CH10090 hermodynamics (art ) Let s remember what we did in CH10089 Enthaly changes during reactions o o o H98 ( reaction) = νi Hf, 98( roducts) νi Hf, 98( reactants) ν i reresents the stoichiometric coefficient. his method can be used to calculate H for any reaction where we have H f,98 data for the comonents. A 5 C! 1
he Second Law of hermodynamics Sontaneous rocesses are those which increase the entroy of the Universe Entroy related to order sontaneous rocesses are those which lead to an increase in disorder. S = S final - S initial S = q rev he hird Law of hermodynamics he entroy of a erfect crystal at zero Kelvin is zero he standard entroy, S 98 of a comound is the molar entroy at 98.15 K and 1 bar ressure
Entroy changes in chemical reactions the standard entroy of reaction as the difference in standard entroy of the roducts from that of the reactants S 98 (reaction) = Σ ν i S 98 (roducts) - Σ ν i S 98 (reactants) A 5 C! Now we need to be able to work at other temeratures he Gibbs function or Gibbs free energy, G. G = H S or G = H - S For a sontaneous rocess at constant, there must be a lowering of the Gibbs free energy or ( G), < 0 3
Entroy as a redictor of chemical reactivity H (g) + O (g) H O (l) S 98 = [ x S 98 (H O)] - [ x S 98 (H ) + S 98 (O )] = [ x 70 ] - [ x 131 + 05 ] = - 37 J K -1 mol -1 here is a large loss of entroy on forming water. WHY? his is for the H / O / H O system. he nd Law refers to the entroy of the Universe increasing as a requirement for sontaneity. S(Universe) = S(system) + S(surroundings) S(surround) = H (surround) / = - H (system) / H 98 for the above reaction (the system) is -571.6 kj. So S(surround) = + 571.6 x 10 3 J / 98 K = + 190 J K -1 S(Universe) = S(system) + S(surroundings) = - 37 + 190 = + 1593 J K -1 Hence, there is an overall increase in the total entroy of the Universe, in accord with the second law. 4
Balloon containing H (gas) + sark! r H (system) = - 571.6 kj HEA r H (surroundings = + 571.6 kj H (g) + O (g) H O (l) 5
he Gibbs function or Gibbs free energy, G. From nd Law, for a sontaneous rocess at constant, : or S(Universe) = S(Sys.) + S(Surr.) > 0 S(Universe) = S(Sys.) + { H(Surr.) / } > 0 We don t want to focus on the Universe every time: Since, H(Surr.) = H(Sys.). S(Sys.) + ( H(Sys.) / ) > 0 So we can write H - S < 0 he value of H - S is the Gibbs function or Gibbs free energy, G. G = H S or G = H - S For a sontaneous rocess at constant, there must be a lowering of the Gibbs free energy or ( G), < 0 6
In general, we can conclude that if a rocess or reaction results in a lowering of the free energy, it will be sontaneous i.e. that it will be favourable and CAN haen. It does not mean that it WILL haen. e.g. at 1 bar and 5 C H (g) + O (g) H O (l) G = - 73. kj mol -1 However, a mixture of hydrogen and oxygen gases in a container will not react until started by a catalyst or source of ignition. he reaction is thermodynamically very favourable but occurs extremely slowly. It is kinetically unfavourable as it has a high activation energy. 7
Key oints so far: he Second Law of hermodynamics: Sontaneous rocesses increase the entroy of the Universe the hird Law of hermodynamics : he entroy of a erfect crystal at zero Kelvin is zero S 98 (reaction) = Σν i S 98 (rod.) - Σ ν i S 98 (react.) Gibbs function or Gibbs free energy, G. An exact definition of G is: G = H S or G = H - S For a sontaneous rocess at constant, ( G), < 0 Derivation of equations is NO imortant. You do need to erform calculations using the equations and interret the results in terms of the chemistry 8
Calculate the Gibbs free energy change at 98 K and estimate the temerature at which the reaction just becomes sontaneous. Fe O 3(s) + 3 C (s) 4 Fe (s) + 3 CO (g) Fe O 3 (s) C (s) Fe (s) CO (g) f H 98 / kj mol -1 : -84. 0 0-393.5 S 98 / J K -1 mol -1 : 87.4 5.7 7.3 13.7 Solution calculate r H 98 and r S 98 o o o r H 98 = ν i f H 98 ( roducts) ν i f H 98 ( reactants) r H 98 = [ 0 + 3 (-393.5)] - [ (-84.) + (0)] = + 467.9 kj mol -1. o o o S = ν S ( roducts) ν S ( reactants) r 98 i 98 i 98 r S 98 = [4 (3.7)+3 (13.7)] - [ 87.4 + 3 5.7] = + 558.4 J mol -1 K -1. calculate r G 98 r G 98 = r H 98 - r S 98 = + 467.9 - (98.15) (558.4 x 10-3 ) kj mol -1. = + 301.4 kj mol -1. he Gibbs free energy change is ositive and so the reaction is non-sontaneous at 98 K. It becomes sontaneous when r G = r H - r S = 0. Substitute the values of r H 98 and r S 98 to find when r G = 0 9
r G = 0 r H - r S = 0 + 467.9 kj mol -1 ( / K) (558.4 x 10-3 ) kj mol -1 K -1 = 0 Rearrange to find + 467.9 kj mol -1 = ( / K) (558.4 x 10-3 ) kj mol -1 K -1 + 467.9kJmol 558.4 10 3 kjmol 1 1 K 1 = ( / K) = 838 K 10
