1 Chemical Equilibriu m Chapter 14 Nitrogen dioxide, NO 2 (a reddish brown gas), is in equilibrium with dinitrogen tetroxide, N 2 O 4 (a colorless gas). Nitrogen dioxide is present in many urban smogs, giving them a characteristic reddish brown color.

2 Chemical reactions often seem to stop before they are complete. When these two reactions—forward and reverse—occur at the same rate, a chemical equilibrium exists. Actually, such reactions are reversible. That is, the original reactants form products, but then the products react with themselves to give back the original reactants.

3 Chemical Equilibrium When compounds react, they eventually form a mixture of products and unused reactants, in a dynamic equilibrium. –A dynamic equilibrium consists of a forward reaction, in which substances react to give products, and a reverse reaction, in which products react to give the original reactants. –Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.

4 –Much like water in a U-shaped tube, there is constant mixing back and forth through the lower portion of the tube. “reactants”“products” –It’s as if the forward and reverse “reactions” were occurring at the same rate. –The system appears to static (stationary) when, in reality, it is dynamic (in constant motion). See Figure 15.3

5 Figure 14.3: Catalytic Methanation Reaction Approaches Equilibrium

6 Chemical Equilibrium For example, the Haber process for producing ammonia from N 2 and H 2 does not go to completion. –It establishes an equilibrium state where all three species are present. (g)2NH )g(H3)g(N 322 

7 A Problem to Consider –The equilibrium amount of NH 3 was given as 0.080 mol. Therefore, 2x = 0.080 mol NH 3 (x = 0.040 mol). Using the information given, set up the following table. (g)2NH )g(H3 )g(N 322  Starting 1.0003.0000 Change -x-3x+2x Equilibrium 1.000 - x3.000 - 3x2x = 0.080 mol

8 A Problem to Consider Using the information given, set up the following table. (g)2NH )g(H3 )g(N 322  Starting 1.0003.0000 Change -x-3x+2x Equilibrium 1.000 - x3.000 - 3x2x = 0.080 mol Equilibrium amount of N 2 = 1.000 - 0.040 = 0.960 mol N 2 Equilibrium amount of H 2 = 3.000 - (3 x 0.040) = 2.880 mol H 2 Equilibrium amount of NH 3 = 2x = 0.080 mol NH 3

9 The Equilibrium Constant Every reversible system has its own “position of equilibrium” under any given set of conditions. –The ratio of products produced to unreacted reactants for any given reversible reaction remains constant under constant conditions of pressure and temperature. –The numerical value of this ratio is called the equilibrium constant for the given reaction.

10 The Equilibrium Constant The equilibrium-constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation. ba dc c ]B[]A[ ]D[]C[ K  –For the general equation above, the equilibrium-constant expression would be: dDcC bBaA  –The molar concentration of a substance is denoted by writing its formula in square brackets.

11 The Equilibrium Constant The equilibrium constant, K c, is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted. –A large K c indicates large concentrations of products at equilibrium. –A small K c indicates large concentrations of unreacted reactants at equilibrium.

12 The Equilibrium Constant The law of mass action states that the value of the equilibrium constant expression K c is constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted. –Consider the equilibrium established in the Haber process. (g)2NH )g(H3)g(N 322 

13 The Equilibrium Constant The equilibrium-constant expression would be –Note that the stoichiometric coefficients in the balanced equation have become the powers to which the concentrations are raised. (g)2NH )g(H3)g(N 322  3 22 2 3 c ]H][N[ ]NH[ K 

14 Equilibrium: A Kinetics Argument If the forward and reverse reaction rates in a system at equilibrium are equal, then it follows that their rate laws would be equal. –If we start with some dinitrogen tetroxide and heat it, it begins to decompose to produce NO 2. –Consider the decomposition of N 2 O 4, dinitrogen tetroxide. (g)2NO )g(ON 242 –However, once some NO 2 is produced it can recombine to form N 2 O 4.

15 Equilibrium: A Kinetics Argument –Call the decomposition of N 2 O 4 the forward reaction and the formation of N 2 O 4 the reverse reaction. (g)2NO )g(ON 242 kfkf krkr –These are elementary reactions, and you can immediately write the rate law for each. ]ON[kRate 42f(forward)  2 2r(reverse) ]NO[kRate  Here k f and k r represent the forward and reverse rate constants.

