Chapter 6 Chemical equilibrium Classification of reaction Chemical reactions are either homogeneous or heterogeneous. In the former case only one phase is present e.g: In the gas phase: H2(g) + I2 → 2 HI(g) In the liquid phase: CH3COCl + CH3OH → CH3COOCH3 + HCI Acetyl Chloride methyl alcohol methylacetate In heterogeneous reactions the mixture is not uniform throughout. e.g: Alumina C2H5OH(l) ⇔ ethanol Chemical equilibrium C2H4(g) + H2O ethane Page 148 On the other hand, when a chemical reaction proceeds in one direction, i.e: from reactant to products, the reaction is known as irreversible reaction or complete reaction. A + B → AB On the contrary, reactions that proceed either in the forward or in the backward (reverse) directions are known as reversible reaction and the equation for the reaction can be written. N2(g) + 3 H2(g) 2 NH3(g) The double arrow () indicates that the equation can be read in either direction. All reversible processes tend to attain a state of equilibrium. For a reversible chemical reaction, an equilibrium state is attained when the rate at which the direct reaction is proceeding equals the rate at which the reverse reaction is proceeding. Chemical equilibrium Page 149 • The Law of Mass Action: Consider a hypothetical reversible reaction A2(g) + B2(g) 2 AB(g) The rate of the forward reaction is rate f = Kf. [A2] [B2] and the rate of the reverse reaction is rate r = Kr [AB]2 At equilibrium ratef = rater Kf [A2] [B2] = Kf [AB] 2 Kf [AB] 2 K r [A 2 ][B2 ] Kf = K which is called the equilibrium constant Kr [AB] 2 Therefore, K = [A 2 ][B2 ] The numerical value of K varies with temperature. It does not vary with changes in the amounts of substances used to establish the equilibrium, changes in pressure, or the presence of catalyst. Chemical equilibrium Page 150 In general, for any reversible reaction aA + bB eE + fF at constant temperature at equilibrium, [E]e [F]f Kc = [A] a [B]b where the quantities written within brackets denote equilibrium molar concentrations (mol/liter). If the general equation were written in reverse form: eE +fF aA + bB the expression for the equilibrium constant (K\) would be: K\ = [A] a [B]b [E] e [F]f Notice that K' is the reciprocal of Kc i.e K\ = 1 Kc * If the law of chemical equilibrium hold for reactions that occur by mechanisms of more than one step 1 : 2 NO2CI(g) 2 NO2(g) + CI2(g) [NO 2 ]2 [Cl2 ] Kc = [NO 2Cl] 2 The reaction is believed to occur by means of a mechanism consisting of two steps: Chemical equilibrium Page 151 1. NO2CI NO2 + Cl 2. NO2CI NO2 + CI2 therefore; and K1 = K2 = KC [NO 2 ] [Cl] [NO 2Cl] [NO 2 ] [Cl2 ] [NO 2 ] [Cl] [NO 2 ] [Cl] [NO 2 ] [Cl2 ] [NO 2 ]2 [Cl2 ] = K1 K2 = [NO 2Cl] [NO 2 ] [Cl] [NO 2Cl] 2 In this case, the equilibrium constant for the overall change is the product of the equilibrium constants of each the steps: KC = K1 K2 For reactions involving gases, the partial pressures of the reactants and products are proportional to their molar concentrations. The equilibrium constant expression for these reactions can therefore be written using partial pressures instead of concentration. For example, N2(g) + 3 H2(g) 2 NH3(g) Chemical equilibrium Page 152 2 pNH3 Kp = PN2 .PH23 We will use the symbol, Kp, to donate equilibrium constants derived from partial pressure and kc to indicate equilibrium constants