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Chemical Equilibrium 1 Equilibrium Initially all liquid Gas only, produced Balance of gas and liquid production 2 Equilibrium When compounds react, they eventually form a mixture of products and unreacted 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. 3 Chemical Equilibrium H2 + I2 < -- > 2HI – Initially only H2 and I2 are present. – The rxn. proceeds only – As HI concentration increases, some HI is able to decompose back into H2 and I2 – Rxn. proceeds < -- > – At some point – The rate of H2 + I2 -- > 2HI equals – The rate of 2HI -- > H2 + I2 Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal. 4 See Le Chatelier Chemical Equilibrium For example, the Haber process for producing ammonia from N2 and H2 does not go to completion. N 2 ( g ) 3H 2 ( g ) 2NH3 (g) – It establishes an equilibrium state where all three species are present. (see Figure 15.3) 6 Chemical Equilibrium Chemical Equilibrium is a fundamentally important concept to master because most chemical reactions fail to go to completion. 7 A Problem to Consider Applying Stoichiometry to an Equilibrium Mixture. – Suppose we place 1.000 mol N2 and 3.000 mol H2 in a reaction vessel at 450 oC and 10.0 atmospheres of pressure. The reaction is N 2 ( g ) 3H 2 ( g ) 2NH3 (g) – What is the composition of the equilibrium mixture if it contains 0.080 mol NH3? 8 A Problem to Consider Using the information given, set up an ICE table. N 2 ( g ) 3H 2 ( g ) 2NH3 (g) Initial 1.000 3.000 0 Change -x -3x +2x Equilibrium 1.000 - 3.000 - 2x = 0.080 mol x 3x – The equilibrium amount of NH3 was given as 0.080 mol. Therefore, 2x = 0.080 mol NH3 (x = 0.040 mol). 9 A Problem to Consider Using the information given, set up the following table. N 2 ( g ) 3H 2 ( g ) 2NH3 (g) Initial 1.000 3.000 0 Change -x -3x +2x Equilibrium 1.000 - 3.000 - 2x = 0.080 mol x 3x Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2 Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2 Equilibrium amount of NH3 = 2x = 0.080 mol NH3 10 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. 11 The Equilibrium Constant H2 + I2 < -- > 2HI Rate of forward rxn: Ratef = kf [H2][I2] Rate of reverse rxn: Rater = kr[HI]2 At equilibrium: kf [H2][I2] =kr[HI]2 Therefore: kf = [HI]2 kr [H][I] Kc = [HI]2 [H][I] 12 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. aA bB cC dD – For the general equation above, the c d [C] [D] equilibrium-constant expression would be: Kc a b [ A ] [B ] 13 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. aA bB cC dD – The molar concentration of a substance c d [C] [D] is denoted by writing its formula in square brackets. Kc a b [ A ] [B ] 14 The Equilibrium Constant The equilibrium constant, Kc, is the value obtained for the equilibrium- constant expression when equilibrium concentrations are substituted. – A large Kc indicates large concentrations of products at equilibrium. – A small Kc indicates large concentrations of unreacted reactants at equilibrium. 15 The Equilibrium Constant The law of mass action states that the value of the equilibrium constant expression Kc is constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted. – Consider the equilibrium established in the Haber process. N 2 ( g ) 3H 2 ( g ) 2NH3 (g) 16 The Equilibrium Constant The equilibrium-constant expression would be 2 [NH 3 ] Kc 3 [N 2 ][H 2 ] – Note that the stoichiometric coefficients in the balanced equation have become the powers to which the concentrations are raised. N 2 ( g ) 3H 2 ( g ) 2NH3 (g) 17 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. 