16 (g)2NO )g(ON 242 –Ultimately, this reaction reaches an equilibrium state where the rates of the forward and reverse reactions are equal. Therefore, 2 2r42f ]NO[k]ON[k  –Combining the constants you can identify the equilibrium constant, Kc, as the ratio of the forward and reverse rate constants. ]ON[ ]NO[ k k K 42 2 2 r f c 

17 Figure 14.4: Temperature Effect on NO 2 -N 2 O 4 Equilibrium Do exercise 14.2 See Problems 14.35 and 36 Do Now Page 585 Example 14.2 a. b. c. d.

18 Obtaining Equilibrium Constants for Reactions Equilibrium concentrations for a reaction must be obtained experimentally and then substituted into the equilibrium-constant expression in order to calculate K c. Consider the reaction below (see Figure 15.5). O(g)H (g)CH (g)H 3 )g(CO 242  –Suppose we started with initial concentrations of CO and H 2 of 0.100 M and 0.300 M, respectively.

19 O(g)H (g)CH (g)H 3 )g(CO 242  Consider the reaction below (see Figure 15.5). –Suppose we started with initial concentrations of CO and H 2 of 0.100 M and 0.300 M, respectively. –When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows. Reactants Products [CO] = 0.0613 M [H 2 ] = 0.1893 M [CH 4 ] = 0.0387 M [H 2 O] = 0.0387 M

20 –The equilibrium-constant expression for this reaction is: 3 2 24 c ]H][CO[ ]OH][CH[ K  –If we substitute the equilibrium concentrations, we obtain: 93.3 )M1839.0)(M0613.0( )M0387.0)(M0387.0( K 3 c  –Regardless of the initial concentrations (whether they be reactants or products), the law of mass action dictates that the reaction will always settle into an equilibrium where the equilibrium-constant expression equals K c.

21 Figure 14.5: Some equilibrium compositions for the methanation reaction

22 –As an example, let’s repeat the previous experiment, only this time starting with initial concentrations of products: [CH 4 ] initial = 0.1000 M and [H 2 O] initial = 0.1000 M O(g)H (g)CH (g)H 3 )g(CO 242  –We find that these initial concentrations result in the following equilibrium concentrations. ReactantsProducts [CO] = 0.0613 M [H 2 ] = 0.1893 M [CH 4 ] = 0.0387 M [H 2 O] = 0.0387 M

23 –Substituting these values into the equilibrium- constant expression, we obtain the same result. 93.3 )M1839.0)(M0613.0( )M0387.0)(M0387.0( K 3 c  –Whether we start with reactants or products, the system establishes the same ratio. Do exercise 14.3See problems 14.43-44 Do Now - Example 14.3

24 Figure 15.5: Some Equilibrium Composition for the Methanation Reaction Do exercise 14.4See problems 14.45-48

25 The Equilibrium Constant, K p In discussing gas-phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities (see Figure 14.6 –It can be seen from the ideal gas equation that the partial pressure of a gas is proportional to its molarity. MRTRT V n P )( 

26 Figure 14.6: The Concentration of a Gas at a Given Temperature is Proportional to the Pressure

27 If we express a gas-phase equilibria in terms of partial pressures, we obtain K p. –Consider the reaction below. O(g)H (g)CH (g)H 3 )g(CO 242  –The equilibrium-constant expression in terms of partial pressures becomes: 3 HCO OHCH p 2 24 K PP PP  In general, the numerical value of K p differs from that of K c.

28 –From the relationship n/V=P/RT, we can show that where  n is the sum of the moles of gaseous products in a reaction minus the sum of the moles of gaseous reactants. n cp )RT(KK   Consider the reaction (g)SO 2 )g(O)g( 2 322  –K c for the reaction is 2.8 x 10 2 at 1000 o C. Calculate K p for the reaction at this temperature.

29 A Problem to Consider –We know that n cp )RT(KK   From the equation we see that  n = -1. We can simply substitute the given reaction temperature and the value of R (0.08206 L. atm/mol. K) to obtain K p. Consider the reaction (g)SO 2 )g(O)g( 2 322 

30 –Since n cp )RT(KK   3.4 K) 1000 08206.0( 108.2K 1- Kmol atmL 2 p    Do exercise 14.5See problems 14.51-54

31 Page 591 123123 1 with A = red 2 with Either 3 with Either

32 Equilibrium Constant for the Sum of Reactions Similar to the method of combining reactions that we saw using Hess’s law in Chapter 6, we can combine equilibrium reactions whose K c values are known to obtain K c for the overall reaction. –With Hess’s law, when we reversed reactions or multiplied them prior to adding them together, we had to manipulate the  H’s values to reflect what we had done. –The rules are a bit different for manipulating K c.

