Chemical Equilibrium Chapter 14 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Last chapter, the rate of a chemical reaction was described as: Dt When A B What happens when the reaction nears completion? A B rate = - D[A] Dt D[B] = = 0 Equilibrium is a state in which there are no observable changes as time goes by.

Physical equilibrium Chemical equilibrium Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Physical equilibrium H2O (l) H2O (g) Chemical equilibrium N2O4 (g) 2NO2 (g) 14.1

forward reaction rate = reverse reaction rate On a molecular level, reactants never stop reacting. So why does the rate of reactant disappearance drop to zero? Rate of products reacting is the same as the rate of reactants reacting. forward reaction rate = reverse reaction rate Consider the following reaction: H2+ I2 2 HI Rate forward = kf [H2][I2] Rate reverse = kr[HI]2 At equilibrium: kf [H2]eq[I2] eq = kr[HI]2 eq Rearranging: kf = [HI]2 eq  K Equilibrium constant kr [H2]eq[I2] eq

N2O4 (g) 2NO2 (g) equilibrium kinetics kinetics kinetics equilibrium Start with NO2 Start with N2O4 Start with NO2 & N2O4 It does not matter where you start, the ratios at equilibrium are always the same (at the same temperature).

constant 14.1

Equilibrium Will N2O4 (g) 2NO2 (g) K = [NO2]2 [N2O4] = 4.63 x 10-3 aA + bB cC + dD Products K = [C]c[D]d [A]a[B]b Law of Mass Action Equilibrium Will Reactants K >> 1 Lie to the right Favor products K << 1 Lie to the left Favor reactants When at equilibrium, the ‘eq’ subscript is assumed. No reaction mechanism is assumed !! 14.1

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N2O4 (g) 2NO2 (g) Kp = NO2 P2 N2O4 P Kc = [NO2]2 [N2O4] In most cases Kc  Kp Concentration units for K’s:

P C P D P A P B aA (g) + bB (g) cC (g) + dD (g) Kp = Kc(RT)Dn Remember: PV = nRT n V PA = RT = [A] RT RT 1 [A] = P C P C P D D [C]C [D]D Kc = Kp = [A]A [B]B P A P B A B Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

Homogeneous Equilibrium CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) [CH3COO-][H3O+] [CH3COOH][H2O] Kc = ‘ [H2O] = constant [CH3COO-][H3O+] [CH3COOH] = Kc [H2O] ‘ Kc = General practice not to include units for the equilibrium constant. 14.2

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2 (g) COCl2 (g) [COCl2] [CO][Cl2] = 0.14 0.012 x 0.054 Kc = = 220 Kp = Kc(RT)Dn Dn = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K Kp = 220 x (0.0821 x 347)-1 = 7.7 14.2

The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm? 2NO2 (g) 2NO (g) + O2 (g) 2 Kp = 2 PNO PO PNO PO 2 = Kp PNO PO 2 = 158 x (0.400)2/(0.270)2 = 347 atm 14.2

They must just be present. Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO3 (s) CaO (s) + CO2 (g) Kc = ‘ [CaO][CO2] [CaCO3] [CaCO3] = constant [CaO] = constant Kc x ‘ [CaCO3] [CaO] Kc = [CO2] = Kp = PCO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. The equilibrium state is not affected by the quantity of solids and liquids, if they are pure. They must just be present. 14.2

does not depend on the amount of CaCO3 or CaO CaCO3 (s) CaO (s) + CO2 (g) PCO 2 = Kp PCO 2 does not depend on the amount of CaCO3 or CaO 14.2

Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction? NH4HS (s) NH3 (g) + H2S (g) Kp = P NH3 H2S P = 0.265 x 0.265 = 0.0702 Kp = Kc(RT)Dn Kc = Kp(RT)-Dn Dn = 2 – 0 = 2 T = 295 K Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4 14.2

‘ ‘ ‘ ‘ ‘ Kc = [C][D] [A][B] Kc = [E][F] [C][D] Kc A + B C + D Kc C + D E + F [E][F] [A][B] Kc = A + B E + F Kc Kc = Kc ‘ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

