Thermodynamic of equilibria

Equilibrium constant - Van't Hoff reaction isotherm - Van't Hoff reaction Isochore - Law of mass action - Le - Chatelier Principle - Thermodynamic interpretation

1. Dr.P.GOVINDARAJ   Associate Professor & Head , Department of Chemistry SAIVA BHANU KSHATRIYA COLLEGE ARUPPUKOTTAI - 626101 Virudhunagar District, Tamil Nadu, India THERMODYNAMICS OF EQUILIBRIA

2. THERMODYNAMICS OF EQUILIBRIA For an equilibrium reaction A + B ⇌ C + D K 1 K 2 The rate of forward reactio n r 1 = k 1 [A][B] The rate of backward reaction r 2 = k 2 [C][ D ] Since at equilibrium r 1 = r 2 k 1 [A][B] = k 2 [C][D] Equilibrium constant

3. =   K C =   Where K C & K P are known as equilibrium constant and P C , P D , P A , P B are partial pressure of product and reactant THERMODYNAMICS OF EQUILIBRIA Rearranging the above equation In terms of partial pressure of the reactant and product the above equation becomes K P =   i.e., The equilibrium constant is the ratio between the concentrations (or) partial pressures of the products and reactants of an equilibrium reaction

4. Thermodynamics interpretation of law of mass action Consider two equilibrium in the box I and box II in which components A, B, C and D are in equilibrium Where ‘C’ terms represent concentration and ‘p’ terms represent Partial pressure THERMODYNAMICS OF EQUILIBRIA A + B ⇌ C + D Statement Law of mass action state that the ratio of the concentrations (or) partial pressures of the products and reactants for an equilibrium reaction is constant

5. Let one mole of ‘A’ is removed from box I by moving the piston infinitesimally slowly outwards and the work done ( W 1 ) for the same in according to first law of thermodynamics is W 1 = RT ln ------(1)   Similarly the work done (W 2 ) in transferring one mole of ‘B’ from box I to box II is W 2 = RT ln ------(2)   Since the concentration of ‘A’ and ‘B’ remains constant in equilibrium, proportionate amounts of ‘C’ and ‘D’ will react to form ‘A’ and ‘B’ in order to maintain the equilibrium So, one mole of ‘C’ is removed from box II to box I, then the work done (W 3 ) is W 3 = RT ln ------(3)   THERMODYNAMICS OF EQUILIBRIA

6. Similarly, the work done (W 4 ) in transferring one mole of ‘D’ from box II to box I is W 4 = RT ln ------(4)   Since this process is an isothermal cyclic process, the total amount of work done is zero i.e., W 1 + W 2 + W 3 + W 4 = 0 i.e., RT ln + RT ln + RT ln + RT ln = 0   ln + ln = ln + ln   ln = ln   THERMODYNAMICS OF EQUILIBRIA

7. =   = ---------(1)   In terms of concentration terms the equation (1) becomes = ---------(2)   Equation (1) and (2) are mathematical statements for law of mass action THERMODYNAMICS OF EQUILIBRIA On removing the ‘ ln ’ terms in both side of the above equation, we get On rearranging the above equation, we get i.e., According to Law of mass action the ratio of the concentrations (or) partial pressures of the products and reactants for an equilibrium reaction is constant

8. Relation between Equilibrium constant & Free energy Consider a reversible reaction a A + b B ⇌ l L + m M ∆G = (G) products – (G) reactants -------(1) Since [ (G) T,P,N = n 1 μ 1 + n 2 μ 2 ….] , equation (1) becomes ∆G = (l μ L + m μ M ) – (a μ A + b μ B ) -------(2) w here μ A , μ B , μ L , μ M are chemical potential of A, B , L and M We know that the chemical potential for a ‘ i ’ th gas in a gaseous mixture is μ i = + RT ln p i -------(3)   THERMODYNAMICS OF EQUILIBRIA Change in free energy for the reversible reaction is

9. Substituting chemical potential of A, B, L, and M in equation (2) ∆G = {l ( + RT ln p L ) + m( + RT ln p M ) } – {a ( + RT ln p A ) + b( + RT ln p B ) }   Rearranging , we get ∆G = {(l + m ) – (a + b ) } + RT ln   ∆G = ∆ G + RT ln J -------(4) where J =   At equilibrium , ∆G = 0 and J becomes equilibrium constant K P then the equation (4,) becomes 0 = ∆ G + RT ln K P ∆G = - RT ln K P -------(5) Equation (5) is the relation between equilibrium constant and free energy and also called as Van’t Hoff reaction isotherm THERMODYNAMICS OF EQUILIBRIA

10. Consider an equilibrium reaction Derivation of Van’t Hoff reaction isotherm A + B ⇌ C+ D ∆G = (G) product – (G) reactant ∆G = (G C + G D ) – (G A + G B ) -------(1) ∆G = ( μ C + μ D ) – ( μ A + μ B ) -------(2) w here μ A , μ B , μ C , μ D are the chemical potential of A, B, C and D Since μ = μ + RT ln p, the equation (2) becomes ∆G = ( + RT ln p C + + RT ln p D ) – ( + RT ln p A + + RT ln p B )   ∆G = ( + ) – ( + ) + RT ln p C + RT ln p D – RT ln p A – RT ln p B   THERMODYNAMICS OF EQUILIBRIA Change in free energy for the equilibrium reaction is Since [ (G) T,P,N = n 1 μ 1 + n 2 μ 2 ….] , equation (1) becomes