33 2.If you multiply each of the coefficients in an equation by the same factor (2, 3, …), raise K c to the same power (2, 3, …). 3.If you divide each coefficient in an equation by the same factor (2, 3, …), take the corresponding root of K c (i.e., square root, cube root, …). 4.When you finally combine (that is, add) the individual equations together, take the product of the equilibrium constants to obtain the overall K c. 1.If you reverse a reaction, invert the value of K c. Equilibrium Constant for the Sum of Reactions

34 For example, nitrogen and oxygen can combine to form either NO(g) or N 2 O (g) according to the following equilibria. NO(g) 2 )g(O)g(N 22  O(g)N )g(O)g(N 22 2 1 2  K c = 4.1 x 10 -31 K c = 2.4 x 10 -18 (1) (2) K c = ? –Using these two equations, we can obtain K c for the formation of NO(g) from N 2 O(g): NO(g) 2 )g(O)g(ON 2 2 1 2  (3) Equilibrium Constant for the Sum of Reactions

35 To combine equations (1) and (2) to obtain equation (3), we must first reverse equation (2). When we do we must also take the reciprocal of its K c value. NO(g) 2 )g(O)g(ON 2 2 1 2  2 )g(O)g(N 22  K c = 4.1 x 10 -31 (1) )g(O (g)N O(g)N 2 2 1 22  K c = (2) (3) 18- 10 4.2 1  13 18 31 c 107.1 4.2 1 ) 1.4()overall(K )(       Equilibrium Constant for the Sum of Reactions Check example page 591

36 Heterogeneous Equilibria A heterogeneous equilibrium is an equilibrium that involves reactants and products in more than one phase. –The equilibrium of a heterogeneous system is unaffected by the amounts of pure solids or liquids present, as long as some of each is present. –The concentrations of pure solids and liquids are always considered to be “1” and therefore, do not appear in the equilibrium expression.

37 Heterogeneous Equilibria Consider the reaction below. (g)H CO(g) )g(OH)s(C 22  –The equilibrium-constant expression contains terms for only those species in the homogeneous gas phase…H 2 O, CO, and H 2. ]OH[ ]H][CO[ K 2 2 c  Do exercise 14.6See problems 14.55-56 Do Example 14.4 Now! Page 593

38 Using the Equilibrium Constant 1. Qualitatively interpreting the equilibrium constant 2. Predicting the direction of reaction 3. Calculating equilibrium concentrations Do Exercise 14.7 Now See Problems 14. 56-60

39 Predicting the Direction of Reaction How could we predict the direction in which a reaction at non-equilibrium conditions will shift to reestablish equilibrium? –To answer this question, substitute the current concentrations into the reaction quotient expression and compare it to K c. –The reaction quotient, Q c, is an expression that has the same form as the equilibrium-constant expression but whose concentrations are not necessarily at equilibrium.

40 Predicting the Direction of Reaction For the general reaction dDcC bBaA  the Q c expresssion would be: ba dc c ]B[]A[ ]D[]C[ Q ii ii 

41 Predicting the Direction of Reaction For the general reaction –If Q c < K c, the reaction will shift right… toward products. –If Q c = K c, then the reaction is at equilibrium. dDcC bBaA  –If Q c > K c, the reaction will shift left…toward reactants.

42 Figure 14.10: Direction of reaction

43 A Problem to Consider –Consider the following equilibrium. –A 50.0 L vessel contains 1.00 mol N 2, 3.00 mol H 2, and 0.500 mol NH 3. In which direction (toward reactants or toward products) will the system shift to reestablish equilibrium at 400 o C? –K c for the reaction at 400 o C is 0.500. (g)2NH )g(H3 )g(N 322  –First, calculate concentrations from moles of substances. 1.00 mol 50.0 L 3.00 mol 50.0 L 0.500 mol 50.0 L 0.0100 M0.0600 M0.0200 M

44 –The Q c expression for the system would be: 3 22 2 3 c ]H][N[ ]NH[ Q  0.0100 M 0.0600 M0.0200 M 1.23 )0600.0)(0200.0( )0100.0( Q 3 2 c  (g)2NH )g(H3 )g(N 322  Because Q c = 23.1 is greater than K c = 0.500, the reaction Will go to the left as it approaches equilibrium. Therefore, Ammonia will dissociate Do exercise 14.8 See problems 14.61-62

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46 Calculating Equilibrium Concentrations Once you have determined the equilibrium constant for a reaction, you can use it to calculate the concentrations of substances in the equilibrium mixture. O(g)H (g)CH (g)H 3 )g(CO 242  – For example, consider the following equilibrium. –Suppose a gaseous mixture contained 0.30 mol CO, 0.10 mol H 2, 0.020 mol H 2 O, and an unknown amount of CH 4 per liter. –What is the concentration of CH 4 in this mixture? The equilibrium constant K c equals 3.92.