Add two reactions => multiply K’s [NO2][O2] K1 O3 + NO NO2 + O2 K1 = [O3][NO] NO2 + O . NO + O2 K2 [NO][O2] K2 = O3 + O . 2 O2 Knet [NO2][O .] K1 Knet = x K2 [NO2][O2] [NO][O2] Knet = [O3][NO] [NO2][O .] [O2]2 Knet = [O3][O .] Add two reactions => multiply K’s 14.2

‘ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4] = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 14.2

Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

Chemical Kinetics and Chemical Equilibrium ratef = kf [A][B]2 A + 2B AB2 kf kr rater = kr [AB2] Equilibrium ratef = rater kf [A][B]2 = kr [AB2] kf kr [AB2] [A][B]2 = Kc = 14.3

Qc < Kc system proceeds from left to right to reach equilibrium The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. IF Qc < Kc system proceeds from left to right to reach equilibrium Qc = Kc the system is at equilibrium Qc > Kc system proceeds from right to left to reach equilibrium 14.4

For the reaction NH3 (g) N2 (g) + 3 H2 (g) Kc = 2.6 x 10-5 at 1270C. The initial concentrations are [N2 ]0 = 0.50 mol/L, [H2]0 = 1.5 mol/L, and [NH3]0 = 0.020 mol/L. Are we at equilibrium? If not, which way will the reaction proceed? Qc = [N2] [H2]3 [NH3] 2 = (0.50)(1.5)3 0.0202 = 4.2 x 103 Is Qc = Kc ? 4.2 x 103  2.6 x 10-5 Compare Qc to Kc, if Qc > Kc then [Products] are too high and the [Reactants] are too low. Thus reaction must cnvert products to reactants. 4.2 x 103 > 2.6 x 10-5

Calculating Equilibrium Concentrations Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. Having solved for x, calculate the equilibrium concentrations of all species. 14.4

ICE At 12800C the equilibrium constant (Kc) for the reaction Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br2 (g) 2Br (g) Let x be the change in concentration of Br2 Br2 (g) 2Br (g) Initial (M) 0.063 0.012 ICE Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x [Br]2 [Br2] Kc = Kc = (0.012 + 2x)2 0.063 - x = 1.1 x 10-3 Solve for x 14.4

Kc = (0.012 + 2x)2 0.063 - x = 1.1 x 10-3 4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x2 + 0.0491x + 0.0000747 = 0  -b ± b2 – 4ac ax2 + bx + c =0 x = 2a x = -0.0105 x = -0.00178 Br2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.063 0.012 -x +2x 0.063 - x 0.012 + 2x At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M At equilibrium, [Br2] = 0.062 – x = 0.0648 M 14.4

Assume zero if not given The equilibrium constant (Kc) for the reaction is 4.00 x 10-4. If the initial concentration of [HCN] = 0.39 mol/L, calculate [C2N2] at equilibrium. 2 HCN (g) H2 (g) + C2N2(g) Let x be the change in concentration of H2 and C2N2 2 HCN (g) H2 (g) + C2N2(g) Assume zero if not given Initial (M) 0.39 x Change (M) -2x x x x Equilibrium (M) 0.39 -2x [H2][C2N2] (x)(x) (0.39 - 2x)2 = 4.00 x 10-4 Solve for x Kc = Kc = [HCN] 2 14.4

x = 0.0075 mol/L = [C2N2] at equilibrium (x)(x) (0.39 - 2x)2 = 4.00 x 10-4   x 0.39 - 2x = 2.00 x 10-2 => x = 0.0200(0.39 - 2x) x = 0.0078 - 0.0400 x 1.04 x = 0.0078 x = 0.0075 mol/L = [C2N2] at equilibrium Use the definition of x to help you answer the question given in the problem.