having molar concentrations in the mass action expression (Kc ≠ Kp) [NH 3 ]2 Kc = [N2 ] [H2 ]3 2 pNH3 Kp = PN2 .PH23 If Kc is a larger number, the forward reaction is fairly complete. If Kc is a small number, the reverse reaction is fairly complete. The Relationship Between Kp and KC: For the general equation, aA + bB eE + fF PEe . PFf Kp = a PA . PBb [E]e [F]f Kc = [A] a [B]b n Concentration has the units mol/L V. Assuming ideal gas behavior, we can use the gas law, PV= nRT Chemical equilibrium Page 153 to obtain the concentration of a gas X [X] = nX P X V RT Where Px is its partial pressure Px = [X] RT [E] e (RT) e [F]f (RT) f PEe . PFf Kp = a PA . PBb [A] a (RT) a [B]b (RT) b [E] e [F]f (RT) (e f )(ab) Kp = a b [A] [B] Kp = Kc. (RT)n(g) Where n(g) is the change in the number of moles of gas when going from reactants to products. ng = (number of moles of gaseous products) -(number of moles of gaseous reactants) Example 6.1: For the reaction N2O4(g) 2 NO2(g) The concentrations of the substances present in an equilibrium mixture at 25°C are [N2O4] = 4.27 x 10-2 mol/L [NO2] = 1.41 x 10-2 mol/L Chemical equilibrium Page 154 what is the value of Kc for this temperature. Solution: [NO 2 ]2 (1.41x10 2 mol/l) 2 4.66x103 mol/l Kc = 2 [N2O4 ] (4.27X10 mol/l) Example 6.2: At 500 K. 1.0 mol of ONCI(g) is introduced into a one - liter container. At equilibrium the ONCI(g) is 9.0% dissociated: 2 ONCI(g) 2 NO(g) + CI2(g) Calculate the value of Kc for equilibrium at 500 K. Solution: The concentration of ONCI(g) is 1 mol/L since ONCl is 9.0% dissociated, Number of moles dissociated = 9 [X] = x1.0 mol 0.09 mol ONCI 100 The concentration of ONCI at equilibrium, therefore, is [ONCI] = 1.0 mol/L - 0.09 mol/L = 0.91 mol/L We can derive the amounts of CI2 from the coefficient of the chemical equation: Chemical equilibrium Page 155 2 ONCI 2NO + CI2 2x 0.09 mol x 0.045 mol 2ONCI 2NO + CI2 at start 1.0 mol/L ----- ------ Change - 0.09 mol/L + 0.09 + 0.045 0.91 0.09 0.045 at equilibrium [NO] 2 [Cl2 ] Therefore, at Kc = [ONCl] 2 (0.09 mol/l) 2 (0.045mol/ l) 4.4x10 4 mol/l 2 (0.9 mol/l) = Example 6.3: For the reaction 2 SO3(g) 2 SO2(g) + O2(g) at 1100 K Kc is 0.0271 mol/L. what is Kp at same temperature. Solution: n = 3-2 =1 Kp = Kc (RT)+1 = 0.0271 mol/L x (0.0821 L. atm / K. mol) (1100 K) = 2.45 atm Chemical equilibrium Page 156 Example 6.4: What is Kc for the reaction? N2(g) + 3 H2(g) 2 NH3(g) At 500°C if Kp is 1.5 x 10-5 / atm-2 at this temperature. Solution: n = 2 - 4 = -2 T = 273 + 500 = 773 K Therefore, Kp = Kc (RT)n Kp = Kc (RT)-2 1.5 x 10 -5 Kc = Kp (RT) = x [0.0821.atm/K.mol x 773K]2 2 atm 2 = (1.5 x I0-5/ atm2) (4.03 x 103 L2. atm2 / mol2) = 6.04 x 10-2 L2 / mol2 Some facts dealing with expressions for equilibrium constants may be summarized as follows: 1- The concentration terms for substances that appear on the right side of the chemical equation are written in the numerator of the expression for kc. The concentration terms for substances that appear on the left side are written in the denominator. Chemical equilibrium Page 157 2- No concentration terms are included for pure solids or pure liquids. 