1. CHEMICAL EQUILIBRIUM

2. THE PROPERTIES OFTHE EQUILIBRIUM 1.Within the natural events, it is the deal point of having the tendency of longing to have either MINIMUM ENTALPHY or MAXIMUM ENTROPY. That is, the equilibrium is the mid-point of two opposite tendencies 2.THE EQUILIBRIUM is DYNAMIC.After being reached to the equilibrium (Temperature and Pressure are held constant and There are no added substances), the concentrations of the reactants and products do not change with time since the forward and reverse reactions rate remain the same. For this reason, we call it dynamic and since it moves in two directions it is called as reversible. 3.Le CHATELİER PRINCIPLE (ACTION ≡ REACTION): Once a reaction has been reached into the equilibrium, it stays at the equlibrium, unless the conditions are changed. The change in temperature or pressure is effective on the equlibrium. When the net reaction proceeds to one side(left or right) by means of explicit factors, the equilibrium can be reached at a specific time again.(As soon as possible it is reached.)

3. THE CONDITION OF DYNAMIC EQUILIBIRIUM 1.When a liquid vaporizes into a closed container, after a time vapor molecules condense to the liquid state at the same rate at which liquid molecules vaporize. Even though molecules continiue to pass back and forth between liquid and vapor, the pressure exerted by the vapor remains constant with time. The vapor pressure of a liquid is a property associated with an equilibrium condition. 2.When a solute dissolves in a solvent, a time is reached when the rate of dissolving is just matched by the rate at which dissolved solute precipitates. The solubility of a solute is a property associated with an equilibrium condition.

4. THE CONDITION OF DYNAMIC EQUILIBIRIUM 3. When an aqueous solution of Iodine is shaken with pure Carbontetrachloride, iodine molecules move into CCl 4 layer. When I 2 molecules pass between two liquids(water and CCl 4 ) at equal rates, the concentration of I 2 in each layer remains constant. The ratio of concentrations of a solute in two immiscible solvents is called the distribution coefficient. The distribution coefficient of a solute between two immiscible solvents is a property associated with an equilibrium condition.

5. FACTORS AFFECTING EQUILIBRIUM The environmental conditions: The change in P,V,n,,T have an effect on it. They create a new equilibirium point. ∑n RT ≠ ∑n PR EFFECTIVE 1.PRESSURE EFFECT: ∑n RT ≠ ∑n PR Δp is EFFECTIVE. ∑n RT = ∑n PR ise Δp is not effective, The pressure leads the net reaction to move towards the side with less amount of substance in moles. 2.EFFECT OF VOLUME: 3.MOLARITY: 4.TEMPERATURE EFFECT: DYNAMIC EQUILIBRIUM

6. 1) H 2 O (s) ⇄ H 2 O (g ) 2) NaCl (s) ⇄ NaCl(aq) 3) I 2 (H 2 O) ⇄ I 2 (CCl 4 ) 4) CO (g) + 2 H 2(g) ⇄ CH 3 OH (g) 5) HA (aq) ⇄ H + + A - EXAMPLES FOR THE DYNAMIC EQUILIBRIUM

7. THE EQUILIBRIUM CONSTANT EXPRESSION CO(g) + 2 H 2 (g) ⇄ CH 3 OH(g) kfkf krkr In the equilibrium f = r k f [CO][H 2 ] 2 = k r [CH 3 OH] [CH 3 OH] [CO][H 2 ] 2 = kfkf krkr Kc = Forward: CO (g) + 2 H 2 (g) → CH 3 OH (g) Reverse: CH 3 OH(g) → CO(g) + 2 H 2 (g) f = k f [CO][H 2 ] 2 r = k r [CH 3 OH] kfkf krkr

8. CO (g) + 2H 2 (g)  CH 3 OH (g) THREE APPROACHES TO EQUILIBRIUM CO(g) H 2 (g) CH 3 OH(g) Experiment 1 initial amounts,mol1,000 0,000 equil. amounts, mol0,9110,8220,0892 equil. concn, mol/L0,09110,08220,00892 Experiment 2 initial amounts,mol0,000 1,000 equil. amounts, mol0,7531,5060,247 equil. concn, mol/L0,07530,1510,0247 Experiment 3 initial amounts,mol1,000 equil. amounts, mol1,3801,7600,620 equil. concn, mol/L0,1380,1760,062

