventilated.

33 EQUILIBRIUM APPLICATIONS: LIME PRODUCTION FROM LIMESTONE

34 EQUILIBRIUM APPLICATIONS: METHANOL PRODUCTION Methanol is an important alcohol used in fuel mixtures, making methyl esters and oxidation to methanol (formaldehyde) to make ureaformaldehyde resin glues. It is manufactured directly from synthesis gas CO(g) + 2H 2 (g) CH 3 OH(g) ( H = -90 kj mol -1 ) Kp = pch3oh pco ph22

35 EQUILIBRIUM APPLICATIONS: METHANOL PRODUCTION The reaction is carried out at 250 o C, over a Cu-ZnO-Al 2 O 3 (alumina) catalyst at a pressure of 5-10 x 106 Pa (5-10MPa, atm). Theoretically the reaction is favored by high pressure (3 gas mol ==> 1 gas mol) and low temperature. In practice a high pressure is used to give an acceptable yield in accordance with Le Chatelier's Principle, but a moderately high temperature plus a catalyst, are employed to get an economic production rate.

36 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS The three raw principal materials are (i) air, (ii) coal-coke/methane or higher hydrocarbons and (iii) water. The nitrogen is obtained from air (80% N2) by two principal methods: Air is cooled and compressed under high pressure to form liquid air (liquefaction). The liquid air is fractionally distilled at low temperature to separate oxygen (used in welding, hospitals etc.), nitrogen (for making ammonia), Noble Gases e.g. argon for light bulbs, helium for balloons). By burning natural gas in air and separating the nitrogen from the water and carbon dioxide combustion products.

37 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS Synthesis gas is the primary source of hydrogen, though 'syn-gas' or 'syngas' is a mixture of hydrogen and carbon monoxide and is made by three main methods. 1. Historically synthesis gas was made by passing steam over white hot coke at o C. Coke is made by heating coal at high temperature and is mainly carbon. This process is still used in many parts of the world and if coal is available in a country without oil, then it will be cheaper than using hydrocarbons from oil.

38 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS C(s) + H 2 O(g) CO(g) + H 2 (g) ( H = +131 kj mol-1) The very high temperature favors the endothermic production of hydrogen and of course considerably increases the rate of reaction. Water and carbon show little reaction at lower temperatures. The reaction can be carried out at normal pressure since higher pressures favour the reverse reaction (1 gas mol <== 2 gas mol).

39 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS 2. Methane can be partially oxidized by reaction with the oxygen in air. CH 4 (g) + 1/2O 2 (g) CO(g) + 2H 2 (g) ( H = -36 kj mol -1 ) Low pressure (1.5 gas mol ==> 3.0 gas mol) and low temperatures (exothermic) favor this reaction. This reaction can be carried out with nearly 100% conversion at atmospheric pressure (so low pressure) at o C using cobalt/nickel based catalysts e.g. Ni-ThO 2 or Ni-SiO 2. Very little water or carbon dioxide is formed if the reaction conditions are carefully controlled.

40 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS These days hydrogen is primarily made by reacting methane (natural gas) and water (steam), and the process is called steammethane reforming. CH 4 (g) + H 2 O(g) 3H 2 (g) + CO(g) ( H = +206 kj mol-1) Kp = ph23 pco pch4 ph2o The reaction is carried out at o C, over a nickel catalyst, at a pressure of 1-2 x 10 6 Pa (1-2MPa, atm).

41 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS This reaction is favoured by high temperature (endothermic, heat absorbing) and low pressure (2 gas moles ==> 4 gas moles). In practice a high temperature is indeed used, and combined with a catalyst, both factors will increase the rate of reaction to ensure the process is economic. However, although a low pressure is advantageous, in practice a moderately high pressure is used, which effectively increases the concentration of the reactants and provides a greater bulk of material passing through the reactor in a given time.

42 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS The yield of hydrogen can be further increased by using the 'shift' reaction. CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) ( H = -41 kj mol-1) Since 2 gas mol ==> 2 gas mol, changing pressure offers no advantage. However, the lower the temperature, the greater the yield of this exothermic reaction. But, too low a temperature leads to an uneconomic low rate of reaction, so research continues into low temperature catalysts The carbon dioxide can be absorbed under pressure into water or potassium carbonate solution.

43 HABER PROCESS:

44 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS The Bosch-Haber synthesis equation for ammonia is N 2 (g) + 3H 2 (g) 2NH 3 (g) ( H = -92 kj mol -1 ) Kp = pnh pn2 ph23 Since a dynamic equilibrium will form, there is no chance of 100% yield even if you use, as you actually do, the theoretical stoichiometric reactant volume/mole ratio of N2 :H2 of 1 : 3! In forming ammonia, heat energy is given out to the surroundings Four moles of 'reactant' gas form two moles of 'product' gas, so there is net decrease in gas molecules on forming ammonia.