Variation of enthaly with temerature What do we already know? Enthaly Heat energy (at constant ressure) We have data at 5 C Heat caacities describe changes in energy with temerature We define a constant ressure heat caacity, C, - the amount of heat energy needed to raise the temerature of 1 mole of substance by 1 K at constant ressure C = dq d = dh d or dh = C d So, H must increase by (C d) if heated by d. dh = 1 C d the Kirchoff Equation 11
Aroximation: treat c as indeendent of for small tem. ranges and use the average value over the range of If a comound has an enthaly H 1 at temerature 1 etc. H H 1 = C ( 1) Particularly useful: standard enthalies of formation are defined at 5 C. Hence, H f, = H f, 98 + C ( - 98.15 ) How can we use this? C for liquid mercury at 5 C is 7.98 J mol -1 K -1. Calculate the enthaly of formation at 50 C. H f, = H f, 98 + C ( - 98.15 ) H f, at 5 C is 0 (by definition). H f, 33 = 0 + (7.98 J mol -1 K -1 ) ( 33.15-98.15 ) K = 699 J mol -1 1
Enthalies of Reaction A model reaction: he enthaly of R at 5 C is R P H f, 98 (R). he enthaly at some other temerature is [ H f, 98 (R) + C (R) ( - 98.15)]. H (reaction) = [ H f, (P) ] [ H f, (R)] = [ H f, 98 (P) + C (P) ( 98.15) ] [ H f, 98 (R) + C (R) ( 98.15)] = [ H f, 98 (P) - H f, 98 (R)} + {[ C (P) - C (R)] ( 98.15)} or H (reaction) = H 98 (reaction) + C ( - 98.15) where C = ν i C (roducts) - ν i C (reactants) so, we can now calculate the enthaly of reaction at any temerature. 13
he comlete combustion of ethane releases 1558.8 kj mol -1 at 5 C. Calculate H (combustion) at 100 C. c / J K -1 mol -1 : C H 6(g) 5.6; O (g) 9.4; CO (g) 37.1; H O (l) 75.3 C H 6(g) + 3½ O (g) CO (g) + 3 H O (l) H 98 = - 1558.8 kj mol -1 C = [ C (CO ) + 3 C (H O)] - [C (C H 6 ) + 3½ C (O )] = [ (37.1) + 3 (75.3)] - [ 5.6 + 3½ (9.4)] = 144.6 J K -1 mol -1 H 373 = H 98 + C = - 1558.8 x 10 3 (J mol -1 ) + (144.6 J K -1 mol -1 )(75 K) = - 1547.9 kj mol -1 (See CH10089 notes for this roblem at 5 C) A gas mixture consisting of 5% N and 75% H (by volume) was assed over a catalyst at a rate of 11 dm 3 min -1 (Gas volumes corrected to 0 C and 1 atm ressure). he reaction took lace at 450 C and 1 bar ressure and comlete conversion to ammonia was achieved. Calculate the rate of heat evolution or absortion. H f, 98 (NH 3 ) - 46.0 kj mol -1 Mean C / J K -1 mol -1 : N 9.7; H 9.3; NH 3 39.7. 14
N (g) + 3 H (g) NH 3(g) H 98 = [ x H f, 98 (NH 3 )] - [ H f, 98 (N ) + 3 H f, 98 (H )] = [ - 9.0 ] - [ 0 ] kj mol -1 H 98 = - 9.0 kj mol -1 Changing to 450 C using the Kirchoff Eqn. H (reaction) = H 98 (reaction) + c ( - 98.15) C = [ x c (NH 3 )] - [ c (N ) + 3 x c (H )} = [ x 39.7] - [ 9.7 + (3 x 9.3)] = - 38. J K -1 mol -1 H 73 = - 9000 Jmol -1 + (- 38. J K -1 mol -1 ) (73.15-98.15)K = - 9000 Jmol -1 + ( - 1635 Jmol -1 ) H 73 = - 108.4 kj mol -1 Hence, at 450 C, 108.4 kj is evolved for each mole of the equation i.e 108.4 kj is evolved er mole of nitrogen which reacts. Since, for a gas, volume number of moles, the 1:3 mixture is in the stoichiometric amount. At 0 C and 1 atm ressure, 1 mole occuies.4 dm 3. hus, the total amount of gas assing over the catalyst amounted to 5 moles er minute, of which 1.5 moles were N. Rate of heat evolution = (mol of N min -1 ) (heat er mole N ) = ( 1.5 mol min -1 ) (- 108.4 kj (mol N -1 ) = - 135.3 kj min -1 he negative sign indicates an exothermic reaction so that heat is evolved. 15
Effect of temerature on entroy From before, we know that: and S q rev = C d = q rev at constant ressure. he entroy at f is related to entroy i by S = dq S rev S f = = i f i c d If we aroximate c is constant over the temerature range involved, then roviding no change of hase occurs: S f = S + c i f i 1 d = S i + c ln f i 16
Given the heat caacities of ice and water at 0 C are. and 4.18 J K -1 g -1 and the enthaly of fusion is 33 J g -1, calculate the entroy change for the freezing of 1 mole of suercooled water at -10 C. Clearly, H O (l), -10 C H O (s), -10 C cannot be carried out in a reversible manner. hus, in order to calculate S, we must searate the rocess into a series of reversible changes and sum S for each. his rocedure is valid since entroy is a state function. o S o S o S o 1 3 water, -10 C water, 0 C ice, 0 C ice, -10 C S= S1+ S + S3 = 4. 18 ln 73-33 + +. ln 63 J K -1 g 63 73 73 1 S = 0.156 + (-1.16) + (-0.08) = -1.14 J K -1 g -1 Hence, S = (18 g mol -1 )(-1.14 J K -1 g -1 ) = - 0.556 J K -1 mol -1 Note: We have a negative entroy change arising from a liquid forming a more ordered solid. 17
Standard or Absolute Entroies For convenience, entroies are listed under standard conditions. he standard entroy, S 98 of a comound is the molar entroy at 98.15 K and 1 bar ressure e,g. Consider a gas at room temerature. How can we find the standard entroy,s 98? 1. Heating from 0K to m. S. Melting at m 3. Heating from m to b S 4. Vaorization at b 5. Heating from b to 98 K S S o 1 m = 0 S b o 3 = o 4 = im o 5 98 = b c o = c H c ( solid) d H fus. m ( liquid) d va. b ( vaour) d S 98 = S 1 + S + S 3 + S 4 + S 5 For liquids at 98 K and 1 bar, omit stes 4 and 5 and relace b with 98 in ste 3 18