18 Calculating Equilibrium Concentrations – For example, consider the following equilibrium. CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – Suppose a gaseous mixture contained 0.30 mol CO, 0.10 mol H2, 0.020 mol H2O, and an unknown amount of CH4 per liter. – What is the concentration of CH4 in this mixture? The equilibrium constant Kc equals 3.92. 19 Calculating Equilibrium Concentrations First, calculate concentrations from moles of substances. CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) 0.30 mol 0.10 mol 0.020 mol ?? 1.0 L 1.0 L 1.0 L 20 Calculating Equilibrium Concentrations First, calculate concentrations from moles of substances. CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) 0.30 M 0.10 M ?? 0.020 M – The equilibrium-constant expression is: [CH 4 ][H 2O] Kc 3 [CO][H 2 ] 21 Calculating Equilibrium Concentrations First, calculate concentrations from moles of substances. CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) 0.30 M 0.10 M ?? 0.020 M – Substituting the known concentrations and the value of Kc gives: [CH 4 ](0.020M ) 3.92 3 (0.30M )(0.10M ) 22 Calculating Equilibrium Concentrations First, calculate concentrations from moles of substances. CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) 0.30 M 0.10 M ?? 0.020 M – You can now solve for [CH4]. ( 3.92)(0.30M )(0.10M )3 [CH 4 ] 0.059 (0.020M ) – The concentration of CH4 in the mixture is 0.059 mol/L. 23 Calculating Equilibrium Concentrations Suppose we begin a reaction with known amounts of starting materials and want to calculate the quantities at equilibrium. 24 Calculating Equilibrium Concentrations Consider the following equilibrium. CO(g ) H 2O(g ) CO2 (g) H 2 (g) • 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 oC. • Kc for the reaction is 0.58 at 1000 oC. 25 Calculating Equilibrium Concentrations – First, calculate the initial molarities of CO and H2O. CO(g ) H 2O(g ) CO2 (g) H 2 (g) 1.000 mol 1.000 mol 50.0 L 50.0 L 26 Calculating Equilibrium Concentrations – First, calculate the initial molarities of CO and H2O. CO(g ) H 2O(g ) CO2 (g) H 2 (g) 0.0200 M 0.0200 M 0M 0M • The starting concentrations of the products are 0. • We must now set up a table of concentrations (starting, change, and equilibrium expressions in x). 27 Calculating Equilibrium Concentrations – Let x be the moles per liter of product formed. CO(g ) H 2O(g ) CO2 (g) H 2 (g) Initial 0.0200 0.0200 0 0 Change -x -x +x +x Equilibrium 0.0200-x 0.0200-x x x – The equilibrium-constant expression is: [CO2 ][H 2 ] Kc [CO][H 2O] 28 Calculating Equilibrium Concentrations – Solving for x. CO(g ) H 2O(g ) CO2 (g) H 2 (g) Initial 0.0200 0.0200 0 0 Change -x -x +x +x Equilibrium 0.0200-x 0.0200-x x x – Substituting the values for equilibrium concentrations, we get: ( x )(x ) 0.58 (0.0200 x )(0.0200 x ) 29 Calculating Equilibrium Concentrations – Solving for x. CO(g ) H 2O(g ) CO2 (g) H 2 (g) Initial 0.0200 0.0200 0 0 Change -x -x +x +x Equilibrium 0.0200-x 0.0200-x x x – Or: 2 x 0.58 (0.0200 x ) 2 30 Calculating Equilibrium Concentrations – Solving for x. CO(g ) H 2O(g ) CO2 (g) H 2 (g) Initial 0.0200 0.0200 0 0 Change -x -x +x +x Equilibrium 0.0200-x 0.0200-x x x – Taking the square root of both sides we get: x 0.76 (0.0200 x ) 31 Calculating Equilibrium Concentrations – Solving for x. CO(g ) H 2O(g ) CO2 (g) H 2 (g) Initial 0.0200 0.0200 0 0 Change -x -x +x +x Equilibrium 0.0200-x 0.0200-x x x – Rearranging to solve for x gives: 0.0200 0.76 x 0.0086 1.76 32 Calculating Equilibrium Concentrations – Solving for equilibrium concentrations. CO(g ) H 2O(g ) CO2 (g) H 2 (g) Initial 0.0200 0.0200 0 0 Change -x -x +x +x Equilibrium 0.0200-x 0.0200-x x x – If you substitute for x in the last line of the table you obtain the following equilibrium concentrations. 0.0114 M CO 0.0086 M CO2 0.0114 M H2O 0.0086 M H2 33 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. 34 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. 