47 Calculating Equilibrium Concentrations –First, calculate concentrations from moles of substances. 0.30 mol 1.0 L 0.10 mol 1.0 L 0.020 mol 1.0 L ?? O(g)H (g)CH (g)H 3 )g(CO 242  ??0.30 M0.10 M0.020 M –The equilibrium-constant expression is: 3 2 24 c ]H][CO[ ]OH][CH[ K 

48 –Substituting the known concentrations and the value of K c gives: 3 4 )M10.0)(M30.0( )M020.0](CH[ 92.3  –You can now solve for [CH 4 ]. 059.0 )M020.0( )M10.0)(M30.0)(92.3( ]CH[ 3 4  –The concentration of CH 4 in the mixture is 0.059 mol/L.

49 Calculating Equilibrium Concentrations Suppose we begin a reaction with known amounts of starting materials and want to calculate the quantities at equilibrium. –Consider the following equilibrium. Suppose you start with 1.000 mol each of carbon monoxide and water in a 50.0 L container. Calculate the molarity of each substance in the equilibrium mixture at 1000 o C. K c for the reaction is 0.58 at 1000 o C.

50 –First, calculate the initial molarities of CO and H 2 O. 1.000 mol 50.0 L 1.000 mol 50.0 L (g)H CO )g(OH)g( 222  0.0200 M 0 M The starting concentrations of the products are 0. We must now set up a table of concentrations (starting, change, and equilibrium expressions in x). Starting0.0200 00 Change-x +x Equilibrium0.0200-x xx

51 –The equilibrium-constant expression is: ]OH][CO[ ]H][CO[ K 2 22 c  Starting 0.0200 00 Change -x +x Equilibrium 0.0200-x xx –Substituting the values for equilibrium concentrations, we get: )x0200.0)(x0200.0( )x)(x( 58.0   (g)H CO )g(OH)g( 222 

52 Calculating Equilibrium Concentrations –Solving for x. x +x 000.0200 Starting x0.0200-x Equilibrium +x -x Change –Or: 2 2 )x0200.0( x 58.0   (g)H CO )g(OH)g( 222  –Taking the square root of both sides we get: )x0200.0( x 76.0  

53 –Rearranging to solve for x gives: –Solving for equilibrium concentrations. Starting 0.0200 00 Change -x +x Equilibrium 0.0200-x xx –If you substitute for x in the last line of the table you obtain the following equilibrium concentrations. 0.0114 M CO 0.0114 M H 2 O0.0086 M H 2 0.0086 M CO 2 Do Exercise 14.10 Look at Problems 14.67-68 0086.0 76.1.00200.0 x   

54 Calculating Equilibrium Concentrations The preceding example illustrates the three steps in solving for equilibrium concentrations. 1.Set up a table of concentrations (starting, change, and equilibrium expressions in x). 2.Substitute the expressions in x for the equilibrium concentrations into the equilibrium-constant equation. 3.Solve the equilibrium-constant equation for the values of the equilibrium concentrations.

55 Calculating Equilibrium Concentrations In some cases it is necessary to solve a quadratic equation to obtain equilibrium concentrations. The next example illustrates how to solve such an equation.

56 Calculating Equilibrium Concentrations –Consider the following equilibrium. Suppose 1.00 mol H 2 and 2.00 mol I 2 are placed in a 1.00-L vessel. How many moles per liter of each substance are in the gaseous mixture when it comes to equilibrium at 458 o C? K c at this temperature is 49.7. HI(g)2 )g(I)g(H 22 

57 Calculating Equilibrium Concentrations –The concentrations of substances are as follows. 2x +2x 02.001.00Starting 2.00-x1.00-xEquilibrium -x Change –The equilibrium-constant expression is: HI(g)2 )g(I)g(H 22  ]I][H[ ]HI[ K 22 2 c 

58 –Substituting our equilibrium concentration expressions gives: )x00.2)(x00.1( )x2( K 2 c   –Solving for x. Starting1.002.000 Change-x +2x Equilibrium1.00-x2.00-x2x –Because the right side of this equation is not a perfect square, you must solve the quadratic equation.