Where there is more than one hase in the solid, the additional S entroy for the hase changes must also be taken into account. S 1 can t be calculated in this form it must be evaluated grahically or numerically. he shaded area is equivalent to the standard entroy at temerature Entroy changes in chemical reactions S 98 (reaction) = Σ ν i S 98 (roducts) - Σ ν i S 98 (reactants) he entroy change at other temeratures can be calculated by S = S + 1 1 c d 19
Using mean heat caacities, S = S + c 1 = S + c 1 1 1 d ln 1 his allows us to calculate entroy changes during reactions at any temerature. Using these values, at any temerature, G = H - S Key oints so far: Calculation of H for reactions at any temerature Calculation of S for reactions at any temerature Definition and measurement of absolute entroies Calculation of G for reactions at any temerature 0
Variation of Gibbs free energy with conditions We begin with a single comonent. From the definition of free energy: G = H - S = (U + PV) S o find the change in Gibbs free energy, dg, caused by small changes in any of the other conditions, we take the total differential dg = du + PdV + VdP - ds - Sd From earlier lectures, du = dq + dw = dq - PdV and ds = dq / or dq = ds du = ds - PdV Substituting this: dg = (ds - PdV) + PdV + VdP - ds - Sd dg = VdP - Sd dg = VdP - Sd the master equation or the fundamental equation - allows us to calculate the effect on G of changing conditions. Note: Strictly, this is for a single comonent. We will look at mixtures with varying comosition later. 1
How can we use the Master equation? Variation of G with temerature dg = VdP Sd at constant ressure: dp = 0 dg = - S d or d G d = P S If we know the entroy of a substance, we can find out how G varies with temerature. an alternative interretation of entroy - the deendence of G For chemical reactions, d G d = P S However, it is usually easier to calculate H and S from data and then use G = H - S
Given the ure comonent data below, calculate the standard free energy change for the following reaction at 5 C and 105 C. MgO (s) + CO (g) Mg (s) + CO (g) MgO (s) CO (g) Mg (s) CO (g) H / kj mol -1 : -601.7-110.5 0-393.5 S / J K -1 mol -1 : 6.9 197.7 3.7 13.7 C / J K -1 mol -1 : 37. 9.1 4.9 37.1 H 98 = [ 0 + (-393.5)] - [ (-601.7) + (-110.5)] = + 318.7 kj mol -1. S 98 = [ 3.7+ 13.7] - [ 6.9 + 197.7] = 1.8 J mol -1 K -1. G 98 = H 98 - S 98 = 318.7 - (98.15)( 1.8 x 10-3 ) kj mol -1. = + 31. kj mol -1. o convert H 98 and S 98 to the higher temerature, we need C. C = [ 4.9+ 37.1] - [ 37. + 9.1] = -4.3 J mol -1 K -1. H 198 = H 98 + C (198-98) = 318.7 + (-4.3 x 10-3 )(1000) = 314.4 kj mol -1. S 198 = S 98 + C ln (198 / 98) = 1.8 + (-4.3)(1.471) = 15.47 J mol -1 K -1. G 198 = H 198 - S 198 = 314.4 - (198.15)(15.47 x 10-3 ) kjmol -1 = + 94.3 kj mol -1. Reduction of a metal oxide with CO is sometimes used to recover the metal from an ore. In this case, is this aroach feasible? 3
Variation of G with ressure dg = VdP Sd At fixed temerature, d = 0. dg = VdP or dg d = V Molar volumes of solids and liquids, and their changes during reactions, are small so G or G changes little with ressure. For an ideal gas. dg d = V = nr If the free energy changes from G 1 to G when the ressure changes from 1 to, G G 1 dg nr 1 = d = nr d 1 1 which leads to G G = nr ln 1 1 4
G is the value at the standard ressure of = 1 bar. G = G + nr ln ( / ) = G + nr ln ( / 1 bar) his is often written as G = G + nr ln () - OK as long as the ressure is measured in units of bar For gases, ressure concentration so this gives a measure of how the free energy deends on concentration for a gas or gaseous reaction. he thermodynamic activity o get a consistent reresentation of the comosition deendence of G in all systems, we define the activity or effective concentration of the comonent. We can use the general equation G = G + nr ln a with the activity, a, and standard states defined. Ideal gases: a = / = 1 bar he effect of ressure on solids and liquids is negligible: G = G for a ure liquid or solid so the activity must be one. a (ure solid) = a (ure liquid) = 1 we can deal with non-ideal gases (but not here.!!) 5
Variation of G with concentration We use the same form for solutions by using molar concentrations so that we adot as our standard state a concentration of 1 mol dm -3. G i = G i + nr ln ( [i] / 1 mol dm -3 ) = G i + nr ln [i] Note: he use of molar concentrations is a sort of ideal aroximation which is valid at low concentrations. We can now take data (available in tables) and calculate the enthaly, entroy and free energy of reactions under a wide range of conditions. But, can we do more than redict sontaneity?? 6