35 Calculating Equilibrium Concentrations – Consider the following equilibrium. H 2 (g ) I 2 (g ) 2HI(g) • Suppose 1.00 mol H2 and 2.00 mol I2 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 oC? • Kc at this temperature is 49.7. 36 Calculating Equilibrium Concentrations The concentrations of substances are as follows. H 2 (g ) I 2 (g ) 2HI(g) Initial 1.00 2.00 0 Change -x -x +2x Equilibrium 1.00-x 2.00-x 2x – The equilibrium-constant expression is: 2 [HI] Kc [H 2 ][I 2 ] 37 Calculating Equilibrium Concentrations The concentrations of substances are as follows. H 2 (g ) I 2 (g ) 2HI(g) Initial 1.00 2.00 0 Change -x -x +2x Equilibrium 1.00-x 2.00-x 2x – Substituting our equilibrium concentration expressions gives: 2 ( 2x ) Kc (1.00 x)(2.00 x) 38 Calculating Equilibrium Concentrations – Solving for x. H 2 (g ) I 2 (g ) 2HI(g) Initial 1.00 2.00 0 Change -x -x +2x Equilibrium 1.00-x 2.00-x 2x – Because the right side of this equation is not a perfect square, you must solve the quadratic equation. 39 Calculating Equilibrium Concentrations – Solving for x. H 2 (g ) I 2 (g ) 2HI(g) Initial 1.00 2.00 0 Change -x -x +2x Equilibrium 1.00-x 2.00-x 2x – The equation rearranges to give: 0.920x 3.00x 2.00 0 2 40 Calculating Equilibrium Concentrations – Solving for x. H 2 (g ) I 2 (g ) 2HI(g) Initial 1.00 2.00 0 Change -x -x +2x Equilibrium 1.00-x 2.00-x 2x – The two possible solutions to the quadratic equation are: x 2.33 and x 0.93 41 Calculating Equilibrium Concentrations – Solving for x. H 2 (g ) I 2 (g ) 2HI(g) Initial 1.00 2.00 0 Change -x -x +2x Equilibrium 1.00-x 2.00-x 2x – However, x = 2.33 gives a negative value to 1.00 - x (the equilibrium concentration of H2), which is not possible. Only x 0.93 remains. 42 Calculating Equilibrium Concentrations Solving for equilibrium concentrations. H 2 (g ) I 2 (g ) 2HI(g) Initial 1.00 2.00 0 Change -x -x +2x Equilibrium 1.00-x 2.00-x 2x – If you substitute 0.93 for x in the last line of the table you obtain the following equilibrium concentrations. 0.07 M H2 1.07 M I2 1.86 M HI 43 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. – See LeChatelier’s Principle animation 44 Removing Products or Adding Reactants Let’s refer an illustration of a U-tube. “reactants” “products” – 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. 45 Removing Products or Adding Reactants Let’s refer back to the illustration of the U-tube in the first section of this chapter. “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. 46 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 15.12). CO + 3H2 CH4+H2O 3 mol 9 mol 3 mol 3 mol 47 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 favor the products. • If the reactants require less volume (that is, fewer moles of gaseous reactant) • decreasing the volume of the reaction vessel would shift the equilibrium to the left (toward reactants). 48 Effects of Pressure Change 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. 49 Effect of Temperature Change Temperature has a significant effect on most reactions (see Figure 15.13). – Reaction rates generally increase with an increase in temperature. Consequently, equilibrium is established sooner. – In addition, the numerical value of the equilibrium constant Kc varies with temperature. 50 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. 