3- The value of kc for a given equilibrium is constant at constant temperature. At a different temperature the value of kc is different. Predicting the Direction of a Relation: For the reaction PCl2(g) PCl3(g) + Cl2 at 250oC Kc = [PCl3 ][Cl2 ] 0.0415mol/ l [PCl5 ] Suppose that a mixture of 1.00 mol of PCI5(g)/ 0.05 mol of PCI3(g)/ and 0.03 mol of CI2(g) is placed in 1.0 L container. Is this an equilibrium system, or will a net reaction occur in one direction or the other? If the mixture is as equilibrium, the concentrations of the three substances, when substituted into the expression for kc, will equal kc. A value obtained by substituting initial concentration into the expression for the equilibrium constant is called a reaction quotient (Q). For the example at hand, Q= [PCl 3 ][Cl2 ] 0.05 x 0.03 0.015 mol/l [PCl5 ] 0.1 Chemical equilibrium Page 158 In this case Q (0.015 mol/L) is smaller than kc (0.0415 mol/L). The system is not at equilibrium. The reaction will proceed from left to right. In general, the values of a Q and the corresponding kc may be related in one of three ways: 1- Q < Kc In this instance, the reaction will move from left to right (the forward direction) to approach equilibrium. 2- Q = Kc The system is in equilibrium. 3- Q > Kc In this instance, the reaction will move from right to left (the reversible direction) to approach equilibrium. Example 6.5: For the reaction 2 SO2(g) + O2(g) 2 SO3(g) at 827°C, kc is 36.9 L / mol. If 0.05 mol of SO2(g), 0.03 mol of O2(g), and 0.125 mol of SO3(g) are mixed in a 1.0 L container at 827°C, in what direction will the reaction proceed? Solution: [SO3 ]2 [0.125mol/ l] 2 208 L/moL Q= 2 2 [SO2 ] [O2 ] [0.05 mol/l] [0.03mol/l ] Chemical equilibrium Page 159 Since Q (208 L/mol) > kc (36.9 L/mol), the reaction will proceed from right to left (SO3 will dissociate). Heterogeneous Equilibria: The concentration of a pure solid or a pure liquid is constant and do not appear in the expression for the equilibrium constant. For example 1- CaCO3(S) CaO(S) + CO2(g) Kc = [CO2] 3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g) 2- [H2 ]4 Kc [H2O] 4 Equilibrium Concentrations Using K Example 6.6: Kc for the HI equilibrium at 425°C is 54.5: H2(g) + I2(g) 2 HI(g) A quantity of HI(g) is placed in a 1.01 container and allowed to come to equilibrium at 425°C What are the concentrations of H2(g) and I2(g) in equilibrium with 0.5 mol/L of HI(g) Solution: Chemical equilibrium Page 160 The [H2] = [I2] at equilibrium since they are produced in equal amounts by decomposition of HI(g) [H2] = [I2] = x [HI] = 0.5 mol/L [HI] 2 Kc 54.5 4 [H2 ] [I2 ] [0.5 mol/l] 2 54.5 X2 (0.5)2 0.00456 mol 2 /L2 x = 54.5 2 x = 0.068 mol/L The equilibrium concentration is: [HI] = 0.5 mol/ L [H2] = [I2] = 0.068 mol/ L Example 6.7: For the reaction H2(g) + CO2(g) H2O(g) + CO(g) kc is 0.771 at 750°C. If 0.01 mol of H2 and 0.01 mol of CO2 are mixed in 1 liter container at 750°C, what are the concentrations of all substances present at equilibrium? Solution: Chemical equilibrium Page 161 If x mol of H2 reacts with x mol of CO2 out of the total amount supplied, x mol H2O and x mol CO will be produced. Hence H2(g) + CO2(g) H2O(g) CO(g) At start 0.01 0.01 mol/L ---- ----- Change mol/L -x -x +x +x 0.01 -x 0.01 -x X X at equilibrium Kc [H2O] [CO] 0.771 [H2 ] [CO2 ] X2 Kc (0.01 - X)2 If we extract the square root of both sides of this equation: X 0.878 0.01 - X X = 0.0878 – 0.878 X X= 0.00468 mol/l At equilibrium, therefore [H2] = [CO2] = 0.01 mol/L - 0.00468 mol/L = 0.0053 mol/L Chemical equilibrium Page 162 [H2O] = [CO] = 0.00468 mol/L Example 6.8: For the reaction C(s) + CO2(g) 2 CO(g) Kp is 167.5 atm at 1000°C. What is the partial pressure of CO(g) in an equilibrium system in which the partial pressure ofCO2(g) is 0.1atm? Solution: 2 PCO Kp 167.5 atm PCO2 2 PCO 167.5 atm 0.1 PCO2= 16.8 PCO= 4.10 atm Example 6.9: Kp for the equilibrium: FeO(s) + CO(g) Fe(s) + CO2(g) at 1000°C is 0.403. If CO(g) at a pressure of 1.0 atm, and excess FeO(s) are placed in a container at 1000°C, what are the pressures of CO(g) and CO2(g) when equilibrium is attained? Chemical equilibrium Page 163 Solution: Let x equal the partial pressure of CO2 when equilibrium is attained FeO(s) + CO(g) Fe(s) + CO2(g) At start Change At equilibrium Kp 1.0 atm --- -x 1.0 - x atm +x x PCO2 0.403 PCO X atm 0.403 1.0 atm - X X = PCO2 = 0.287 atm 1.0 – X = Pco = 0.713 Le Chatelier's Principle: Le chatelier's principle predicts how a system in equilibrium will respond to changes in experimental conditions (such as temperature or pressure), which states Chemical equilibrium that a system in Page 164 equilibrium reacts to a stress in a way that counteracts the stress and establishes a new equilibrium state: 1- Concentration changes: If the concentration of substance is increased, the equilibrium will shift in a way that will decrease the concentration of the substance that was added. H2(g) + I2(9) 2 HI(g) e.g: Increase H2 or I2 → shift to to formation of HI Removal of H2 or I2 ← Reaction shift to decomposition of HI. 2- Pressure changes: Increasing the pressure causes a shift in the direction that will decrease the number of moles of gas. 2 SO2(g) + O2(g) 2 SO3(g) 3 moles 2 moles When the pressure on an equilibrium mixture is increased (or the volume of the system decreased), the position of equilibrium shifts to the right. In this way the system counteracts the change, and vice versa. Chemical equilibrium Page 165 For reactions in which n = 0, pressure changes have no effect on the position equilibrium. e.g: N2(g) + O2(g) 2 NO(g) For heterogeneous equilibrium the effect of pressure is predicated by counting the number of moles of gas indicated on each side of the equation. 3- Temperature changes: For the reaction N2(g) + 3 H2(g) 2 NH3(g) H = - 92.4 KJ Since H is -ve, the reaction to the right evolves heat N2(g) + 3 H2(g) 2 NH3(g) + 92.4 KJ The highest yields of NH3 will be obtained at the lowest temperatures and high pressures. Also consider the reaction CO2(g) + H2(g) CO(g) + H2O(g) H = + 41.2KJ Since H is + ve , we can write the equation 41.2 KJ + CO2(g) + H2(g) —— CO(g) + H2O(g) Increasing the temperature always favors the endothermic change, and decreasing the temperature always favors the exothermic change. Chemical equilibrium Page 166 The numerical value of the equilibrium constant charges when the temperature is changed. 4- Effect of catalyst: The addition of a catalyst causes a system to achieve equilibrium faster but does not alter the position of equilibrium. 5- Addition of an inert gas: If an inert gas is introduced into a reaction vessel containing other gases at equilibrium, it will not affect the position of equilibrium because it will not alter the partial pressures or the concentrations of any of the substances already present. Chemical equilibrium Page 167