Free energy and equilibrium he free energy of the system will change throughout a reaction deending on the amounts of reactants and roducts. It will reach a minimum at some comosition (100% reactant no reaction or 100% roduct comlete reaction or any value in between) At the minimum, the reaction will have no tendency to go anywhere and so will sto. It has reached equilibrium. R = reactants P = roducts E = equilibrium In (a), G(roducts) < G(reactants). G is negative so the reaction goes forward. he oosite is true for (c). For (b), G would be negative if the system started at either end so the reaction would roceed until oint E. If the comosition changes, G would increase. hus, the reaction cannot change from this comosition - it has come to equilibrium. 7
8
his schematic aroach allows us to define one of the fundamental statements of chemistry A system will come to equilibrium when it has reached its minimum free energy. We need a quantitative measure of the comosition of a reaction mixture. he easiest uses the Law of Mass Action to define the Reaction Quotient, Q For our general reaction, α A + β B γ C + δ D Q a a = ( C ) ( D ) α ( a ) ( a ) A γ B δ β When the system reaches equilibrium, Q eqm γ ( ac) eqm( ad) eqm = = α β ( a ) ( a ) A eqm δ B eqm K eqm K eqm is therefore the equilibrium constant Q = ν ( a ( roducts) ) ν ( a ( reac tan ts) ) ν is the stoichiometric coefficient. he symbol Π is shorthand for the roduct of 5 ( ) = 1 ai a 1 a a 3 a 4 a 5 9
Different tyes of equilibrium constant can be defined: If K eqm is calculated from gas ressures - K. If calculated from concentrations - K c. Examles. N (g) + 3 H (g) NH 3(g) K = ( NH ) 3 eqm 3 H eqm N eqm ( ) ( ) CaO (s) + CO (g) CaCO 3(s) K = ( acaco ) 3 a = 1 ( ) ( ) ( ) ( ) = 1 1 ( ) CO eqm CaO CO eqm CO eqm HCOOH (aq) + C H 5 OH (aq) HCOOC H 5(aq) + H O (l) K c HCOOC H eqm H O = [ 5 ] [ ] [ HCOOH] [ C H OH] eqm 5 eqm eqm CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) K c = + ( eqm) ( eqm) [ CH COO ] [ H ] 3 [ CH COOH] 3 ( eqm) Although we write ress. or conc. in the exressions, they are in fact ratios e.g. / 1 bar, c / mol dm -3 etc. so K eqm is dimensionless and has no units. 30
he definitions of Q and K eq allow us to redict the direction of a chemical reaction. We can say that: If G < 0 or If G > 0 or If G = 0 or Q < K eqm the forward reaction roceeds Q > K eqm the reverse reaction roceeds Q = K eqm the system is at equilibrium But there s more.. Relation between G and K eqm Consider the reaction (ν I = 1) A + B C + D G(react.) = G (rod) - G (react) = (G C + G D ) - (G A + G B ) Let s start by assuming: G A = G + R ln a A etc G(react.) = {(G C + R ln a C ) + (G D + R ln a D )} - {(G A + R ln a A ) + (G B + R ln a D )} ={(G C +G D )-(G A +G B )} + R{(ln a C + ln a D ) - (ln a A +ln a D )} 31
a = {(G C + G D ) - (G A + G B )} + R ln CaD aaab or, G(reaction) = G (reaction) + R ln Q At equilibrium, G is a minimum, G(reaction) = 0 and Q = K eq. 0 = G (reaction) + R ln K eqm K eqm = ex G o R At equilibrium, G for the reaction under the equilibrium conditions will be zero; G which is defined under standard conditions will not in general equal zero Any stoichiometry or reaction leads to the above exression. NB: he above definition is not quite strictly true - but it is close enough (for now..) 3
Gibbs Free Energy and Reactivity We know that if G < 0, K eqm > 1 so the forward reaction takes lace. he roducts will be resent in larger concentration than the reactants at equilibrium. Can we quantify this? Define a comlete reaction as a yield of 99.99% roducts. Amount of roducts 99. 99 K eqm = = = 9999 10 4 Amount of reactants 0. 01 For K eqm = 10 4 at 98 K, G must be -.8 kj mol -1 so any reaction with a more negative value will go to comletion. If G > +.8 kj mol -1 the reaction will not occur. Between these limits, a variety of behaviour results. K eqm 10 4 1 10-4 G / -3 0 + 3 kj mol -1 COMPLEE FAVOURABLE UNFAVOURABLE NO REACION REACION REACION REACION hese limits are temerature deendent because of the - S contribution to the Gibbs free energy. 33
Pause for breath where are we? Let us reca. We can take data in terms of S 98, H f,98 and C for the comonents in a reaction, calculate H and S at any tem., calculate G and hence the K eqm. We can redict the course of the reaction and the osition of equilibrium. his gives us the maximum ossible yield of roducts. Worked Examle: Ag CO 3(s) he next ste is to quantify the yields of roducts and reactants at equilibrium. 34
