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

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Presentation on theme: "Chemical Equilibrium Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."— Presentation transcript:

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

2 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 H 2 O (l) Chemical equilibrium N2O4 (g)N2O4 (g) H 2 O (g) 2NO 2 (g) At equilibrium, the conversion is still going on, but the two rates are equal. A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products.

3 K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Equilibrium Will K = [C] c [D] d [A] a [B] b aA + bB cC + dD For a reaction at particular temperature, the equilibrium constant K: Here we first discuss the equilibrium constant of forward Rxn.

4 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 ] Kp =Kp = NO 2 P2P2 N2O4N2O4 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) In most cases K c  K p K c : concentration in molarity, mol/L P= (n/V)RT At constant temp., the pressure is related to the concentration, mol/L. Kp applies to gaseous reaction. For Gaseous Rxn: K c = K p (RT) -  n or

5 The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = atm and P NO = atm? 2 2NO 2 (g) 2NO (g) + O 2 (g) 14.2 K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm

6 The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = M, [Cl 2 ] = 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 ] = x = 220 K p = K c (RT)  n  n = 1 – 2 = -1 R = T = = 347 K K p = 220 x ( x 347) -1 =

7 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] 14.2 The equilibrium constant of solvent (usually water) does not appear in equilibrium constant expression. For Liquid Rxn:

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9 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 p = P CO 2 The concentrations of pure solids, pure liquids and solvent do not appear in the equilibrium constant expressions. P CO 2 does not depend on the amount of CaCO 3 or CaO

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11 Consider the following equilibrium at 295 K: The partial pressure of each gas is 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= x = K p = K c (RT)  n K c = K p (RT) -  n  n = 2 – 0 = 2 T = 295 K K c = x ( x 295) -2 = 1.20 x

12 A + B C + D C + D E + F A + B E + F K c = ‘ [C][D] [A][B] K c = ‘ ‘ [E][F] [C][D] [E][F] [A][B] K c = KcKc ‘ KcKc ‘‘ KcKc KcKc ‘‘ KcKc ‘ 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 Multiple equilibrium

13 N 2 O 4 (g) 2NO 2 (g) = 4.63 x 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 Forward RxnReverse Rxn A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products. Equilibrium Constant for a reversible reaction

14 14.3 Chemical Kinetics and Chemical Equilibrium A + 2B AB 2 kfkf krkr rate f = k f [A][B] 2 rate r = k r [AB 2 ] Equilibrium rate f = rate r k f [A][B] 2 = k r [AB 2 ] kfkf krkr [AB 2 ] [A][B] 2 =K c = Elementary reaction:

15 What does equilibrium constant tell us? K c helps us to predict the direction in which a reaction mixture will proceed to achieve equilibrium and to calculate the concentrations of reactants and products once the equilibrium has been achieved.

16 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 14.4 Predict the reaction direction

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18 The equilibrium constant (K c ) for the reaction Is 1.1 x If the initial concentrations are [Br 2 ] = M and [Br] = 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) x-x+2x x x [Br] 2 [Br 2 ] K c = ( x) x = 1.1 x Solve for x Calculating Equilibrium Concentrations

19 K c = ( x) x = 1.1 x x x = – x 4x x = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) x-x+2x x x x = x = At equilibrium, [Br] = x = Mor M At equilibrium, [Br 2 ] = – x = M 14.4

20 Factors that affect chemical equilibrium

21 Le Châtelier’s Principle Changes in Concentration ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left aA + bB cC + dD Add Remove 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.

22 Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas Change pressure will not affect the concentration of reacting species in condensed phase, e.g. liquid and solid. In general, the increase of pressure favors the net reaction that decreases the total number of moles of gases and decrease of pressure favors the net reaction that increases the total number of moles of gases

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24 Le Châtelier’s Principle Changes in Temperature ChangeExothermic Rx Increase temperatureK decreases Decrease temperatureK increases Endothermic Rx K increases K decreases 14.5 Increase temperature favors an endothermic reaction, and the temperature decrease favors an exothermic reaction.

25 uncatalyzedcatalyzed 14.5 Adding a Catalyst Catalyst lowers E a for both forward and reverse reactions. does not change equilibrium constant K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

26 ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyesno Volumeyesno Temperatureyes Catalystno 14.5


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