Thermodynamics of Reactive Systems The Equilibrium Constant

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1 Lecture 27 Thermodynamics of Reactive Systems The Equilibrium Constant A. K. M. B. Rashid rofessor, Department of MME BUET, Dhaka Today s Topics The Equilibrium Constant Free Energy and Equilibrium Constant Temperature Dependency of Equilibrium Constant The Uses of Equilibrium Constant 1

2 The Equilibrium Constant The law of mass action The rate of any chemical reaction is proportional to the active masses of the reacting substances By active masses, it means the activities of the solid or liquid reacting substances, or partial pressures in the case of gaseous substances. If pure solids and liquids are involved in a reaction, their active masses, or activities, can be regarded as unity. mm + nn = xx + yy Rate of forward reaction = k 1. (a M ) m. (a N ) n Rate of reverse reaction = k 2. (a X ) x. (a Y ) y where a j etc. are the activities of element j at any state of the reaction, and k 1 and k 2 are proportionality constants. At equilibrium, the rates of forward and reverse reactions are equal. k 1. = k 2. where a j is the concentration of element at equilibrium. K K = 1 = K 2 where K is the equilibrium constant of the reaction. K a = The suffice e used in the above equation indicates that the activities are of equilibrium values. e 2

3 The Relationship Among Equilibrium Constants K a = For gaseous components, K p = (p X ) x. (p Y ) y (p M ) m. (p N ) n ; K c = (c X ) x. (c Y ) y (c M ) m. (c N ) n p k = partial pressure of component k c k = molar concentration of component k For ideal gas, p k = (n k /V) RT = c k RT K p = K c (RT) Dn Dn = (x +y) (m+n) Free Energy and Equilibrium Constant mm + nn = xx + yy DG = (xm X + ym Y ) (mm M + nm N ) m k = m o k + RT ln a k DG = (xm o X + ym o Y) (mm o M + nm o N) + RT (x ln a X + y ln a Y ) (m ln a M + n ln a N ) DG = DG o + RT ln DG = Chemical Affinity or Gibbs free energy change of a reaction ne 3

4 DG = DG o + RT ln ne Q = Activity Quotient = ne DG = DG o + RT ln Q If the reactants and products are in equilibrium with each other, DG = 0, and 0 = DG o + RT ln DG o = RT ln K a and K a = exp DGo RT e Example 8.6 Calculate the equilibrium constant for the reaction <NiO> + (H 2 ) = <Ni> + (H 2 O) at 750 C from the following data: <Ni> + 1/2(O 2 ) = <NiO> ; DG o = T cal (H 2 ) + 1/2(O 2 ) = (H 2 O) ; DG o = T cal ANSWER (+) <NiO> = <Ni> + 1/2(O 2 ) ; DG o = T cal (H 2 ) + 1/2(O 2 ) = (H 2 O) ; DG o = T cal <NiO> + (H 2 ) = <Ni> + (H 2 O) ; DG o = T cal At 750 C (or, 1023 K), DG o = (1023) = cal/mol K a = exp ( cal/mol) (1.987 cal/deg/mol) x (1023 K) =

DG = DG o + RT ln Q and DG o = RT ln K a DG = RT ln (Q/K a ) vant s Hoff Isotherm Knowing the value of DG o and the activities of reactants and products at any given condition, the free energy change

5 DG = DG o + RT ln Q and DG o = RT ln K a DG = RT ln (Q/K a ) vant s Hoff Isotherm Knowing the value of DG o and the activities of reactants and products at any given condition, the free energy change for the equation at that condition, DG, can be calculated, and the thermodynamics feasibility of the reaction can be predicted. Q-scale Q/K a > 1 DG > 0 roducts decompose Q/K a = 1 Q/K a < 1 K DG = 0 Equilibrium DG < 0 roducts form FIGURE 8.1 Sketch of the scale for the DG and (Q/K a ) showing the spontaneous change of a chemical reaction. The importance of van t Hoff isotherm an be understood by considering the thermodynamic possibility of the reduction of alumina by carbon at 1500 C: Al 2 O 3 + 3C = 2Al + 3CO; DG o 1773 = J DG 1773 = DG o RT ln (a Al ) 2. (p CO ) 3 (a Al2O3 ). (a C) 3 DG 1773 = ln (p CO ) 3 So it is seen that if p CO >1, clearly there is no possibility of a spontaneous change because DG 1773 is positive. However, if the partial pressure of CO gas can be reduced, for example, by placing the system in a vacuum, the value of DG will decrease. p CO, atm DG 1773, J/mol Thus, reaction becomes thermodynamically possible if p CO is lowered sufficiently. 5