Silver carbonate decomoses on heating. Calculate the equilibrium constant at 110 C for the reaction. Ag CO 3(s) Ag O (s) + CO (g) AgCO 3(s) Ag O (s) CO (g) H / kj mol -1 : -501.4-9.07-393.5 S / J K -1 mol -1 : 167.3 11.7 13.7 C / J K -1 mol -1 : 109.6 68.6 37.1 H 98 = [ (-9.07) + (-393.5)] - [ (-501.4) ] = + 79.03 kj mol -1. S 98 = [ 11.7+ 13.7] - [ 167.3] = 168.0 J mol -1 K -1. C = [ 68.6 + 37.1] - [ 109.6] = -3.9 J mol -1 K -1. H 383 = 79.03 + (-3.9 x 10-3 )(383.15-98.15) = 78.70 kj mol -1 S 383 = 168.0 + (-3.9) ln (383.15/98.15) = 167.0 J mol -1 K -1. G 383 = 78.7 - (383.15)(167.0 x 10-3 ) = 14.71 kj mol -1. K o 3 G 14.71x 10 = ex = ex = 9.87 x 10 (8.314)(383.15) R -3 Say a samle of Ag CO 3(s) is to be dried in an oven. Q for this reaction is simly (CO ). If the conditions in the oven are such that the artial ressure of CO is > 9.9 x 10-3 (or 1% by volume) then the silver carbonate will not decomose. 35
Relating the Equilibrium Constant to Exeriment ake an examle:- gas hase reaction: N O 4(g) NO (g) Introduce 1 mol of N O 4 into a container and allow to reach eqm at a total ressure of P tot At the start of the reaction: 1 mole N O 4, 0 moles NO If α moles of N O 4 have reacted when the reaction comes to equilibrium, (1 - α) mol of N O 4 remain, α mol of NO have been formed. he total number of moles of gas resent at equilibrium is (1 - α) + α = (1 + α) mol. α is better defined as the fraction of moles of reactant which react and is often termed the degree of dissociation. We now know the number of moles of each comonent and the total ressure so we can work out an exression for the equilibrium constant! For gases acting ideally, the artial ressure,, is given by Dalton s Law: = mole fraction total ressure = x P tot. x is given by: N o of m oles of com onent x = otal N o. of m oles resent 36
From now on, we need some algebra. For: N O 4(g) NO (g) K ( N O ) ( x N O P to t ) = = = ( ) ( x P ) N O 4 N O to t 4 α 1+ α 1 1 + α α P P to t to t K = 4α Ptot ( 1+ α) 4α P = tot ( 1 α)( 1+ α) P ( 1 α)( 1+ α) tot or K = 4 α P to t (1 α ) Hence, if we calculate K we can calculate α at any tem. and ressure. From this, the maximum amount of roduct can be readily found. BU: each different reaction stoichiometry leads to a different exression for K (or K c as aroriate)so each must be worked out searately no standard formulae to learn!!. 37
Many stoichiometries give rise to quadratic or cubic equations. Numerical solution of the equations is easy. o get an idea, we often assume that the degree of reaction is small so that α << 1 so that (1 - α) 1. e.g. for the reaction above, K 4 α P and so α K 4 P tot. tot his gives a useful first estimate. Worked Examle: Formulate exressions α 38
Formulate exressions in terms of the degree of dissociation for the equilibrium constants for the following reactions. (a) (b) (c) Ni(CO) 4(g) Ni (s) + 4 CO (g) HgO (s) Hg (l) + O (g) CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq) (a) If we start with 1 mole Ni(CO) 4(g) and α react, then at eqm., 1-α moles remain. α moles of Ni (s) together with 4α moles of CO form. When considering K, only gases contribute to the artial ressures. he total number of gas moles = (1-α) + 4α = (1+3α) K ( 1) ( CO ) = = Ni( CO) 4 4 4 4α ( 1) 1+ 3α 1 α P 1+ 3α P tot 4 tot 4 3 tot 4 3 tot ( 4α) ( 1+ 3α) P 56α P = = ( 1 α)( 1+ 3α) ( 1 α)( 1+ 3α) (b) he only gas involved is O. hus, since the activity of solids is 1 and the only gas resent is oxygen, 4 3 ( a ) ( ) (1) ( ) K = Hg O O = = ( a HgO ) (1) P tot (c) If the initial concentration is c mol dm -3 and α is the fraction which reacts, at equilibrium there will be (αc) mol dm -3 of H +, (αc) mol dm -3 of CH 3 COO - and (1-α)c mol dm -3 of CH 3 COOH. hus + H COO K c = [ ] [CH 3 ] COOH = ( α c ) ( α c) α c = [ CH ] ( 1 α) c 1 α 3 39
Effect of conditions on reaction yields and K eqm Effect of ressure K eqm deends on G which is only valid at 1 bar ressure and hence K eqm is a thermodynamic constant which does not vary at constant temerature. Just for emhasis: K eqm does not vary with Pressure / Concentration at constant temerature his does not mean that ressure has no effect on the osition of equilibrium. e.g. for N O 4 NO ( NO ) K = K = ( ) N O 4 4 α P to t (1 α ) If K is constant, if P tot changes α changes to comensate. 40
Effect of temerature Since G = H - S = - R ln K eqm, o o o ln K eqm = G = H S + R R R or S l n K eqm = R o H R o his is a variation of the Van t Hoff equation. - If H and S are indeendent of temerature then ln K eqm = ( constan t) so there is a linear relation between ln K eqm and (1/). H o 1 R Alternatively, if K eqm is known at two temeratures 1 and, ln K K eqm eqm ( ) ( ) = H 1 1 R o 1 1 If we have an exothermic reaction, H is negative, then K eqm will decrease as increases. he oosite is true for an endothermic reaction. 41