51 Effect of Temperature Change Exothermic A + B → C + D + heat (-)∆H How would adding heat effect the equilibrium? • 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 Kc. 52 Effect of Temperature Change Endothermic A + B + heat → C + D (+)∆H How would adding heat effect the equilibrium? • Increasing temperature would be analogous to adding more reactant, causing the equilibrium to shift right. • This change results in more product at equilibrium, and a larger numerical value for Kc. 53 Effect of Temperature Change In summary: – For an endothermic reaction (DH positive) the amounts of products are increased at equilibrium by an increase in temperature (Kc is larger at higher temperatures). – For an exothermic reaction (DH is negative) the amounts of reactants are increased at equilibrium by an increase in temperature (Kc is smaller at higher temperatures). 54 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 Kc. 55 Obtaining Equilibrium Constants for Reactions Consider the reaction below CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – Suppose we started with initial concentrations of CO and H2 of 0.100 M and 0.300 M, respectively. 56 Obtaining Equilibrium Constants for Reactions Consider the reaction below (see Figure 15.5). CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows. Reactants Products [CO] = 0.0613 M [CH4] = 0.0387 M [H2] = 0.1893 M [H2O] = 0.0387 M 57 Obtaining Equilibrium Constants for Reactions Consider the reaction below CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – The equilibrium-constant expression for this reaction is: [CH 4 ][H 2O] Kc 3 [CO][H 2 ] 58 Obtaining Equilibrium Constants for Reactions Consider the reaction below CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – If we substitute the equilibrium concentrations, we obtain: (0.0387M )(0.0387M ) Kc 3 3.93 (0.0613M )(0.1839M ) 59 Obtaining Equilibrium Constants for Reactions Consider the reaction below CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – 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 Kc. 60 Obtaining Equilibrium Constants for Reactions Consider the reaction below CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – As an example, let’s repeat the previous experiment, only this time starting with initial concentrations of products: [CH4]initial = 0.1000 M and [H2O]initial = 0.1000 M 61 Obtaining Equilibrium Constants for Reactions Consider the reaction below CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – We find that these initial concentrations result in the following equilibrium concentrations. Reactants Products [CO] = 0.0613 M [CH4] = 0.0387 M [H2] = 0.1893 M [H2O] = 0.0387 M 62 Obtaining Equilibrium Constants for Reactions Consider the reaction below CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – Substituting these values into the equilibrium- constant expression, we obtain the same result. (0.0387M )(0.0387M ) Kc 3 3.93 (0.0613M )(0.1839M ) – Whether we start with reactants or products, the system establishes the same ratio. (see Figure 15.5). 63 The Equilibrium Constant, Kp In discussing gas-phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities – It can be seen from the ideal gas equation, PV = nRT that the partial pressure of a gas is proportional to its molarity. n P ( )RT MRT V 64 The Equilibrium Constant, Kp If we express a gas-phase equilibria in terms of partial pressures, we obtain Kp. – Consider the reaction below. CO(g ) 3 H 2 (g) CH4 (g) H 2O(g) – The equilibrium-constant expression in terms of partial pressures becomes: PCH PH O Kp 4 2 3 PCO PH 2 65 The Equilibrium Constant, Kp In general, the numerical value of Kp differs from that of Kc. – From the relationship n/V=P/RT, we can show that Dn K p K c ( RT) where Dn is the sum of the moles of gaseous products in a reaction minus the sum of the moles of gaseous reactants. Animation: Pressure and Concentration of a Gas). 66 A Problem to Consider Consider the reaction 2SO 2 (g ) O 2 (g ) 2 SO 3 (g) – Kc for the reaction is 2.8 x 102 at 1000 oC. Calculate Kp for the reaction at this temperature. 