9. CO (g) + 2H 2 (g)  CH 3 OH (g) THREE APPROACHES TO EQUILIBRIUM [CH 3 OH] [CO][H 2 ] 2 K c(1) =14.5 M -2 K c(2) =14.4 M -2 K c(3) =14.5 M -2 [CH 3 OH] [CO][H 2 ] K c = [CH 3 OH] [CO].2[H 2 ] 0.596 M -1 1.09 M -1 1.28 M -1 1.19 M -1 2.17 M -1 2.55 M -1 CO(g) + 2 H 2 (g) ⇄ CH 3 OH(g) k1k1 k -1

10. Relationship of K c to the Balanced Chemical Equation a A + b B …. ⇄ g G + h H …. The Equi. constant = K c = [G] g [H] h [A] a [B] b …. We must always make certain that the expression for K c matches the corresponding balanced equation. In doing so, reverseinvert When we reverse an equation, we invert the value of K c multiply powerWhen we multiply the coefficients in a balanced equation by a common factor(2,3…), we raise the equilibrium constant to the corresponding power(2,3…) divide rootWhen we divide the coefficients in a balanced equation by a common factor, we take the corresponding root of the equilibrium constant(i.e. Square root, cube root,…) A) N 2 (g) + ½O 2 ⇄ N 2 O(g) K cA = 2.7 x 10 +18 = B)N 2 (g) + O 2 ⇄ 2 NO(g) K cB = 4.7 x 10 -31 C) N 2 O(g) + ½O 2 ⇄ 2 NO(g)K c = ? Kc=Kc= [N 2 O][O 2 ] ½ [NO] 2 = [N 2 ][O 2 ] ½ [N 2 O][N 2 ][O 2 ] [NO] 2 =K cB K cA = 1.7 x 10 -13 [N 2 ][O 2 ] [NO] 2 = [N 2 ][O 2 ] ½ [N 2 O]

11. EXAMPLES 1. CO (g) + 2 H 2 (g) ⇄ CH 3 OH (g) K c = 14,5 In the above reaction [CO] = 1,03 M and [CH 3 OH] = 1,56 M ; [H 2 ] = ? [H 2 ] = 0,322 M. 2. The synthesis reaction of NH 3 (g), and its K o c value is as follows: N 2 (g) + 3 H 2 (g) ⇄ 2 NH 3 (g) K o c =3,5x 10 8 NH 3 (g) ⇄ ½ N 2 (g) + 3/2 H 2 (g) K o c =? Solution: 2 NH 3 (g) ⇄ N 2 (g) + 3 H 2 (g) K’ c = 1/3,5 x 10 8 =2,8 x 10 -9 In order to obtain the demanded equation, the coefficients are divived by 2. Since, the square root of K’ c is calculated.. K’’ c = 5,3 x10 -5 [CH 3 OH] [CO] [H 2 ] 2 1,56 1,03 x [H 2 ] 2 K c = = = 14,5 

12. Equilibria Involving Gases: The Equilibrium Constant, K P Mixtures of gases are as much solutions as are mixtures with a liquid solvent. partial pressures ideal gasesWe derive an equilibrium constant expression,K p by using the partial pressures of gases in atm in the equation of ideal gases: 2 SO 2 (g) + O 2 (g) ⇄ 2 SO 3 (g) K c = [SO 2 ] 2 [O 2 ] [SO 3 ] 2 [SO 3 ]= V n SO 3 = RT P SO 3 [SO 2 ]= V n SO 2 = RT P SO 2 [O 2 ] = V nO2nO2 = RT PO2PO2

13. The Equilibrium Constant K P = K c (RT) Δn 2 SO 2 (g) + O 2 (g) ⇄ 2 SO 3 (g) K c = [SO 2 ] 2 [O 2 ] [SO 3 ] RT P SO 3 2 RT P SO 2 RT PO2PO2 = 2 = P SO 3 2 P SO 2 PO2PO2 2 K c = K P (RT)K P = K c (RT) -1 K P = K c (RT) Δngas Δn gas =2-(2+1)=-1

14. Equilibria Involving Pure Liquids and Solids Even though solids and liquids participate in a chemical reaction, they remain pure in a heterogenous mixture, that is the concentrations of pure solids and liquids can not vary: The concentrations terms for pure solids and liquids do not appear in equilibrium constant expressions.The concentrations terms for pure solids and liquids do not appear in equilibrium constant expressions. C (s) + H 2 O (g) ⇄ CO (g) + H 2(g) K c = [H 2 O] [CO][H 2 ] = PH2OPH2O P CO. P H 2 / (RT) 1 C(s) does not appear in the equilibrium equation.