Le Chatelier s rincile he above rovides the exlanation of a rincile suggested by Le Chatelier on the basis of emirical observations. If a system is subjected to a constraint, it will react so as to minimise the effect of the constraint. e.g. increased reduced the yield of NO from N O 4. If we increase the ressure of a reactant, the reaction moves toward the side of the equation with the lowest number of moles of gas i.e. if the ressure is raised, the system reacts by minimising the ressure of gas roduced. If is increased in an exothermic reaction, the system will react so as to absorb the extra energy and so change to minimise the roduction of energy by the reaction. he reaction therefore roceeds to a lesser extent so that the yield is lowered. Worked Examle: Acetic acid dimerizations he only way to become comfortable with this material and the methods which have been described is to attemt LOS of roblems (after working through the text. 4
In the vaour, acetic acid artially associates into dimers. At a total ressure of 0. bar, acetic acid is 9% associated at 5 C and 8% associated at 45 C. Calculate the enthaly and entroy changes for the association reaction. What will be the effect on the dissociation of changing the total ressure? he reaction can be reresented as HOAc (HOAc) If we begin with 100 mol and 9% dimerises, 8 mol remain at eqm. he 9 mol which react give rise to 46 mol of dimer. he total number of moles resent is therefore 54. Hence, at 5 C, K is given by K ( HOAc ) x HOAc ( ) tot ( 46 / 54) 0. = = = = ( ) ( x ) {(8 / 54) 0.} HOAc HOAc tot 194. 1 Using the data at 45 C gives K = 37.33. G = -13.06 kj mol -1 at 5 C and -9.576 kj mol -1 at 45 C. K ln K eq eq ( ( 1 ) ) = H R o 1 1 1 ; ln 37.33 = 194.1 o H 8.314 1 98.15 1 318.15 Hence, H = -65.0 kj mol -1. Using G = H - S gives S = -174. J K -1 mol -1. As exected, it is an endothermic reaction since the extent of dimerisation decreases as the temerature rises. he dimerisation must lead to an increase in order, hence the negative entroy change. 43
Alying Le Chatelier s rincile, an increase of ressure should favour the side of the reaction with the lower number of gas moles i.e. favour the dimerisation. x ( HOAc ) ( HOAc ) tot ( HOAc ) K = = = ( HOAc ) ( x HOAc tot ) ( x HOAc ) i.e. if P tot increases, x (HOAc) must also go u to comensate and kee K constant x tot An alternative aroach to calculating the values would be to write the two equations: G (5 C) = -13.06 kj mol -1 G (45 C) = - 9.58 kj mol -1 = H - (98.15) S = H - (318.15) S and solve simultaneously assuming that H and S do not vary with temerature. 44
Chemical otential Earlier, we made the assumtion that in a mixture G A = G + R ln a A - suggests the free energy of the comonent is indeendent of all the other comonents in the mixture. OK for ideal gases but not for other systems. e.g. think about 1 mole of water. Will the free energy change will be the same if it is added to a large amount of water, or benzene, or ethanol or sulhuric acid? We need the free energy change when a comonent is added to the system with all other conditions and amounts of other comonents ket equal. his is given by the artial differential of G with resect to that comonent. G A G = n A P,, n B G A is the artial molar free energy of comonent A and reresents the free energy change when 1 mole of A is added to a system at constant conditions. the chemical otential, given the symbol µ. For a ure substance, µ A corresonds to the molar Gibbs free energy. 45
If we have a large amount of a substance, B, and we add an amount dn A moles of A, then the free energy change at constant,, dg, is dg = µ A dn A For a multicomonent system, this can be generalised to ( ) dg = µ dn, i i i We can now write the fundamental equation of thermodynamics as dg = V dp S d+ µ i dni i Why is µ imortant? G = G + nr ln ( / ). If we consider solely comonent A in the mixture, then G = n A G + R ln n A µ A = µ A + R ln ( A / A ) A A µ A is the standard chemical otential at the standard artial ressure of A = 1 bar. his is identical with the free energy of 1 mole of ure A 46
Also, since dg = 0 at equilibrium. ( ) dg = µ dn, i i i or 0 = (µ A (-dn)) + µ B dn µ A dn = µ B dn or µ A = µ B this is an examle of a general requirement for chemical equilibrium. A reaction comes to equilibrium at constant temerature and ressure under conditions where the chemical otential of the reactants equals that of the roducts. Why the name - chemical otential?. At a higher altitude a body has a greater driving force to fall to earth - we say it has a higher gravitational otential. he body will sontaneously move from high gravitational otential to one low. In an electric circuit a high voltage - or electric otential - rovides a greater driving force for the electrons to move from high otential to low. 47
he chemical otential can be thought of as the driving force for a chemical rocess. µ is equivalent to the sloe of the Gibbs free energy versus comosition curve. he steeer the curve - hence higher µ - the greater the tendency for the system to move toward equilibrium where µ is zero. Investigation of many chemical systems involves calculating the free energy / chemical otential and seeing how this varies with the conditions - more later.. 48