Assume zero if not given When solid NH4HS is stored is a sealed container, ir decomposed to some extent to NH3 and H2S gases. Given: and Kc = 1.2 x 10-4 at 21.8°C. Determine [NH3] and [H2S] at equilibrium. NH4HS (s) NH3 (g) + H2S (g) Let x be the change in concentration of NH3 and H2S NH4HS (s) NH3 (g) + H2S (g) Assume zero if not given Initial (M)  x Change (M) -x x x x Equilibrium (M) ( - x) [NH3] [H2S] (x)(x) = 1.2 x 10-4 Solve for x Kc = Kc = “1” x = 0.011 mol/L = [NH3] = [H2S ] at equilibrium

Given some information about the equilibrium state, we can calculate the remainder of the information. What is the partial pressure of NH3 in equilibrium with 0.733 atm H2 and 0.527 atm N2, given: N2 (g) + 3 H2 (g) 2 NH3 (g) and KP = 1.45 x 10-5 at 25°C. Determine [NH3] and [H2S] at equilibrium. N2 (g) + 3 H2 (g) 2 NH3 (g) Equilibrium 0.527 atm 0.733 atm x x2 PNH3 2 KP = = = 1.45 x 10-5 PN2 PH23 (0.527)(0.733)3 Solve for x x = 1.73 x 10-3 atm = PNH3 at equilibrium

Assume zero if not given Initial (M) 1.000 atm x x Change (M) -x At 250°C, PCl5 is introduced into a vessel at an initial pressure of 1.000 atm. When equilibrium is established for the following reaction, the total pressure increses to 1.714 atm. What is KP? PCl5 (g) PCl3 (g) + Cl2 (g) Assume zero if not given Initial (M) 1.000 atm x x Change (M) -x Equilibrium 1.000 - x x x PPCl3 PCl2 Dalton’s Law PTOT = PPCl3 + PCl2 + PPCl5 KP = PPCl5 1.714 atm = (1.000 - x) + x + x 1.714 atm = 1.000 + x (x) (x) KP = (1.000 - x) Solve for x x = 0.714 atm KP = (0.714 atm)(0.714 atm) = 1.78 = KP (1.000 - 0.714) atm PPCL5 = (1.000-0.714) atm = 0.286 atm at equilibrium

Must increase PCl5 and decrease PCl3 Add 0.250 atm Cl2. What will happen? PCl5 (g) PCl3 (g) + Cl2 (g) 0.714 atm (0.714 + 0.250) atm Initial (M) 0.268 atm - x - x Change (M) x Must increase PCl5 and decrease PCl3 Equilibrium 0.268 + x 0.714 - x 0.964 -x (0.714 - x)(0.964 -x) PPCl3 PCl2 KP = = = 1.78 PPCl5 (0.268 + x) 0.688 - 0.714x - 0.964x + x2 = 0.509 + 1.78x x2 -3.458x + 0.179 = 0 use quadratic equation x = 3.258 3.4582 - (4)(1)(1.179) = 3.458 11.24 (2)(1) 2 0.0527 PPCl3 = 0.714-0.0527 = 0.661 atm PCl2 = 0.964-0.0527 = 0.911 atm 3.451 does not make sense PPCl5 = 0.268+0.0527 = 0.339 atm KP = (0.661 atm)(0.911 atm) = 1.78 0.339 atm

Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Changes in Concentration N2 (g) + H2 (g) NH3 (g) Add NH3 Equilibrium shifts left to offset stress 14.5

Le Châtelier’s Principle Changes in Concentration continued Add Add aA + bB cC + dD Remove Remove Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left 14.5

Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas 14.5

Le Châtelier’s Principle Changes in Temperature Change Exothermic Rx Endothermic Rx Increase temperature K decreases K increases Decrease temperature K increases K decreases Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner 14.5

Catalyst lowers Ea for both forward and reverse reactions. uncatalyzed catalyzed Catalyst lowers Ea for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. 14.5

Le Châtelier’s Principle Change Equilibrium Constant Change Shift Equilibrium Concentration yes no Pressure yes no Volume yes no Temperature yes yes Catalyst no no 14.5