Chemical reaction equilibria

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Nigel Eric Potter
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1 Chemical reaction equilibria

Chemical reaction equilibria in metallurgical processes and the conditions that maintain equilibrium are important to obtain maximum efficiency from production processes

liquid phases iron and slag in the blast furnace consist of solutions of Fe, C, Si, Mn, P and SiO 2, Al 2 O 3, CaO, FeO respectively Flue gases typically contain CO, CO

2 Chemical reaction equilibria in metallurgical processes and the conditions that maintain equilibrium are important to obtain maximum efficiency from production processes For example, steel production takes place in a blast furnace that is aimed to collect liquid iron, slag and flue gases formed as a result of reaction with C and CO he liquid phases iron and slag in the blast furnace consist of solutions of Fe, C, Si, Mn, P and SiO 2, Al 2 O 3, CaO, FeO respectively Flue gases typically contain CO, CO 2 and N 2 as main components Iron oxide is reduced by CO to metallic iron while impurities in liquid iron are subjected to reaction with gaseous oxygen in converting stage

3 Consider a general reaction in equilibrium: aa + bb cc + dd he general criterion for equilibrium under constant and P is G = 0 G = G products G reactants = cg C + dg D ag A bg B he complete differential of G in terms of and P is dg = G P dp + dg = VdP Sd Consider the reaction in a mixture of ideal gases at constant temperature he change in Gibbs free energy of each ideal gas component as a function of its pressure is given as G i P i = V i dg i = RdP i P i G P d

4 dg i = RdP i P i G i = G i o + R ln P i P i o he change in free energy of the system at constant temperature is the sum of the free energy change of its components ng = n i G i d ng = n i dg i + G i dn i ng = Rn i P i dp i + G i dn i Since mole number and pressure of ideal gases are proportional, n i /P i is constant and since the total pressure of the system is constant, dp i = 0 G = G i dn i

5 In the case of system equilibrium G = G i dn i = 0 G = G i o dn i + R ln(p i dn i ) he stoichiometric coefficients a, b, c, d of each component in the ideal gas mixture can be used to represent dn i : cg C o + dg D o ag A o bg B o + R ln P C c + R ln P D d + R ln P A a + R ln P B b = 0 where G o = cg C o + dg D o ag A o bg B o G o + R ln P C c P D d P A a P B b = 0 Absolute Gibbs free energy is computed for gass phases as: G i = G i o + R ln P i

6 he equation for gas phases can be written as G = G o + R ln P C c P D d Q R is called the reaction quotient Q R = K when G = 0 P A a P B b = Go + R ln Q R G = 0 = G o + R ln K G o is readily given in literature for most compounds at SP

7 he relationship between DG o and K at 298 K REVERSE REACION FORWARD REACION Example - Estimate DG o for the decomposition of NO 2 at 25 o C At 25 o C and 1.00 atmosphere pressure, K =4.3x10-13

8 G = R ln Q R R ln K = R ln Q R K G can be calculated for any temperature, since G o = H o S o G = H o C P d S o C P d where C P = a + b + c 2 and C P = a + b + c where a, b, c = a, b, c 2 products a, b, c reactants G o is the free energy change that would accompany the complete conversion of all reactants, initially present in their standard states, to all products in their standard states DG is the free energy change for other temperatures and pressures

9 G = R ln Q R R ln K = R ln Q R K DG has a very large positive or negative value if Q R and K are very different he reaction releases or absorbs a large amount of free energy DG has a very small positive or negative value if Q R and K are close he reaction releases or absorbs a small amount of free energy

10 Example -he equilibrium constant at different temperatures for the following reaction is given: SO 3 (g) = SO 2 (g) + ½ O 2 (g) K= 900K K= 1000K K= 1100K Estimate the enthalpy change of the reaction at 1000K and the equilibrium composition at the same temperature

11 Example - Consider the equilibria in which two salts dissolve in water to form aqueous solutions of ions: NaCl(s) Na+(aq) + Cl-(aq) ΔH soln(nacl)= 3.6 kj/mol, ΔS soln(nacl)= 43.2 J/mol.K AgCl(s) Ag+(aq) + Cl-(aq) ΔH soln(agcl)= 65.7 kj/mol, ΔS soln(nacl)= 34.3 J/mol.K a) Calculate the value of ΔG at 298 K for each of the reactions. How will ΔG for the solution process of NaCl and AgCl change with increasing? What effect should this change have on the solubility of the salts? b) Is the difference between two free energies primarily due to the enthalpy term or the entropy term of the standard free-energy change? c) Use the values of ΔG to calculate the K values for the two salts at 298 K d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part c? e)how will ΔG for the solution process of these salts change with increasing? What effect should this change have on the solubility of the salts?