59 –Solving for x. –The equation rearranges to give: 2x +2x 02.001.00Starting 2.00-x1.00-xEquilibrium -x Change HI(g)2 )g(I)g(H 22  000.2x.3x920.0 2  –The two possible solutions to the quadratic equation are: 0.93x and 33.2x 

60 –Solving for x. –However, x = 2.33 gives a negative value to 1.00 - x (the equilibrium concentration of H 2 ), which is not possible. 2x +2x 02.001.00Starting 2.00-x1.00-xEquilibrium -x Change HI(g)2 )g(I)g(H 22  remains. 0.93Only x  –If you substitute 0.93 for x in the last line of the table you obtain the following equilibrium concentrations. 0.07 M H 2 1.07 M I 2 1.86 M HI Do Exercise 14.11 Look at Problems 14.69-70

61 Concept Check Page 601

62 Le Chatelier’s Principle Obtaining the maximum amount of product from a reaction depends on the proper set of reaction conditions. –Le Chatelier’s principle states that when a system in a chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the equilibrium will shift in a way that tends to counteract this change.

63 Changing the Reaction Conditions Le Chatelier’s Principle 1.Change concentrations by adding or removing products or reactants. 2.Changing the partial pressure of gaseous reactants or products by changing the volume. 3.Changing the temperature. 4.(Catalyst)

64 Removing Products or Adding Reactants Let’s refer back to the illustration of the U-tube in the first section of this chapter. –It’s a simple concept to see that if we were to remove products (analogous to dipping water out of the right side of the tube) the reaction would shift to the right until equilibrium was reestablished. “reactants”“products” –Likewise, if more reactant is added (analogous to pouring more water in the left side of the tube) the reaction would again shift to the right until equilibrium is reestablished.

65 Effects of Pressure Change A pressure change caused by changing the volume of the reaction vessel can affect the yield of products in a gaseous reaction only if the reaction involves a change in the total moles of gas present See Figure 14.12

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67 Effects of Pressure Change If the products in a gaseous reaction contain fewer moles of gas than the reactants, it is logical that they would require less space. So, reducing the volume of the reaction vessel would, therefore, favor the products. Conversely, if the reactants require less volume (that is, fewer moles of gaseous reactant), then decreasing the volume of the reaction vessel would shift the equilibrium to the left (toward reactants).

68 Literally “squeezing” the reaction will cause a shift in the equilibrium toward the fewer moles of gas. It’s a simple step to see that reducing the pressure in the reaction vessel by increasing its volume would have the opposite effect. In the event that the number of moles of gaseous product equals the number of moles of gaseous reactant, vessel volume will have no effect on the position of the equilibrium. Do Exercise 14.13 Look at Problems 14.75-76 See Example 14.10

69 Effect of Temperature Change Temperature has a significant effect on most reactions (See Figure 14.13) –Reaction rates generally increase with an increase in temperature. Consequently, equilibrium is established sooner. –In addition, the numerical value of the equilibrium constant K c varies with temperature.

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71 Effect of Temperature Change Let’s look at “heat” as if it were a product in exothermic reactions and a reactant in endothermic reactions. We see that increasing the temperature is analogous to adding more product (in the case of exothermic reactions) or adding more reactant (in the case of endothermic reactions). This ultimately has the same effect as if heat were a physical entity.

72 Effect of Temperature Change For example, consider the following generic exothermic reaction. Increasing temperature would be analogous to adding more product, causing the equilibrium to shift left. Since “heat” does not appear in the equilibrium- constant expression, this change would result in a smaller numerical value for K c. negative) is H( heat""products reactants  

73 Effect of Temperature Change For an endothermic reaction, the opposite is true. Increasing temperature would be analogous to adding more reactant, causing the equilibrium to shift right. This change results in more product at equilibrium, amd a larger numerical value for K c. positive) is H( products reactantsheat""  

74 Effect of Temperature Change In summary: –For an exothermic reaction (  H is negative) the amounts of reactants are increased at equilibrium by an increase in temperature (K c is smaller at higher temperatures). –For an endothermic reaction (  H positive) the amounts of products are increased at equilibrium by an increase in temperature (K c is larger at higher temperatures). Do Exercises 14.14 and 15Look at Problems 14.77-78, 81-82

75 At Higher T

76 Effect of a Catalyst A catalyst is a substance that increases the rate of a reaction but is not consumed by it. –It is important to understand that a catalyst has no effect on the equilibrium composition of a reaction mixture (see Figure 15.15).(see Figure 15.15). –A catalyst merely speeds up the attainment of equilibrium.

77 Oxidation of Ammonia Using a copper catalyst Results in N 2 and H 2 O Using a platinum catalyst Results in NO and H 2 O

78 Operational Skills Applying stoichiometry to an equilibrium mixture Writing equilibrium-constant expressions Obtaining the equilibrium constant from reaction composition Using the reaction quotient Obtaining one equilibrium concentration given the others Solving equilibrium problems Applying Le Chatelier’s principle