45 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS So applying the equilibrium rules, the formation of ammonia is favored by: (a) Using high pressure because you are going from 4 to 2 gas molecules (the high pressure also speeds up the reaction because it effectively increases the concentration of the gas molecules) and allows a greater bulk flow rate material through the reactor, but very high pressure means more dangerous and more costly engineering to address the health and safety issues involved. (b) Carrying out the reaction at a low temperature, because it is an exothermic reaction favored by low temperature, but this may produce too slow a rate of reaction,

46 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS So, the idea is to use a set of optimum conditions to get the most efficient yield of ammonia and this may well, as in this case, be a low % yield (e.g. 8% conversion) but fast!!! Described below are the conditions to give the most economic production of ammonia. These arguments make the point that the yield* of an equilibrium reaction depends on the conditions used

47 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS

48 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS The above chart completely illustrates and vindicates the predictions from Le Chatelier's Principle. In industry moderate-high pressures of x 10 6 Pa (2.5-25MPa, atm) in line with the theory but to high to raise H&S engineering costs. Theoretically a low temperature would give a high yield of ammonia BUT... Nitrogen is very stable molecule and not very reactive i.e. chemically inert and the rate of reaction at high % yield low temperatures is much to slow to be economic.

49 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS To speed up the reaction an iron oxide catalyst (Fe 3 O 4 ) is used as well as a higher temperature (e.g o C). The higher temperature is an economic compromise, i.e. it is more economic to get a low yield very fast, than a high yield very slowly! Note: a catalyst does NOT affect the yield of a reaction, i.e. the equilibrium position BUT you do get there faster!

50 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS At the end of the process, ammonia is liquified and is so can be removed and stored in cylinders. Any unreacted nitrogen or hydrogen (NOT liquified), is recycled back through the reactor chamber, nothing is wasted! Nitrogen (bp: -196 o C) and hydrogen (bp: -252 o C) have much lower boiling points than ammonia (bp: -33 o C). To sum up: A low % yield of ammonia is produced quickly at moderate high temperatures and pressures, and is more economic than getting a higher % equilibrium yield of ammonia at a more costly high pressure or slower lower temperature reaction.

51 EQUILIBRIUM APPLICATIONS: MANUFACTURE OF SULFURIC ACID The Contact Process of sulphur trioxide production must be economically efficient for the manufacture of the important industrial chemical sulphuric acid. In the Contact Process reactor the sulphur dioxide is mixed with air (the required stoichiometric volume/mole SO2:O2 ratio is 2:1, in practice 1-2:1 is used) and the mixture passed over a catalyst of vanadium(v) oxide V at a relatively high temperature of about 450 C and at a pressure of between 1-2 atm.

52 EQUILIBRIUM APPLICATIONS: MANUFACTURE OF SULFURIC ACID (1) In the reactor the sulphur dioxide is oxidised in the reversible exothermic reaction. 2SO 2 (g) + O 2 (g) 2SO 3 (g) ( H = -196 kj mol-1) Kp = pso pso22 po2 The reaction forms sulphur trioxide and the equilibrium is very much to the right hand side because

53 EQUILIBRIUM APPLICATIONS: MANUFACTURE OF SULFURIC ACID Despite the reaction being exothermic a relatively high temperature is used which favours the reverse reaction R to L, from the energy change equilibrium rule, i.e. increasing temperature shifts the equilibrium in the endothermic direction. However the value of Kp is high enough to give a 99% yield. The reaction is favoured by high pressure (pressure equilibrium rule, 3 => 2 gas molecules), but only a small increase in pressure is used to give high yields of sulphur trioxide, because the right hand side is energetically very favourable (quite exothermic and high Kp)

54 EQUILIBRIUM APPLICATIONS: MANUFACTURE OF SULFURIC ACID The use of the V2O5 catalyst ensures a fast reaction without having to use too a higher temperature which would begin to favor the left hand side too much (energy change equilibrium rule), but remember a catalyst does not affect the % yield or equilibrium concentration of SO 3, you just get there more economically faster. Multiple reactor beds are used to ensure the maximum % conversion and heat exchange systems are used to control the temperature, and pre-heat incoming reactant gases. Good anti-pollution measures need to be in place since the sulphur oxides are harmful and would cause local acid rain! To help this situation AND help the economics of the process the residual SO 2 is kept to the minimum by the reaction conditions describe above.

55 EQUILIBRIUM APPLICATIONS: MANUFACTURE OF SULFURIC ACID (2) The sulphur trioxide is dissolved in concentrated sulphuric acid to form fuming sulphuric acid (oleum). SO 3 (g) + H 2 SO 4 (l) ==> H 2 S 2 O 7 (l) (3) Water is then carefully added to the oleum to produce concentrated sulphuric acid (98% H 2 SO 4 ). H 2 S 2 O 7 (l) + H 2 O(l) ==> 2H 2 SO 4 (l) If the sulphur trioxide is added directly to water an acid mist forms which is difficult to contain because the reaction to form sulphuric acid solution is very exothermic with a big K value!

56 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS

57 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS

58 EQUILIBRIUM APPLICATIONS: SYNTHESIS OF AMMONIA - THE HABER PROCESS