12 Effect of pressure on equilibrium Although equilibrium constant is independent of pressure, Le Chetelier s principle states that an increase in total pressure at constant temperature will shift the equilibrium in the direction which decreases the number of moles of gaseous species in the system K = P C c d P D P a b A P = (X CP) c (X D P) d B (X A P) a (X B P) b K is not affected by changes in pressure, but consists of two terms; K X and P: K = K X P (c+d a b) Change in pressure may have effect on K X, quotient of mole fractions depending on the values of a, b, c, and d If c+d>a+b, increasing pressure decreases K X, reaction shifts towards reactants c+d=a+b, pressure does not affect K X c+d<a+b, K X is proportional to pressure, reaction shifts towards products with increasing K X

13 Effect of temperature on equilibrium At equilibrium, G o = R ln K G o = H o + R ln K = H o ln K = Ho R 2 Go P R ln K P ln K 1 = Ho R Van t Hoff equation For the case of H o > 0, temperature increase shifts the reaction towards products For the case of H o < 0, temperature increase shifts the reaction towards reactants ln K exothermic Slope>0 1/ Slope<0 endothermic

14 ln K 1 = Ho R

15 Recall that G = H S G o = H o + Go Since Go P P = S, Multiplying both sides by d and dividing by 2, H o d 2 = Go d 2 G o d 2 = Ho d 2 + Go 2 dgo 2 = d Go, H o 2 P = d G o d Gibbs- Helmholtz Eqn Example Determine the heat exchange between system and surroundings for the following reaction in order to keep the temperature of the system constant at 1300 K P 4 g 2P 2 g G o = ln 50.1

16 Oxygen pressure dependence of spontaneity of oxidation reactions he spontaneity of any process at constant and P is dependent on the change in the Gibbs free energy of the system: G can be calculated for any temperature since G = G o + R ln Q G o = H o S o G = H o C P d S o C P d where C P = a + b + c 2 and C P = a + b + c 2 where a, b, c = a, b, c products a, b, c reactants G = H o a + b + c 2 d S o a + b + c d Plotting the G o values of similar oxidation reactions as a function of and comparing their relative reactivities would be useful for engineering complex systems like furnace charge, if it was possible to express G o of any reaction by a simple 2-term fit such as G o = A + B

17 he following grouping lead to a condensed representation of G o which can further be simplified G = H o a + b2 2 c S o a ln + b c Replacement of the upper and the lower limits yields G = 0 = G o I o + I 1 a ln b 2 2 c 2 where I o = H o 298 a298 + b c 298 I 1 = a S o a ln b298 c abulated thermochemical data such as H o 298, S o 298, C P for a specific reaction are replaced into the general equation for G o to obtain the variation of the spontaneity with temperature Alternatively experimental variation of G o with can be calculated from the measured oxygen partial pressure P O2 (eqm) that is in equilibrium with a metal and metal oxide using equation: G o = R ln P O2 (eqm)

18 Ellingham diagram

19 Example - Will the reaction 4Cu(l) + O 2 (g) = 2Cu 2 O(s) go spontaneously to the right or to the left at 1500K when oxygen pressure is 0.01 atm? Cu(s) S 298 =33.36 J/molK, C p = J/molK ΔH m = J/mole at 1356K Cu(l) C p = J/molK Cu 2 O(s) H 298 = J/mol S 298 =93.14 J/molK, C p =83.6 J/molK O 2 (g) S 298 = J/molK, C p =33.44 J/molK

20 Determining the composition of reaction system under equilibrium aa(g) + bb(g) cc(g) + dd(g) Consider the reacting A, B to produce C and D K = P C c P D d P A a P B b he partial pressures of the components are expressed as a function of the total P: P A = n A. P n A + n B + n C + n D where n A is the mole number of A under equilibrium Equilibrium constant can be represented as K = n C c n D d n A a n B b P n A + n B + n C + n D c+d (a+b)

21 Suppose the reaction reaches equilibrium after a while and x moles of A is converted to products hen n A =Moles of unreacted A = 1 x a n B = Moles of unreacted B = 1 x b n C = Moles of formed C = x. c n D = Moles of formed D = x. d and K = (x. c) c (x. d) d a ax a (b bx) b P 1 x a + b + x c + d c+d (a+b) If equilibrium temperature and the standard free energy change at that temperature are given, the fraction x can be conveniently determined since G = G o + R eqm ln K = 0 G o = R eqm ln (x. c) c (x. d) d a ax a (b bx) b P 1 x a + b + x c + d c+d (a+b)

22 Example Determine the equilibrium composition of the system when 1 mole of P 4 reacts to form P 2 at 1300 K G o = ln 50.1 P 4 g 2P 2 g G o = R eqm ln (x. c) c (x. d) d a ax a (b bx) b P 1 x a + b + x c + d c+d (a+b)

4. CHEMICAL REACTION EQUILIBRIA

4. CHEMICAL REACTION EQUILIBRIA 4. CHEMICAL REACION EQUILIBRIA 4.1. Introduction In materials processing or materials in service, chemical reactions take place. For example, in the production of steel, liquid iron containing impurities