Equilibrium slideshttp:\\asadipour.kmu.ac.ir1

Equilibrium The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate slideshttp:\\asadipour.kmu.ac.ir N 2 O 4 (g) 2 NO 2 (g)

Equilibrium A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant slideshttp:\\asadipour.kmu.ac.ir N 2 O 4 (g) 2 NO 2 (g)

Equilibrium slideshttp:\\asadipour.kmu.ac.ir

Equilibrium The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate law: Rate = k f [N 2 O 4 ] Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate law: Rate = k r [NO 2 ] slideshttp:\\asadipour.kmu.ac.ir N 2 O 4 (g) 2 NO 2 (g)

Equilibrium The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] = slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Manipulating Equilibrium Constants 2NO 2 Cl(g)  2NO 2 (g)+Cl 2 (g) K eq =[NO 2 ] 2 [Cl 2 ]/[NO 2 Cl] 2 R=K[NO2Cl] 1=NO2Cl  NO2+Cl K1=[NO2][Cl]/[NO2Cl] 2=NO2Cl+Cl  NO2+Cl2 K2=[NO2][Cl2]/[NO2Cl][Cl] K=K1*K2=K eq =[NO 2 ] 2 [Cl 2 ]/[NO 2 Cl] slideshttp:\\asadipour.kmu.ac.ir The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

Equilibrium The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] = slideshttp:\\asadipour.kmu.ac.ir N 2 O 4 (g) 2 NO 2 (g)

Equilibrium The Equilibrium Constant(K c ) To generalize this expression, consider the reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD slideshttp:\\asadipour.kmu.ac.ir

Equilibrium The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate slideshttp:\\asadipour.kmu.ac.ir

Equilibrium What Are the Equilibrium Expressions for These Equilibria? slideshttp:\\asadipour.kmu.ac.ir

Equilibrium The Equilibrium Constant(K p ) Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C ) c (P D ) d (P A ) a (P B ) b slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Relationship between K c and K p From the ideal gas law we know that Rearranging it, we get PV = nRT P = RT nVnV slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Relationship between K c and K p Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes Where K p = K c (RT)  n  n = (moles of gaseous product) − (moles of gaseous reactant) slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature no matter what the initial concentrations of NO 2 and N 2 O 4 are slideshttp:\\asadipour.kmu.ac.ir N 2 O 4 (g) 2 NO 2 (g)

Equilibrium Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Equilibrium Can Be Reached from Either Direction It does not matter whether we start with N 2 and H 2 or whether we start with NH 3. We will have the same proportions of all three substances at equilibrium slideshttp:\\asadipour.kmu.ac.ir

Equilibrium What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium slideshttp:\\asadipour.kmu.ac.ir

Equilibrium What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium slideshttp:\\asadipour.kmu.ac.ir

Equilibrium K & Time of Equilibrium The size of K and the time required to reach equilibrium are not directly related! slideshttp:\\asadipour.kmu.ac.ir20

Equilibrium slideshttp:\\asadipour.kmu.ac.ir21

Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction = K c = = at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g) slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. K c = = at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 2 N 2 O 4 (g) 4 NO 2 (g) slideshttp:\\asadipour.kmu.ac.ir

Equilibrium The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression K c = [Pb 2+ ] [Cl − ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq) slideshttp:\\asadipour.kmu.ac.ir

Equilibrium As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. CaCO 3 (s) CO 2 (g) + CaO (s) slideshttp:\\asadipour.kmu.ac.ir K = [CO 2 ]

Equilibrium Homogeneous Equilibria Homogeneous equilibria are equilibria in which all substances are in the same state. N 2(g) + 3H 2(g)  2NH 3(g) H 2(g) + F 2(g)  2HF (g) slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Heterogeneous Equilibria Write equilibrium expressions for the following: 2HOH (l)  2H 2(g) + O 2(g) 2HOH (g)  2H 2(g) + O 2(g) K = [H 2 ] 2 [O 2 ] slideshttp:\\asadipour.kmu.ac.ir CuSO 4. 5 HOH (s)  CuSO 4(s) + 5HOH (g) K = [HOH] 5

Equilibrium slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Equilibrium Calculations A closed system initially containing x 10 −3 M H 2 and x 10 −3 M I 2 At 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 −3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (g) 2 HI (g) slideshttp:\\asadipour.kmu.ac.ir

Equilibrium What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change At equilibrium 1.87 x slideshttp:\\asadipour.kmu.ac.ir