15. Burnt Lime- CaCO 3 (s)  CaO(s) + CO 2 (g) CaCO 3 (s) ⇄ CaO(s) + CO 2 (g) K c = [CO 2 ]K P = P CO 2 (RT)

16. Examples 1. 2 SO 2 (g) + O 2 ( g) ⇄ 2SO 3 (g) K c =2,8 x 10 2 Complete the calculation of Kp =? T=1000K Solution: K p = K c (RT) -1 = 2,8 x 10 2 (0,08206 x 1000) -1 = 3,4 2. : H 2 S (g) + I 2 (k) ⇄ 2 HI (g) + S (k) at 60 o C the partial pressures of the gases in the equilibrium are P HI =0,00365 atm and P H 2 S =0,996 atm. Calculate the K p value of the reaction? Solution: Pure solids do not appear in the equilibrium constant expression K p = = = 1,34 x 10 -5 3 : The steam-iron process is used to generate H 2 (g) mostly for use in hydrogenating oils. Iron metal and steam (H 2 O(g) ) react to produce Fe 3 O 4 (s) and H 2. Write expressions for K c, K p for this reversible reaction. Solution: 3Fe(s) + 4H 2 O(g) ⇄ Fe 3 O 4 (s) + 4H 2 (g) K c =[H 2 ] 4 / [H 2 O] 4 Δn=0 K p = P 4 H 2 / P 4 H 2 O = K c (RT) 0 = K c (P HI ) 2 (P H 2 S ) (3,65 x 10 -3 ) 2 9,96 x 10 -1

17. SIGNIFICANCE OF THE MAGNITUDE OF AN EQUILIBRIUM CONSTANT

18. The Reaction Quotient, Q: Predicting the Direction of A Net Reaction A very large numerical value of K c or K p signifies that the forward reaction goes to completion or very nearly so. A very small numerical value of K c or K p, signifies that the forward reaction does not occur to any significant extent. If the numerical value of K c or Kp in the range from 10 -10 to 10 10, a reaction is most likely to reach a state of equilibrium in which significant quantities of both reactants and products are present.. In some equilibrium calculations it is helpful to determine the direction of net change as a first step: reaction quotient, Q cFor any set of initial concentrations in a reaction we can set up a ratio of concentrations called the reaction quotient, Q c Q c = K c CO(g) + 2 H 2 (g) ⇄ CH 3 OH(g) k1k1 k -1 Q c = [A] t m [B] t n [G] t g [H] t h

19. Example for predicting the Direction of a Net Reaction To increase the yield of H 2 (g) in the reaction of C(s) and H 2 O(g), an additional reaction called « water-gas shift reaction» is generally used. For the water-gas shift reaction at about 1100 K K c =1,00. The following amount of substances are brought together and followed to react at 1100 K: 1,00 mol CO, 1,00 mol H 2 O, 2,00 mol CO 2 and 2,00 mol H 2. Compared to their initial amounts, which of the substances will be present in a greater amount and which, in a lesser amount, when equilibrium is established? CO (g) + H 2 O (g) ⇄ CO 2 (g) + H 2 (g) Solution: Q c = = = 4,00 [CO 2 ] [H 2 ] (2,00/V)(2,00/V) [CO] [H 2 O] (1,00/V)(2,00/V) Because Q c > K c, a net reaction occurs to the left. When equilibrium is established, the amounts of H 2 O and CO will be greater than initially and the amounts of CO 2 and H 2 will be less.

20. ALTERING EQUILIBRIUM CONDITIONS: LE CHÂTELIER PRINCIPLE When an equilibrium system is subjected to a change in temperature, pressure, or concentration of a reacting species, the system reacts in a way that partially offsets the change while reaching a new state of equilibrium. Q = = K c [SO 2 ] 2 [O 2 ] [SO 3 ] 2 Q > K c 2 SO 2 (g) + O 2 (g) ⇄ 2 SO 3 (g) k1k1 k -1 K c = 2,8 x 10 2 1000 K

21. EXAMPLES 1. Predict the effect of adding more H 2 (g) to a constant volume equilibrium mixture of N 2, H 2 and NH 3 N 2 (g) + 3 H 2 (g) ⇄ 2 NH 3 (g) Solution: Solution: The addition of H 2 stimulates a shift of the equilibrium condition to the“right” However, only a portion of the added H 2 is consumed in this reaction. When equilibrium is reestablished, there will be more H 2 then present originally. The amount of NH 3 will be also greater, but the amount of N 2 will be smaller. Some of the original N 2, must be consumed in converting some of the added H 2 to NH 3.