6 Example 8.7 Would an atmosphere containing 15 % CO 2, 5 % CO and 80 % N 2 oxidise nickel at 1000 K? Given data: <Ni> + 1/2(O 2 ) = <NiO> ; (CO) + 1/2(O 2 ) = (CO 2 ) ; K 1 = 5.76x10 7 at 1000 K K 2 = 1.68x10 10 at 1000 K ANSWER In the given atmosphere the following oxidation reaction of nickel could occur: <Ni> + (CO 2 ) = <NiO> + (CO) Nickel would be oxidised if DG = RT ln (Q/K) for the above oxidation reaction is negative. In the equation, DG is negative if Q/K is less than one. Now, given that, K = (p CO / p CO2 ) e = K 1 /K 2 = 3.43x10-3 at 1000 K and Q = (p CO / p CO2 ) actual = (0.05/0.15) = at 1000 K Hence, Q/K = Now, since Q/K >1, DG will be positive for the reaction. Thus, the reaction would not occur, that is, nickel would not be oxidised. Example 8.8 Calculate the equilibrium extent of decomposition of nitrogen tetroxide due to the chemical reaction (N 2 O 4 ) = 2(NO 2 ) at 25 C and 1 atm. What will be the equilibrium composition of the gas mixture? Given data: DG o 298 = 4644 J/mol. ANSWER The equilibrium constant for the reaction K = (p NO2 ) 2 / p N2O4 ) = exp (-DG o 298 / RT) (p NO2 ) 2 / p N2O4 = Considering that all the gases are in their ideal conditions, changing the partial pressure to mole fraction, (X NO2 ) 2 / X N2O4 =

7 Let a be the equilibrium extent of the decomposition of N 2 O 4 according to the reaction (N 2 O 4 ) = 2(NO 2 ) No. of moles before reaction No. of moles at equilibrium Equilibrium mole fraction, X N 2 O 4 1 (1-a) X N2O4 = (1-a) / (1+a) NO 2 0 2a X NO2 = 2a / (1+a) Total moles 1 (1+a) (X NO2 ) 2 / X N2O4 = a 2 / (1 a 2 ) = ; a = 0.19 Hence, about 19 % decomposition of N 2 O 4 would occur and the equilibrium composition of the gas mixture would be: X N2O4 = (1-a) / (1+a) = 0.68, or 68% X NO2 = 2a / (1+a) = 0.32, or 32% Temperature Dependence of Equilibrium Constant DG o = R T ln K Upon differentiation with respect to temperature at constant = R ln K RT ln K Now multiplying both side of the equation with T T = RT ln K RT 2 ln K T = DG o RT 2 ln K 7

8 Now, the Gibbs-Helmholtz equation states that DG o = DH o TDS o = DH o + T (DG DG o DH o = T o ) dg = -SdT+ Vd -S = ( G/) Now since ln K T DG o DH o = DG o RT 2 ln K = DG o RT 2 ln K = DH o RT d ln K = DH or, o dt 2 RT 2 This equation is known as van't Hoff isochore, and it shows the relationship between the equilibrium constant and temperature of the reaction. d ln K = DH o RT 2 K grows with increasing temperature in processes attended by the absorption of heat ( H o >0). If the standard reaction is exothermic, H o <0, then K grows with diminishing temperature. dt Assuming that DH 0 is constant over the temperature range, integration of the equation will result lnk = DH o RT + constant So if the standard enthalpy change for a reaction and the equilibrium constant at one particular temperature is known, calculation of equilibrium constant for the reaction at some other temperature can be calculated by using this equation. 8

9 Next Class Lecture 28 Thermodynamics of Reactive Systems Applications of the Ellingham Diagram 9

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