Kinetics and hermodynamics of Equilibrium We are concentrating on thermodynamic asects of equilibrium. Perhas easier is the kinetic aroach. his also emhasises the dynamic nature of chemical equilibrium. he reaction does not sto when equilibrium is reached but the overall roortion of reactants and roducts does not change. If we have the general reaction A + B C the rate of the forward reaction is given by k f [A] [B] the rate of the reverse reaction is k b [C]. At equilibrium, these rates are equal so that: k [ C] f eqm kf[ A] eqm[ B] eqm = kb[ C] eqm or = = k [ A] [ B] b eqm eqm K c hus, the ratio of the forward and reverse rate constants also gives the equilibrium constant. An analogous argument using artial ressures holds for K. We must also consider the kinetics in determining the otimum conditions for a articular reaction. e.g. with an exothermic reaction, Le Chatelier s rincile shows the yield would be maximised by reducing the temerature. However, this will slow down the reaction. A balance must be worked out between these factors. 49
he standard enthaly of formation of water is -41.8 kj mol -1. Use mean heat caacities and standard entroies to estimate the ercentage dissociation of water vaour at 000 C and 0.01 bar ressure. he reaction of interest is the dissociation of water H O (g) H (g) + ½O (g) 5 C, H 98 = H f,98 = +41.8 kj mol -1 S 98 = [ 130.7 + ½(05.1)] - [ 188.3 ] = 44.95 J mol -1 K -1. o convert H 98 and S 98 to the higher temerature, we need C. C = [ 8.8 + ½(9.4)] - [33.6] = 9.9 J mol -1 K -1. At 000 C, 73 K. H 73 = H 98 + C (73-98) = 41.8 + ( 9.9 x 10-3 )(1975) = 61.35 kj mol -1. S 73 = S 98 + C ln (73 / 98) = 44.4 + (9.9) ln(7.68) = 65.06 J mol -1 K -1. G 73 = H 73 - S 73 = 61.35 - (73)(65.06 x 10-3 ) kj mol -1. = + 113.46 kj mol -1. K = ex (- G /R) = ex (-113.46 x 10 3 / 73 R) =.47 x 10-3 bar If we start with 1 mol of water and the degree of dissociation is α, at equilibrium there will be α mol of H, ½α mol of O and (1-α) mol of H O. 50
K H O = ( )( ) = ( ) H O 1 ( x )( x ) H tot O tot ( x ) H O tot 1 = 1 3 1 ( α) (0.5 α) (1 + 0.5 α) 0. 5α 3 1 = (1 + 0.5 α) (1 + 0.5 α) (1- α) (1- α) (1 + 0.5 α) 1 If we now make the assumtion that α is small, then α <<1, (1-α) (1 + 0.5 α) = 1 so that 3 1 3 3 0. 5 α K = = 0.7071 (0.1) α =. 47 x 10 1 Hence α 0.00653 or 0.65%. Note: As well as the mathematical aroximation in the final ste, the result is an aroximation in that the mean heat caacities used are unlikely to give an accurate result over such a large temerature range. 51
Electrochemical cells An imortant class of reaction not so far discussed is that of electron transfer or redox reactions. Essentially the chemistry of electron transfer reactions If ure Zn metal is laced in a solution of CuSO 4 (aq), the zinc raidly coats with a brown deosit of Cu metal. he chemical reaction taking lace is Zn (s) + Cu + (aq) Cu (s) + Zn + (aq) his can be slit into two half-reactions Zn (s) - e Zn + (aq) Oxidation Cu + (aq) + e Cu (s) Reduction Reactions involving electron gain - reductions Reactions involving electron loss - oxidations. 5
here must be a net balance of electrons and so both must take lace concurrently, hence the name redox reaction. If we kee the two half reactions searate (but linked together) electrons can be made to flow around an external circuit. he otential difference between the electrodes rovides the driving force for the electron flow. If we use a voltmeter with a very high resistance, no electrons can flow so we just measure the e.m.f. (or otential or voltage) of the cell. Under these conditions, the cell acts reversibly. here are many tyes of half-cell and electrochemical reaction we will only consider a short introduction to the toic from a thermodynamic oint of view. Read.. 53
Standard electrode otentials he voltage generated clearly deends on the nature of the two half cells. Again, we need to fix a standard. We set to zero, the otential for a half-cell involving the reaction H + (aq) + e ½H (g) at 98 K, 1 bar of hydrogen ressure and 1 mol dm -3 concentration of H +. Note: Again the definition of 1 mol dm -3 as the standard concentration is an aroximation and it should be defined as unit activity. Construction of a cell in which one half consists of this Standard Hydrogen Electrode, S.H.E. allows the otential for any other half-cell to be determined. Extensive lists of these values have been ublished. (For examle, see your Inorganic textbook and Atkins) Electrode otentials measured under these standard conditions are given the symbol E while E reresents the voltage generated when the cell oerates under conditions other than those secified above. 54