22. 2CO(g) + O 2 (g) ⇄ 2CO 2 (g) Predict the effect of adding of O 2 (g), while the system is in equilibrium ? SolutionSolution :Addition of O 2 (g) stimulates the reaction to the right

23. 2. EXAMPLE : CaCO 3 (s) ⇄ Ca O (s) + CO 2 (g) When equilibrium is established, in which way does the addition of the compounds CaO(s), CaCO 3 (s), CO 2 (g) influence the system ? Explain for each compound the effects briefly SolutionSolution : Addition of CaO(s) and CaCO 3 (s) does not change the equilibrium, however Addition ofCO 2 (g) leads the net reaction to proceed to the left.

24. EFFECT OF CHANGES IN PRESSURE OR VOLUME IN EQUILIBRIUM We can add a gaseous reactant or product to an equilibrium mixture in which we change the equilibrium conditions(either to the left or to the right ). no effectWe can add an inert gas to the constant volume reaction mixture. This has the effect of increasing the total pressure but the partial pressures of the reacting species are all unchanged. As a result the inert gas added in this way has no effect on the equilibrium condition volume.We can change the pressure of a system by changing its volume. K c = [SO 2 ] 2 [O 2 ] [SO 3 ] V n SO 3 2 V n SO 2 V nO2nO2 = 2 = V n SO 3 2 n SO 2 nO2nO2 2

25. EFFECT OF CHANGES IN VOLUME ON EQUILIBRIUM nDnD nCnC d K c = [C] c [D] d [G] g [H] h = V (c+d)-(g+h) nGnG c g nHnH h When the volume of an equilibrium mixture of gases is reduced, a net reaction occurs in the direction producing a smaller number of moles of gases. When the volume is increased, a net reaction occurs in the direction producing a larger number of moles of gases. V-ΔnV-Δn nGnG g nHnH h = nDnD nCnC d c

26. EXAMPLE An equilibrium mixture of gases N 2 (g), H 2 (g) NH 3 (g) is transfered from a 1,50 L flask to a 5,00 L flask. In which direction does a net reaction occur to restore equilibrium? N 2 (g) + 3 H 2 (g) ⇄ 2 NH 3 (g) Solution: When the gaseous mixture is transferred to the larger flask, the partial pressure of each gas and the total pressure drop. Some of the NH 3 decomposes back to N 2 and H 2. A net reaction occurs to the left(reverse direction) in restoring equilibrium. We would come to the same conclusion if we decreased the total pressure instead of increasing the volume.

27. EFFECT OF TEMPERATURE AND CATALYST ON EQUILIBRIUM Effect of Catalyst A catalyst in a reaction mixture speeds up both the forward and reverse reactions. Equilibrium is achieved more rapidly but. –THE EQUILIBRIUM AMOUNTS are UNCHANGED by the catalyst. Raising the temperature of an equilibrium mixture shifts the equilibrium condition in the direction of the ENDOTHERMIC reaction. Lowering the temperature causes a shift in the direction of the EXOTHERMIC reaction Van’t Hoff equation

28. EXAMPLE 2 SO 2 (g) + O 2 (g) ⇄ 2 SO 3 (g) ∆H= -180kJ Will the amount of SO 3 (g) formed from given amounts of SO 2 (g) and O 2 (g) be greater at high or low temperatures? Solution: Raising the temperature favors the endothermic reaction. To favor the exothermic reaction requires that we lower the temperature. Therefore, an equilibrium mixture would have a higher concentration of SO 3 at lower temperatures.

