Chemical Equilibrium What is Chemical Equilibrium? A state in which the concentrations of reactants and products remain constant because the rates of the forward and reverse reactions are equal. The Macroscale View of Equilibrium A ⇔ B If we start with A, [A] decreases and [B] increases. Eventually, concentrations don’t change anymore. Equilibria are Dynamic •  All equilibria are dynamic! •  A and B do not stop reacting. •  No net change in concentrations when equilibrium is established. Chemical equilibrium is dynamic Dynamic equilibrium: both processes are happening Static equilibrium: both processes have stopped Dynamic equilibrium At what time is equilibrium achieved? Between time = 0 and equilibrium What is happening to the rate of reaction? The Nanoscale View of Equilibrium A ⇔ B ratef = kf[A] rater = kr[B] At equilibrium, rate of forward reaction equals rate of reverse reaction. ratef = rater kf[A] = kr[B] Change in rate of reaction in time At equilibrium the forward and reverse rates are equal Equilibrium - Direction of Approach N2 + 3 H2 starting with N2 and H2 2 NH3 starting with NH3 from Chemistry, the Central Science, 8th Ed. (Prentice-Hall 2000) The Equilibrium Constant aA + bB equilibrium constant cC + d D [C]c [D] d Kc = [A]a [B] b equilibrium constant expression Kc Example N2 + 3 H2 2 NH3 [NH3]2 Kc = [N2] [H2]3 Kc are often expressed without units. When needed, simplify the mol dm-3 units This example units are mol-2 dm6 Heterogeneous Equilibria •  Pure solids or liquids do not appear in equilibrium expressions. ! their “concentrations” cannot change •  For CaCO3 (s) CaO (s) + CO2 (g) [CaO][CO2] Kc = = [CO2] [CaCO3] What is the equilibrium expression for the decomposition of ammonium carbamate, NH4CO2NH2, that occurs according to this equation: NH4CO2NH2(s) ! 2 NH3(g) + CO2(g) (A) K = [NH3][CO2] (B) K = [NH3]2[CO2] (C) K = [NH3][CO2] / [NH4CO2NH2] (D) K = [NH3]2[CO2] / [NH4CO2NH2] Consider the reaction: A2 + B2 ! 2 AB The reactants are mixed together and the following is a plot of the reactant and the product concentrations with time: what is the value of the equilibrium constant? [AB]2 K= [ A] [ B] [4]2 K= = 0.44 6 6 [ ][ ] Kp for Gases N2 (g) + 3 H2 (g) Kp = 2 NH3 (g) PNH3 PN2 2 PH2 3 Pressures are partial pressures of the various gases Selected Ks at 25°C Magnitude of K equilibrium lies to right product-favored equilibrium lies to left reactant-favored from Chemistry, the Central Science, 8th Ed. (Prentice-Hall 2000) Direction of RXN and K Kc = 9.60 for the reaction: N2 (g) + 3 H2 (g) 2 NH3 (g) [NH3]2 Kc = 3 = 9.60 [N2] [H2] If reaction written in reverse: 2 NH3 (g) N2 (g) + 3 H2 (g) Kc' = [N2] [H2]3 [NH3]2 = 1 = 0.104 9.60 At 985°C, the equilibrium constant for the reaction, H2(g) + CO2(g) ⇔ H2O(g) + CO(g) Kc = 1.63. What is the equilibrium constant for the reverse reaction? a) 1.63 d) 0.613 b) 0.815 e) 1.00 c) 2.66 Multiplying RXNs and K If a RXN is multiplied by a factor n, then Kc' = (Kc)n If the NH3 reaction is multiplied by 2: 2 N2 (g) + 6 H2 (g) 4 NH3 (g) [NH3]4 2 2 (K ) Kc' = (9.60) = = = 92.2 c 2 6 [N2] [H2] Heterogeneous Equilibria from Chemistry, the Central Science, 8th Ed. (Prentice-Hall 2000) Reaction Quotient = Q •  Q has same mathematical form as equilibrium constant expression, but uses initial or current concentrations instead of equilibrium concentrations. •  If Q < K, RXN proceeds to the right. •  If Q > K, RXN proceeds to the left. •  If Q = K, RXN is at equilibrium. Q and K Q Example For the reaction: N2 (g) + 3 H2 (g) 2 NH3 (g) Kc = 51 at 472°C If a 5.00 L container is charged with 1.00 mol of each species at 472°C, which way will the reaction proceed? Calculating Equilibrium Concentrations •  •  •  •  Write a balanced equation. Set up an equilibrium table ICE table. Use x to represent a change in concentration. Express equilibrium concentrations in terms of x. •  Solve for x. •  Solve for unknowns. •  Check. Calculating Equilibrium Concentrations: ICE Example: in the reaction below, the reaction begins with 2 moles of H2 and 2 moles of F2. At equilibrium, the amount of the acid HF is measured to be 0.60 moles. Calculate the equilibrium amounts of the reactants and calculate Kc. H2(g) + F2(g) ⇔ 2 HF(g) initial 2 2 0 change - 0.3 - 0.3 + 0.6 equilibrium 1.7 1.7 0.6 Then calculate Kc Kc = [HF]2 ______ [H2][F2] [0.6]2 ______ = [1.7][1.7] = 0.125 In this case the units cancel all away! LeChatelier’s Principle If a system is at equilibrium and the conditions are changed so that it no longer is at equilibrium, the system will react to reach a new equilibrium in a way that partially counteracts the change (i.e. the equilibrium shifts.) LeChatelier’s Example For the reaction: N2 (g) + 3 H2 (g) 2 NH3 (g) Kc = 