he e.m.f. of the cell, E, tells us about the tendency for electrons to flow around the circuit - the same as the tendency for the redox reaction to occur - so it must be related to the reaction free energy e.g. higher E = more tendency to haen Higher E will oxidise lower E (or lower E will reduce higher) For the two half-reactions Zn (s) - e Zn + (aq) Cu + (aq) + e Cu (s) E = +0.76 V E = +0.34V For the overall reaction, we add these together Zn (s) + Cu + (aq) Cu (s) + Zn + (aq) E = (+0.76+ 0.34)= 1.11 V Note: In texts, the E values are always listed as standard reduction otentials so that you will see Zn + (aq) + e Zn (aq) E = - 0.76 V. For the oosite reaction, we switch the sign of the otential. he same rinciles hold if conditions other than standard are used. 55
Electrode Potentials and free energy changes he tendency for a reaction to haen is correlated by G so there must then be a direct connection between E and G. If n electrons are transferred in a reaction, then the total charge er mole, Q, is ni. I is the Faraday constant which is the total charge on 1 mole of electrons - 96485 C mol -1. When a charge Q moves through a otential difference E, the work done is given by w elec Q Q final = E dq initial If the otential difference, or voltage, is constant, the work done in transferring a charge ni is simly w elec = - n I E At constant,, this work is done as a result of the lowering of the system free energy hus, we can write G = - n I E In the articular case of standard conditions and all comonents in their standard states, then G = - n I E Since E and G, have oosite sign, if a cell has a ositive cell otential the reaction taking lace will be sontaneous. 56
hermodynamic measurements by electrochemistry Electrochemistry allows us to make thermodynamic measurements on systems which would be difficult to erform calorimetry e.g. biochemical systems. We already know that dg d = S or P S o d G = d o S o = ni o de d so that, as long as we are oerating at constant, the temerature variation of the cell voltage allows us to measure the entroy change for the reaction. Once G (= -n I E ) and S are known, H can be calculated. 57
An electrochemical cell with the following reaction generates a standard electrode otential of 1.055 V at 5 C and 1.015 V at 0 C. Calculate the standard enthaly, entroy and free energy changes for the reaction at 5 C. Zn (s) + AgCl (s) ZnCl (aq) + Ag (s) he standard free energy change at 98 K follows directly from G = - nie = -() (96485) (1.055) = -03.58 kj mol -1 We will make the aroximation that the voltage is a linear function of temerature. he standard entroy arises from the temerature deendence o S = I de 1055. 1015. n = ( ) (96485) = 308. 75 J K mol d 98. 15 7315. o -1-1 H can then be found H = G + S = (-03.58 x 10 3 ) + (98.15)(308.75) = - 111.53 kj mol -1 58
Concentration deendence of electrode otentials We saw earlier that the free energy change for reactions in solution can be described by G = G + R ln Q Substituting for the free energy in terms of electrode otentials, n I E = - n I E + R ln Q or his is known as the Nernst equation. E o R = E n I ln Q For examle, what is the otential of a cell for the following reaction with a coer sulhate concentration of 0.05 mol dm -3 and zinc sulhate of 0.1 mol dm -3? We know from above that E = 1.1 V for the reaction Zn (s) + Cu + (aq) Cu (s) + Zn + (aq) Alying the Nernst equation, remembering that the activity of a solid is 1, E E = E o = 1.1- R ln Q ni = E R a ln ni a 1 1 (8.314JK mol )(98K) 1 ()(96485Cmol ) o Cu Cu ln a + a Zn + Zn (1)(0.1) (0.005)(1) Hence, E = 1.061 V 59
When the cell reaction reaches equilibrium, Q = K eqm and electron flow stos and E = 0. hus, it follows that R E o = ln K eq n I his rovides a very convenient method for estimating the equilibrium constants and standard free energy changes since E values are available for a very wide range of reactions. Alications of electrochemical cells Many and various: Batteries, energy storage, thermodynamic measurements, study of reactions such as corrosion, analytical chemistry If we can arrange an electrochemical cell such that we constant activity of all comonents excet H + then we can write E cell = (const.) (const.) ln (const.) [H + ] In other words, the otential deends only on the log of [H + ]. his forms the basis of a H meter (see Atkins for details). H = - log 10 [H + ] Similar ion selective electrodes are available for a wide range of cations and anions. 60
Where could we go from here We have discussed a small number of alications of equilibrium constants and electrode otentials. he aim was to introduce the basic rinciles. here are many other areas of chemistry where related rinciles can be alied and you should be aware of these from reading textbooks. hese include: acid-base equilibria, solubility of salts and other comounds. Atkins (Elements III) Chater 9 Exercises: 10, 1, 4, 3, 9, 30 61