Equilibrium [HI] Increases by 1.87 x M [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change+1.87 x At equilibrium 1.87 x slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change-9.35 x x At equilibrium 1.87 x slideshttp:\\asadipour.kmu.ac.ir

Equilibrium We can now calculate the equilibrium concentrations of all three compounds… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change-9.35 x x At equilibrium 6.5 x x x slideshttp:\\asadipour.kmu.ac.ir

Equilibrium …and, therefore, the equilibrium constant Kc =Kc = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x ) 2 (6.5 x )(1.065 x ) slideshttp:\\asadipour.kmu.ac.ir

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Equilibrium The Reaction Quotient (Q) To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium slideshttp:\\asadipour.kmu.ac.ir

Equilibrium If Q = K, Q>K & Q<K slideshttp:\\asadipour.kmu.ac.ir

Equilibrium slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.” slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Effects of Changes on the System 1.Concentration : The system will shift away from the added component. 2.Temperature : K will change depending on the endothermic or exothermic reactions. 3.Pressure : a. Addition of inert gas b. Decreasing the volume slideshttp:\\asadipour.kmu.ac.ir does not affect the equilibrium position. shifts the equilibrium toward the side with fewer gaseous molecules.

Equilibrium Production of NH 3 If H 2 is added to the system, N 2 will be consumed and the two reagents will form more NH slideshttp:\\asadipour.kmu.ac.ir N 2(g) + 3H 2(g)  2NH 3(g)

Equilibrium Effect of Changes in Concentration on Equilibrium The addition of M N 2 in position I → the position II slideshttp:\\asadipour.kmu.ac.ir N 2(g) + 3H 2(g)  2NH 3(g) position I 0.399M M M M M M position II =

Equilibrium Changes in Concentration Predict the effect of the changes listed to this equilbrium: As 4 O 6(s) + 6C (s)  As 4(g) + 6CO (g) a) addition of carbon monoxide b) addition or removal of C (s) or As 4 O 6(s) c) removal of As 4(g) a) left b) none c) right slideshttp:\\asadipour.kmu.ac.ir

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Equilibrium The Effect of Container Volume on Equilibrium If the size of a container is changed, the concentration of the gases change. N 2(g) + 3H 2(g)  2NH 3(g). A smaller container shifts to the right Four gaseous molecules produce two gaseous molecules. A larger container shifts to the left two gaseous molecules produce four gaseous molecules slideshttp:\\asadipour.kmu.ac.ir

Equilibrium The Effect of Container Volume on Equilibrium Predict the direction of shift for the following equilibrium systems when the volume is reduced: a) P 4(s) + 6Cl 2(g)  4PCl 3(l) b) PCl 3(g) + Cl 2(g)  PCl 5(g) c) PCl 3(g) + 3NH 3(g)  P(NH 2 ) 3(g) + 3HCl (g) a) right b) right c) none slideshttp:\\asadipour.kmu.ac.ir

Equilibrium The Effect of Changes in Pressure CaCO 3 (s)  CaO(s) + CO 2 (g) slideshttp:\\asadipour.kmu.ac.ir

Equilibrium The system of N 2, H 2, and NH 3 are initially at equilibrium. When the volume is decreased, the system shifts to the right -- toward fewer molecules slideshttp:\\asadipour.kmu.ac.ir N 2(g) + 3H 2(g)  2NH 3(g)

Equilibrium Effect of Temperature on Equilbirum For an exothermic reaction, if the temperature increases, K decreases. N 2(g) + 3H 2(g)  2NH 3(g) + 92 kJ For an endothermic reaction, if the temperature increases, K increases. CaCO 3(s) kJ  CaO (s) + CO 2(g) slideshttp:\\asadipour.kmu.ac.ir

Equilibrium The Effect of Changes in Temperature Q+Co(H 2 O) 6 2+ (aq) + 4 Cl (aq) CoCl 4 (aq) + 6 H 2 O (l) slideshttp:\\asadipour.kmu.ac.ir +Q-Q

Equilibrium slideshttp:\\asadipour.kmu.ac.ir

Equilibrium Catalysts increase the rate of both the forward and reverse reactions slideshttp:\\asadipour.kmu.ac.ir Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

Equilibrium The Haber Process This apparatus helps push the equilibrium to the right by slideshttp:\\asadipour.kmu.ac.ir N 2(g) + 3H 2(g)  2NH 3(g) + 92 kJ Heating Catalyzing Cooling Pressure Refrigerating Removing

Equilibrium Offer to U slideshttp:\\asadipour.kmu.ac.ir54