AP Chemistry Notes Chapter 15 Chemical EquilibriumAP化學(xué)筆記15章化學(xué)平衡

1、AP Chemistry Notes: Chapter 13 Chemical EquilibriumMany reactions do not run to completion. But, reaction concentrations may cease to change. The system contains a mixture of reactants and products.A chemical equilibrium has been reached.There are physical equilibria as well: ex. constant vapor pres

2、sure in a closed container with a volatile liquid, or solution saturation.13.1 The Equilibrium ConditionA chemical equilibrium is a dynamic state. The forward reaction has not stopped. Rather, the reverse reaction rate has caught up. Both rates are now equal, so overall concentrations cease to chang

3、e.Forward reaction: A à Br = kfAReverse reaction: B à Ar = krBBeginning with all A and no B, A decreases as does the forward rate. B increases as does the reverse rate. Eventually, the rates become equal and kfA = krB.If we rearrange the equation, we get kf/kr = B/A = a constant.The Equili

4、brium Constant is a ratio of product to reactant concentrations.Reversibility is indicated by a set of double arrows: A BEx. N2(g) + 3H2(g) 2NH3(g)13.2 The Equilibrium ConstantThe equation above is for the Haber process for synthesizing ammonia from its elements. It is a tremendously important indus

5、trial reaction and is used to produce the majority of the worlds fixed nitrogen. The principal use is for agricultural fertilizers, although there are lots of others as well.The interesting thing about equilibria and reversible reactions, is that the equilibrium state can be reached from either dire

6、ction. We can start with all reactants, all products, or a mixture of the two and the system will still come to equilibrium with a fixed value for the constant.In the Haber process, N2(g) and H2(g) are mixed at about 200 atmospheres and 500ºC. The forward reaction does not go to completion. Aga

7、in, equilibrium can be reached from either direction. See the diagrams below: By changing the starting concentrations and examining the concentrations at equilibrium, we can learn more about the equilibrium system.In 1864, Guldberg and Waage postulated their Law of Mass Action. Rather than a long st

8、atement, a simple example is best: for jA + kB lC + mDKc = Here, = molar concentration, however, equil. constant expressions may be written with pressures for gases as well.Usually, equilibrium equations are written with the products of interest on the right. However, the equations are reversible. R

9、eversing an equilibrium equation results in the reciprocal of K (1/K).Write K for this reaction: 4 NH3(g) + 7 O2(g) à 4 NO2(g) + 6 H2O(g)Kc = Special Notes:· Do not confuse equilibrium constant “K” with the rate constant "k."· Products are always on top, reactants are always

10、 on the bottom.· The value of K at a given temperature does not depend on initial concentrations.· The Law of Mass Action applies only to systems at equilibrium.· There may also be other substances present, but as long as they do not consume the materials in question, they do not affe

11、ct K (ex. catalysts or inert gases.)· K does vary with temperature.· K is typically expressed without units. We are usually more interested in the value of the number than in the units.Table 13.1 on P. 586 in your text shows the results of several experiments with the Haber process. Note t

12、he values of K in each experiment. What do you conclude from this information?When reactant and product materials are expressed using molar concentrations ( ), Kc is used.13.3 Equilibrium Expressions Involving PressureIt is also possible to write K expressions using the partial pressure values for g

13、aseous equilibria.The symbols look like this: PNH3 where P stands for pressure (usually in atomospheres) and the subscript is for the gas in question.In the space below, write the equilibrium expressions for the above Haber process in terms of pressure, and using Kp instead of Kc.· It is possib

14、le to relate Kc to Kp through a simple equation: Kp = Kc(RT) Dn · where Dn is the change in the number of moles of gas: total moles product gases - total moles reactant gases, · T is Kelvin temp. and · R is the universal gas constant in the appropriate units.13.4 Heterogeneous Equilib

15、riaIn homogeneous equilibria, like the examples we have seen so far, all reactants and products are in the same phase.In heterogeneous equilibria, the materials are not all in the same phase. As it turns out, only gases and dissolved materials are really subject to any appreciable concentration chan

16、ge as reactions occur.Solids and liquids by comparison have concentrations that are so high, that they are essentially constant. Therefore, solids and liquids are left out of equilibrium constant expressions. Either they are present, in which case equilibrium can be reached, or they are not.Ex. CaCO

17、3(s) CaO(s) + CO2(g)Kc = CO2 or Kp = PCO213.5 Applications of Equilibrium ConstantsKnowing the equilibrium constant for a system gives the chemist valuable information about several things:1. the tendency for the reaction to occur at all (but not how fast it may occur),2. how much product will form

18、at a given temperature, and3. whether a given set of concentrations will see the system at equilibrium or not.Pp. 591-2 in your text have some representative pictures of how an equilibrium constant is calculated.The Extent of a ReactionIf the numerical value of K is small (<<1) the reaction li

19、es far to the left, that is, there are more reactants in the system than products, or we say, "The reactants are favored."Conversely, if K is large (>>1), the products are favored.Ex. production of phosgene: CO(g) + Cl2(g) COCl2(g) Kc = 4.57x 109Ex. N2(g) + O2(g) 2NO(g) Kc = 1 x 10-3

