Presentation on theme: "C h a p t e r 13 Chemical Equilibrium. The Equilibrium State Chemical Equilibrium: The state reached when the concentrations of reactants and products."— Presentation transcript:
The Equilibrium State Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time. 2NO 2 (g)N2O4(g)N2O4(g) BrownColorless
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 H 2 O (l) Chemical equilibrium N2O4 (g)N2O4 (g) H 2 O (g) 2NO 2 (g) colorless brown
N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium Demo
Using Equilibrium Constants01 We can make the following generalizations concerning the composition of equilibrium mixtures: If K c > 10 3, products predominate over reactants. If K c is very large, the reaction is said to proceed to completion. If K c is in the range 10 –3 to 10 3, appreciable concentrations of both reactants and products are present. If K c < 10 –3, reactants predominate over products. If K c is very small, the reaction proceeds hardly at all.
Using the Equilibrium Constant K c = 4.2 x 10 -48 2H 2 (g) + O 2 (g)2H 2 O(g) (at 500 K) K c = 2.4 x 10 47 2H 2 O(g)2H 2 (g) + O 2 (g) (at 500 K) K c = 57.0 2H I (g)H 2 (g) + I 2 (g) (at 500 K)
N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N 2 O 4 ] aA + bB cC + dD K = [C] c [D] d [A] a [B] b Law of Mass Action
N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N 2 O 4 ] 2NO 2 (g) N 2 O 4 (g) K = [N 2 O 4 ] [NO 2 ] 2 ‘ = 1 K = 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.
K c = Write the Equilibrium Constant K c If you multiply both side of an equilibrium reaction by n the equilibrium constant should be raised to the power of n. N 2 (g) + 3H 2 (g)2NH 3 (g) [NH 3 ] 2 [N 2 ][H 2 ] 3 = 4NH 3 (g)2N 2 (g) + 6H 2 (g) [N 2 ] 2 [H 2 ] 6 [NH 3 ] 4 = K c ´´ 1 KcKc 2NH 3 (g)N 2 (g) + 3H 2 (g) [N 2 ][H 2 ] 3 [NH 3 ] 2 K c = ´ 2
Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P2P2 N2O4N2O4 P In most cases K c K p aA (g) + bB (g) cC (g) + dD (g) K p = K c (RT) n n = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)
Homogeneous Equilibrium CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) K c = ‘ [CH 3 COO - ][H 3 O + ] [CH 3 COOH][H 2 O] [H 2 O] = constant K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] =K c [H 2 O] ‘ General practice not to include units for the equilibrium constant. The concentration of pure liquids are not included in the expression for the equilibrium constant.
The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = 0.14 0.012 x 0.054 = 220 K p = K c (RT) n n = 1 – 2 = -1 R = 0.0821T = 273 + 74 = 347 K K p = 220 x (0.0821 x 347) -1 = 7.7
The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = 0.400 atm and P NO = 0.270 atm? 2 2NO 2 (g) 2NO (g) + O 2 (g) K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO 2 2 2 POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm
Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = ‘ [CaO][CO 2 ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K c x ‘ [CaCO 3 ] [CaO] K p = P CO 2 The concentration of pure solids and pure liquids are not included in the expression for the equilibrium constant.
P CO 2 = K p CaCO 3 (s) CaO (s) + CO 2 (g) P CO 2 does not depend on the amount of CaCO 3 or CaO
Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate K p and K c for the reaction? NH 4 HS (s) NH 3 (g) + H 2 S (g) K p = P NH 3 H2SH2S P= 0.265 x 0.265 = 0.0702 K p = K c (RT) n K c = K p (RT) - n n = 2 – 0 = 2 T = 295 K K c = 0.0702 x (0.0821 x 295) -2 = 1.20 x 10 -4
Writing Equilibrium Constant Expressions The concentrations of the reacting species in the liquid phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids and pure liquids, 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.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.
N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium Demo Review
The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium K c = [NO 2 ] 2 [N 2 O 4 ] Q c = [NO 2 ] 0 2 [N 2 O 4 ] 0
Use reaction quotient to predict the direction of shift when the volume is halved in the following equilibrium: N 2 O 4 (g) æ 2 NO 2 (g), Consider the reaction: N 2 O 4 (g) æ 2 NO 2 (g), taking place in a cylinder with a volume = 1 unit.
Le Châtelier’s Principle10 The Volume is then halved, which is equivalent to doubling the pressure. Since Q > K, the [product] is too high and the reaction progresses in the reverse direction.
At 1280 0 C the equilibrium constant (K c ) for the reaction Is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let (x) be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x [Br] 2 [Br 2 ] K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 Solve for x
K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac 2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x x = -0.00178x = -0.0105 At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.063 – x = 0.0648 M
Calculating Equilibrium Concentrations 1.Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3.Having solved for x, calculate the equilibrium concentrations of all species.
Predicting the direction of a Reaction The reaction quotient (Q c ) is obtained by substituting initial concentrations into the equilibrium constant. Predicts reaction direction. Q c > K c System proceeds to form reactants. Q c = K c System is at equilibrium. Q c < K c System proceeds to form products.
Le Châtelier’s Principle01 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.
Le Châtelier’s Principle02 Concentration Changes: The concentration stress of an added reactant or product is relieved by reaction in the direction that consumes the added substance. The concentration stress of a removed reactant or product is relieved by reaction in the direction that replenishes the removed substance.
Le Châtelier’s Principle: Haber process N 2 (g) + 3 H 2 (g) æ 2 NH 3 (g)Exothermic Cat: iron or iron ruthenium
Le Châtelier’s Principle06 The reaction of iron(III) oxide with carbon monoxide occurs in a blast furnace when iron ore is reduced to iron metal: Fe 2 O 3 (s) + 3 CO(g) æ 2 Fe(l) + 3 CO 2 (g) Use Le Châtelier’s principle to predict the direction of reaction when an equilibrium mixture is disturbed by: (a) Adding Fe 2 O 3 (b) Removing CO 2 (c) Removing CO
Le Châtelier’s Principle07 Volume and Pressure Changes: Only reactions containing gases are affected by changes in volume and pressure. Increasing pressure = Decreasing volume PV = nRT tells us that increasing pressure or decreasing volume increases concentration.
Le Châtelier’s Principle08 N 2 (g) + 3 H 2 (g) æ 2 NH 3 (g) K c = 0.291 at 700 K
Le Châtelier’s Principle11 Does the number of moles of reaction products increase, decrease, or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume. 1. PCl 5 (g) æ PCl 3 (g) + Cl 2 (g) 2. CaO(s) + CO 2 (g) æ CaCO 3 (s) 3. 3 Fe(s) + 4 H 2 O(g) æ Fe 3 O 4 (s) + 4 H 2 (g)
Temperature Changes: Changes in temperature can change the equilibrium constant. Endothermic processes are favored when temperature increases. Exothermic processes are favored when temperature decreases. Le Châtelier’s Principle13
Le Châtelier’s Principle14 Example: The reaction N 2 (g) + 3 H 2 (g) æ 2 NH 3 (g) which is exothermic by 92.2 kJ.
Le Châtelier’s Principle15 In the first step of the Ostwald process for synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction: 4 NH 3 (g) + 5 O 2 (g) æ 4 NO(g) + 6 H 2 O(g) ∆H° = –905.6 kJ How does the equilibrium amount vary with an increase in temperature?
Le Châtelier’s Principle16 The following pictures represent the composition of the equilibrium mixture at 400 K and 500 K for the reaction A(g) + B(g) æ AB(g). Is the reaction endothermic or exothermic?
Effect of Catalysis, Reduction of Activation Energy : No effect