11. ∆G = ( G ) products – ( G ) reactants + RT ln   ∆G = ∆G + RT ln --------(3)   At equilibrium , ∆G = 0, (Equilibrium constant), equation (3) becomes   ∆G + RT ln K P = 0 ∆G = - RT ln K P --------( 4 ) Substituting (4) in (3) , we get ∆G = - RT ln K P + RT ln   -∆G = RT ln K P - RT ln --------( 5 )   THERMODYNAMICS OF EQUILIBRIA Since [(G ) T,P,N = n 1 + n 2 ….] , equation (2) becomes  

12. For general reaction n 1 A + n 2 B + …. ⇌ n 3 C + n 4 D …. Equation (4) becomes -∆G = RT ln K P - RT ln   -∆G = RT ln K P – RT --------(6)   E quation (4) & (6) are known as Van’t Hoff reaction isotherm THERMODYNAMICS OF EQUILIBRIA ∆G = - RT ln K P & -∆G = RT ln K P – RT  

13. T he Van’t Hoff reaction isotherm is Derivation of Van’t Hoff reaction isochore -∆G = RT ln K P – RT -------(1)   In terms of concentration, equation (1) becomes -∆G = RT ln K c – RT -------(2)   At constant volume equation (2) modifies to - ∆A = RT ln K c – RT   Differentiating w.r.t temperature at constant volume V = RT + R ln K c – ) – -------(3)   THERMODYNAMICS OF EQUILIBRIA

14. Since concentration ‘C’ is not a function of temperature . The term = 0, equation (3) becomes   V = RT + R ln K c –   - = RT ) + RT ln K c –   - T V = RT 2 ) + RT ln K c – -------(4)   - T V = RT 2 ) – -------(5)   Since – , equation (4) becomes   THERMODYNAMICS OF EQUILIBRIA

15. A = E + T V   Gibb’s Helmholtz equation is - T V A --------(6)   Comparing equation (5) and (6) we get Since E is the heat absorbed (q v ) at constant volume i.e., E = q v , equation(7) becomes   A = RT 2 ) –   = RT 2 )   = ) --------(7)   THERMODYNAMICS OF EQUILIBRIA = )   d ( ln K c ) = dT -------( 8 )  

16. Integrate the equation (7) with in the limits =     l n   2.303 log --------(9)   Equation (9) & (8) are Van’t Hoff reaction isochore THERMODYNAMICS OF EQUILIBRIA where Kc 1 and Kc 2 are equilibrium constants at the temperatures T 1 and T 2

17. Thermodynamic interpretation for Le - Chatelier’s Principle Le – C hatelier’s Principle It is stated that when a change is applied on an equilibrium system , the equilibrium s hifts in such a direction in order to nullify the effect of change. Example Increase of external pressure will cause the equilibrium to shift in the direction which will bring about a lowering of pressure. PCl 5 ( g) ⇌ PCl 3 ( g) + Cl 2 ( g) i.e., Increase of pressure on the above equilibrium system causes the equilibrium to shift from right to left THERMODYNAMICS OF EQUILIBRIA

18. Consider an equilibrium reaction w here ‘ε’ is the extent of the reaction Thermodynamic treatment for Le - Chatelier’s Principle aA + bB ⇌ mM + nN ∆G of the reaction is the function of T, P and ε i.e., ∆G = f ( T, P, ε ) ------(1) d(∆G) = P, ε dT + T, ε dP + T , P dε ------( 2 )   Put P,T = ∆ G in equation (2) we get   d(∆G) = T,P P, ε dT + T, P T,ε dP + T,P T,P dε   THERMODYNAMICS OF EQUILIBRIA On doing partial differentiation, equation (1) becomes

19. d(∆G) = P, ε T,P dT + T, ε T,P dP + T,P dε -------(3)   Substitute P = - S and T = V in equation (3), we get   d(∆G) = - T,P dT + T,P dP + T,P dε -------( 4 )   d(∆G) = - ∆S dT + ∆V dP + T,P dε -------( 5 )   Substitute - T,P = ∆S and T,P = ∆V in equation (4), we get   At equilibrium ∆G = 0 and ∆S = as per th e equation ∆G = ∆ H - T∆S, so that the equation (5) becomes   - dT + ∆V dP + dε = 0 -------(6)   THERMODYNAMICS OF EQUILIBRIA w here T,P =  

20. At constant Temperature, dT = 0 and the equation (6) becomes ∆V dP + dε = 0 -------(7)   In the above equilibrium reaction ∆V is positive i.e ., [V product > V reactant ] , then the RHS in equation (8) would be positive, so that } T would be negative   i.e., Increase in pressure ( dP > 0 ) in the above equilibrium will de crease ε ( dε < ), so that equilibrium shifted towards the reactant side . On the other hand decrease in pressure ( dP < ) will in crease ε ( dε > ), so that the equilibrium shifted towards the product side THERMODYNAMICS OF EQUILIBRIA On rearranging the equation (7), we get ∆V = - [ ] T -------( 8 )  

21. Conclusion An increase in pressure would shift the equilibrium towards the low volume side of the reaction (i.e., reactant side in the above equilibrium), where as a decrease in pressure would shift the equilibrium towards the high volume side(i.e., product side in the above equilibrium). This is one of the statements of Le - Chatelier’s Principle THERMODYNAMICS OF EQUILIBRIA

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