20、0K must be calculated using experimental values for concentrations.Also, if K is known, equilibrium s can be calculated if just one value is known. This is a major type of problem for this unit and they get really fun!Sometimes we do not know all of the concentrations of products and reactants at eq

21、uilibrium, so if we know initial concentrations and K, we can calculate the others. To do this we use an ICE Table.ICE stands for Initial, Change, and Equilibrium. These tables are very commonly used in this unit and for all types of equilibrium problems. You should become very familiar with them.IC

22、E tables look like this: Reactants ProductsSee example problem 13.11 on p. 599 regarding use of an ICE table.Predicting reaction direction: The Reaction QuotientIf a system is not at equilibrium, it is possible to project which way the reaction will go to reach equilibrium (toward the product side o

23、r toward the reactant side.)This is a simple process: Just write the K expression and plug in the current concentration or pressure values. Calculate the resulting value, which is called the Reaction Quotient (Q).Q is calculated exactly the same way as K.Then compare Q with the actual K value at tha

24、t temperature. IfQ > K, there is too much product and the equilibrium will shift to the left ßQ = K, the system is at equilibrium and will not shift.Q < K, there is too much reactant and the equilibrium will shift to the right àDetermining Equilibrium Concentrations (or "And You

25、 Thought There Wasnt a Use for the Quadratic Equation!")It is possible using the equilibrium constant expression to calculate the concentrations of reactants and/or products at equilibrium.Here is an example problem from the book:Kc = HI2 / H2I2 H2 (g) + I2 (g) 2HI(g)At 448ºC, Kc has a val

26、ue of 50.5.Substituting the Equil. concentration values into Kc: (2x)2 / (1.000 - x)(2.000 - x) = 50.5If this is beginning to look suspiciously like something from a recent math class, its because IT IS!4x2 = 50.5(x2 - 3.000x + 2.000) à 46.5x2 - 151.5x + 101.0 = 0x = -(-151.5)± (-151.5)2 -

27、 4(46.5)(101.0)= 2.323 or 0.9352(46.5)Typically one of the options from the quadratic equation will be automatically eliminated. In this case it is the 2.323 value, since 1.000 M -2.323 gives a negative concentration value, which is impossible. So we choose x = 0.935.H2 = 1.000 - 0.935 = 0.065 MI2 =

28、 2.000 - 0.935 = 1.065 MHI = 2x = 1.870 MThe results can be double checked by plugging the values above into the equilibrium expressions. If the resulting answer is 50.5, the problem has been done correctly.On the bright side, it will be only a rare case when you will actually need to use the quadra

29、tic equation to solve problems of this type. We will talk about why in the near future.13.6 Solving Equilibrium ProblemsProcedure for solving most equilibrium problems follows this pattern:1. Write a balanced equation for the equilibrium reaction.2. Write the equilibrium expression (K) using the law

30、 of mass action.3. List the initial concentrations of reactants and products.4. If K is known, calculate Q to see which direction the system will shift to reach equilibrium. (If any reactant or product has a concentration = 0, the system will automatically shift to produce that substance and there i

31、s no need to calculate Q).5. Draw an ICE table and define the change needed to reach equilibrium. Apply the change to the initial concentrations and find the final (equilibrium) concentrations.6. Substitute the equilibrium concentration values into the K expression and solve for the unknown.7. Doubl

32、e check your answers by substituting them into the K expression and making sure they give the correct K value.13.7 Le Chateliers Principle: Factors Affecting EquilibriumLe Chats Principle simply stated is: When subjected to a stress, an equilibrium system will shift in order to relieve that stress.P

33、ut more simply still, an equilibrium system will shift toward a decrease and away from an increase.What constitutes a stress?· A change in a reactant or product concentration. An equilibrium will shift to use up something that is being added, or to replace something that is being lost.· A

34、pressure change in systems containing gases. If pressure is increased by reducing volume, the equilibrium will shift toward the side with the fewest moles of gas to relieve the pressure. If P is decreased, a shift toward the side with more gas moles will occur.*Note: the addition of an inert gas doe

35、s not alter the partial pressures of any of the reacting gases (even though the total pressure increases) so the equilibrium is not affected.· A temperature change. All reactions are exothermic in one direction and endothermic in the other. Raising the temperature favors the endothermic reactio

36、n and hinders the exothermic one. Remember, temperature is the only factor that affects the value of K.Energy can be added to any equation and treated just like any other component with respect to Le Chats Principle when predicting shifts in response to stress.Table 13.2 from your text nicely illust

37、rates how temperature and pressure affect the Haber process.Which conditions produce the highest yield of ammonia?N2(g) + 3H2(g) 2NH3(g) + EnergyHere, raising the temperature causes a shift away from the energy, or to the left. NH3 decreases and N2 and H2 both increase.The high pressure favors the s

38、ide of the equilibrium with the fewest moles of gas (fewer molecules produce less pressure).Be sure you can predict not only which way an equilibrium will shift in response to a stress, but what will happen to the s of all materials as a result of the shift.Effect of a CatalystCatalysts affect forward and reverse reactions equally and so adding a catalyst a