29. Example Dinitrogentetroxide, N 2 O 4 (l), is an important component of rocket fuels. For example as an oxidizer of liquid hydrazine in the Titan rocket. At 25˚C N 2 O 4 exists as a colorless gas, which partially dissociates into NO 2, a red brown gas. The color of an equilibrium mixture of these two gases depends on their relative proportions Equilibrium is established in the reaction N 2 O 4 (g) ⇄ 2 NO 2 (g),the quantities of the two gases present in a 3L vessel are 7,64 g N 2 O 4 and 1,56 g NO 2. What is the value of K c o for this reaction? Solution:For both N 2 O 4 and NO 2,we need the conversions g mol mol/L [N 2 O 4 ] = 7,64 g /92,01 g/molx3L= 0,0277 M [NO 2 ] = 1,56 g / 46,01 g/molx3L=0,0113 M K c = [NO 2 ] 2 / [N 2 O 4 ] = (0,0113) 2 / 0,0277 = 4,61 x 10 -3

30. EXERCISE Equilibrium involving SO 2 (g), O 2 (g) and SO 3 (g) is important in sulfuric acid production. When a 0,0200 mol sample of SO 3 (g) is introduced into an evacuated 1,52 L vessel at 900 K, 0,0142 mol SO 3 is found to be present at equilibrium. What is the value of K p for the decomposition of SO 3 (g) at 900 K? 2 SO 3 (g) ⇄ 2 SO 2 (g) + O 2 (g) Solution:Let us first determine K c and then convert to K p by using the equation 2 SO 3 (g) ⇄ 2 SO 2 (g) + O 2 (g) initial amounts 0,02 mol 0,00 mol 0,00 mol change -0,0058 mol + 0,0058 mol +0,0029 mol equil. amounts 0,0142 mol 0,0058 mol 0,0029 mol equil. conc. 0,0142mol/1,52 L 0,0058mol/1,52 L 0,0029mol/1,52L [SO 3 ]=9,34x10 -3 [SO 2 ]=3,8x10 -3 [O 2 ]=1,9x10 -3 K c =[SO 2 ] 2 *[O 2 ] / [SO 3 ] 2 = 0,0038) 2 *(0,0019) / (0,00934) 2 = 3,1 x 10 -4 K p =Kc(RT) ∆n =3,1 x 10 -4 (0,0821 x 900) (2+1)-2 = 2,3 x 10 -2

31. EXERCISE Ammonium hydrogen sulfide, NH 4 HS(s), (used as a photographic developer) is unstable and dissociates at room temperature.. NH 4 HS (s) ⇄ NH 3 (g) + H 2 S (g) K p (atm) = 0,108 (25 o C de) A sample of NH 4 HS(s), is introduced into an evacuated flask at 25 o C ? What is the total gas pressure at equilibrium? Solution: K p =(P NH 3 )(P H 2 S ) = (P NH 3 )(P NH 3 ) = (P NH 3 ) 2 = 0,108 P NH 3 = 0,329 atm P H 2 S = P NH 3 = 0,329 atm P toplam = P NH 3 + P H 2 S = 0,329 + 0,329 = 0,658 atm

32. EXERCISE An 0,024 mol sample of N 2 O 4 (g) is allowed to come to equilibrium with NO 2 (g) in an 0,372 L flask at 25˚C. Calculate (a) the amount of N 2 O 4 (g) present at equilibrium and (b) the percent dissociation of the N 2 O 4 N 2 O 4 (g) ⇄ 2 NO 2 (g) Kc = 4,61 x 10 -3 (25 o C) Solution: a) χ = the number of moles of N 2 O 4 that dissociate. N 2 O 4 (g) ⇄ 2 NO 2 (g) initial amounts 0,0240 mol 0,00 mol change -χ mol + 2χ mol equil. amounts (0,0240-χ) mol 2χ mol equil. conc. [N 2 O 4 ]= (0,0240-χ)/0,372 [NO 2 ]=2χ/0,372 Kc== 4,61 x 10 -3 =[NO 2 ] 2 /[N 2 O 4 ]= (2χ / 0,372) 2 : (0,0240-χ/0,372)=4χ 2 : 0,372 (0,0240-χ) 4χ 2 = 4,12 x 10 -5 - (1,71 x 10 -3 ) χ  χ 2 + (4,28 x 10 -4 ) χ - 1,03 x 10 -5 =0  χ = 0,003 mol N 2 O 4 The amount of N 2 O 4 at equilibrium =(0,0240-χ) = 0,024-0,003=0,0210 mol N 2 O 4 b) the percent dissociation % of N 2 O 4 = 0,003*100/0,024= 12,5