67 A Problem to Consider Consider the reaction 2SO 2 (g ) O 2 (g ) 2 SO 3 (g) – We know that Dn K p K c ( RT) From the equation we see that Δn = -1. We can simply substitute the given reaction temperature and the value of R (0.08206 L.atm/mol.K) to obtain Kp. 68 A Problem to Consider Consider the reaction 2SO 2 (g ) O 2 (g ) 2 SO 3 (g) – Since Dn K p K c ( RT) K p 2.8 10 2 Latm (0.08206 mol K 1000 K) 3.4 -1 69 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 Kc values are known to obtain Kc 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 DH’s values to reflect what we had done. – The rules are a bit different for manipulating Kc. 70 Equilibrium Constant for the Sum of Reactions 1. If you reverse a reaction, invert the value of Kc. 2. If you multiply each of the coefficients in an equation by the same factor (2, 3, …), raise Kc 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 Kc (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 Kc. 71 Building Equations with K values Weak acids and bases are assigned a Ka or Kb values based on the degree to which they ionize in water. Larger K values indicate a greater degree of ionization (strength). Ka and Kb, along with other K values that we will study later (Ksp, KD, Kf) are all manipulated in the same manner. 72 Building Equations with K values When equations are added K values are multiplied. MnS ↔ Mn+2 + S-2 K= 5.1 x 10-15 S-2 + H2O ↔HS- + OH- K= 1.0 x 10-19 2H+ + HS- OH- ↔ H2S +H2O K= 1.0 x 10-7 MnS + 2H+ ↔ Mn+2+ H2 K= 5.1 x 10-41 73 Building Equations with K values When equations are reversed the K values are reciprocated. Al(OH)3 ↔ Al+3 + 3OH- K = 1.9 x 10-33 3OH- + Al+3 ↔ Al(OH)3 K = 1/1.9 x 10-33 = 5.2 x 1032 74 Building Equations with K values When equations are multiplied the K values are raised to the power. NH3 + H2O ↔ NH4++ OH- K = 1.8 x 10-5 2NH3 + 2H2O ↔ 2NH4++ 2OH- K = (1.8 x 10-5)2 = 3.24 x 10-10 75 Building Equations with K values When equations are divided the root of the K values are taken. 2HPO4-2 ↔ 2H+ + 2PO43- K = 1.3 x 10-25 HPO4-2 ↔ H+ + PO43- K = √1.3 x 10-25 = 3.6 x 10-13 76 Equilibrium Constant for the Sum of Reactions For example, nitrogen and oxygen can combine to form either NO(g) or N2O (g) according to the following equilibria. (1) N 2 (g ) O 2 (g ) 2 NO(g) Kc = 4.1 x 10-31 (2) N 2 (g ) 1 O 2 (g ) 2 N 2O(g) Kc = 2.4 x 10-18 – Using these two equations, we can obtain Kc for the formation of NO(g) from N2O(g): (3) N 2O(g ) 1 O 2 (g ) 2 2 NO(g) Kc = ? 77 Animation: Equilibrium Decomposition of N2O4 (Click here to open QuickTime animation) Return to Slide 15 100 Figure 15.5: Some equilibrium compositions for the methanation reaction. Return to Slide 35 101 Figure 15.5: Some equilibrium compositions for the methanation reaction. Return to Slide 21 102 Figure 15.5: Some equilibrium compositions for the methanation reaction. Return to Slide 22 103 Figure 15.5: Some equilibrium compositions for the methanation reaction. Return to Slide 23 104 Figure 15.5: Some equilibrium compositions for the methanation reaction. Return to Slide 24 105 Figure 15.5: Some equilibrium compositions for the methanation reaction. Return to Slide 25 106 Figure 15.5: Some equilibrium compositions for the methanation reaction. Return to Slide 26 107 Figure 15.5: Some equilibrium compositions for the methanation reaction. Return to Slide 27 108 Figure 15.6: The concentration of a gas at a given temperature is proportional to the pressure. Return to Slide 43 109 Animation: Pressure and Concentration of a Gas (Click here to open QuickTime animation) Return to Slide 28 110 Figure 15.12 A-C Slide 21 111 Figure 15.13: The effect of changing the temperatur e on chemical equilibrium. Photo courtesy of American Color. Return to Slide 24 112 Figure 15.15: Oxidation of ammonia using a copper catalyst. Photo courtesy of James Scherer. Return to Slide 85 113