51 at 472°C If the reaction is at equilibrium, what happens if additional H2 is added? Shifting Equilibrium H H C CH3 H C CH3 C CH3 CH3 C H LeChat’s - Changing Concentrations disturbance effect add reactant(s) shift right add product(s) shift left remove reactant(s) shift left remove product(s) shift right The value of K does not change for these disturbances. LeChat’s - Changing V and P Only when gases are present! •  PV = nRT •  If volume reduced, pressure increases. •  How will system “relieve” P increase? ! by shifting equilibrium to reduce the number of moles of gas •  K does not change for these disturbances. LeChat’s PV Example N2O4 (g) 2 NO2 (g) Effect of changing pressure or volume •  Increasing pressure or decreasing volume System will react towards the side with fewer moles of gas •  Decreasing pressure or increasing volume System will react towards the side with more moles of gas The sketches below show the pressures, P, of these three gases as a function of time during an experiment. The system is initially at equilibrium. At some time, t, extra I2 is added. Which of the sets of curves shows how the system will respond to this situation? LeChat’s - Adding an Inert Gas •  Adding an inert (nonreactive) gas can increase total pressure. •  But, partial pressures (concentrations) will not change. •  Equilibrium will not shift and K does not change. LeChat’s - Changing Temperature •  T is the only disturbance that will change K! •  If endothermic, consider heat as a “reactant”. •  If exothermic, consider heat as a “product”. LeChat’s - Changing T – Endothermic reactions reactants + heat products If T !, heat !, equilibrium shifts right, K !. If T ", heat ", equilibrium shifts left, K ". LeChat’s - Changing T Exothermic reactions reactants products + heat If T !, heat !, equilibrium shifts left, K ". If T ", heat ", equilibrium shifts right, K !. LeChat’s - Changing T Example • For the reaction: N2 (g) + 3 H2 (g) 2 NH3 (g) #Ho = -92.38 kJ •  If T !, equilibrium shifts left, K ". What About Catalysts? Catalysts... •  speed up rate at which equilibrium is reached. •  do not affect equilibrium position. •  Catalysts affect rate of both forward and reverse reaction •  do not affect value of K. Catalysts lower activation energy of both the forward and the reverse reaction For this reaction at equilibrium, which changes will increase the quantity of Fe(s)? Fe3O4(s) + 4 H2(g) 3 Fe(s) + 4 H2O(g) "H > 0 1. increasing temperature 2. decreasing temperature 3. adding Fe3O4(s) (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1,2, and 3 Haber-Bosch Process N2 (g) + 3 H2 (g) 2 NH3 (g) N2(g) + 3 H2(g) ⇔ 2 NH3(g) "H = -92.2 kJ How would changing these conditions increase yield of ammonia? 1. Adding or removing the reactants N2 and H2 2. Adding or removing the product NH3 3. Increasing or decreasing temperature? 4. Increasing or decreasing pressure? 5. Adding a catalyst? N2(g) + 3 H2(g) ⇔ 2 NH3(g) "H = -92.2 kJ How would changing these conditions would increase the rate of reaction? 1. Adding or removing the reactants N2 and H2 2. Adding or removing the product NH3 3. Increasing or decreasing temperature? 4. Increasing or decreasing pressure? 5. Adding a catalyst? Haber Process In practice 1. A temperature of ! 450˚C. The high temperature is favorable for kinetics but not for the equilibrium 2. Ammonia is removed as it is produced, pushing equilibrium to the right 3. Pressure of ! 200 – 250 atm. Pressure increases concentration of the gases, favorable for kinetics, and favoring the right side of the reaction. Too much pressure involves excessive costs 4. An iron catalyst is used. Helps kinetics, does not affect equilibrium. Allows for a lower temperature, which helps equilibrium and costs. The Contact process 1. One step in the production of sulfuric acid, one of the most manufactured compounds in the world. S + O2 " SO2 Contact process 䚚 2 SO2(g) + O2(g) ⇔ 2 SO3(g) SO3(g) + H2O " H2SO4 The Contact Process 2 SO2(g) + O2(g) ⇔ 2 SO3(g) #H = -196 kJ mol-1 This process has the same issues as the Haber process, with high temperature required for kinetics, but unfavorable for equilibrium In practice 1. A temperature of ! 450˚C. The high temperature is favorable for kinetics but not for the equilibrium 2. Sulfur trioxide is removed as it is produced, pushing equilibrium to the right 3. Pressure of only 1 – 2 atm. Good yields are possible at a fairly low pressure. Pressure increases concentration of the gases, favorable for kinetics, and favoring the right side of the reaction. Too much pressure involves excessive costs 4. A catalyst of V2O5 is used. Helps kinetics, does not affect equilibrium. Allows for a lower temperature, which helps equilibrium and costs.