1. Ch 17 Gaseous Chemical Equilibrium: The Extent of Chemical Reactions 17.1 The N 2 O 4 -NO 2 equilibrium system 17.2 Expressing Equilibria with Pressure Terms: Relation between K c and K p 17.3Determination of K 17.4Application of the Equilibrium Constant 17.5Effect of Changes in Conditions on an Equilibrium System LeChatelier’s Principle Dynamic nature of the equilibrium state Equilibrium constant Equilibrium quotient

2. Kinetics applies to the speed of a reaction, the concentration of product that increases (or of reactant that decreases) per unit time. Equilibrium applies to the extent of a reaction, the concentration of product that has appeared after an unlimited time, or once no further change occurs. At equilibrium: rate forward = rate reverse A system at equilibrium is dynamic on the molecular level; no further net change is observed because changes in one direction are balanced by changes in the other.

3. Reaching equilibrium on the macroscopic and molecular levels. N 2 O 4 (g) 2NO 2 (g) colorless. N 2 O 4 -NO 2 System Partial pressure of N 2 O 4 drops and forward reaction slows down. Reddish-brown.

4. The change in Q during the N 2 O 4 -NO 2 reaction. The partial pressure of N 2 O 4 drops at first. At meanwhile, the partial pressure of NO 2 rises from zero. Finally, the both partial pressure (P eq ) levels off at the equilibrium value. These pressures do NOT change because the forward and reverse reactions occur at the same rate. The quotient Q= is defined as the equilibrium constant. The equilibrium is unitless and depends on the temperature instead of the initial concentration. Partial pressure is proportional to the concentration Partial pressure P NO2 2 P N2O4

5. Initial and Equilibrium Concentration Ratios for the N 2 O 4 -NO 2 System at 200 0 C(473 K) Experiment Initial [N 2 O 4 ][NO 2 ] Ratio(Q) Equilibrium Ratio(K) [N 2 O 4 ] eq [NO 2 ] eq 10.10000.0000 3.57x10 -3 0.193 20.10000.0000 ∞ 9.24x10 -4 9.83x10 -3 30.0500 2.04x10 -3 0.146 40.02500.07500.008332.75x10 -3 0.170 [NO 2 ] 2 [N 2 O 4 ] [NO 2 ] eq 2 [N 2 O 4 ] eq 10.4 10.5 N 2 O 4 (g) 2NO 2 (g) Q is denoted as the Reaction Quotient. K is denoted the equilibrium constant.

6. The range of equilibrium constants small K The reaction goes little toward the forward reaction before reaching the equilibrium. No reaction large K The reaction reaches the equilibrium with little reactant remaining. Completion of reaction intermediate K At equilibrium, both reactants and products present.

7. Expression of the Equilibrium constant In many cases, the reactants do NOT go to completion, but stop before all reactants are consumed. Equilibrium is a process when the amount of reactants and products stops changing with time. For a specific reaction : a A + b B  c C + d D Equilibrium constant can be expressed as : K = Capital letters A, B, C, D – stand for chemical species Superscript letters a, b, c, d – denote stoichiometry coefficients [ ] square bracket – concentration in mol/L related to the standard state [A] a [B] b [C] c [D] d ([A]/[A] st ) a ([B]/[B st ]) b ([C]/[[C st ]) c ([D]/[D st ]) d This is also known as the LAW OF MASS ACTION.

8. Standard state: For solution: 1 M gas: 1 bar (1 bar = 10 5 Pa) NOT 1 atm Solid/liquid : pure solid and liquid, its concentration = 1 For example : Na(s) + H 2 O(l)  NaOH (aq) + H 2 (g) K = = [NaOH]  [H 2 ] [Na ]  [H 2 O] 1 [NaOH]  [H 2 ][NaOH]  P H2  1

9. PROBLEM: SOLUTION: Writing the Reaction Quotient from the Balanced Equation Write the reaction quotient, Q c, for each of the following reactions: (a) The decomposition of dinitrogen pentoxide, N 2 O 5 ( g ) NO 2 ( g ) + O 2 ( g ) (b) The combustion of propane gas, C 3 H 8 ( g ) + O 2 ( g ) CO 2 ( g ) + H 2 O( g ) PLAN:Be sure to balance the equations before writing the Q c expression. 42(a) N 2 O 5 ( g ) NO 2 ( g ) + O 2 ( g ) Q c = [NO 2 ] 4 [O 2 ] [N 2 O 5 ] 2 345 (b) C 3 H 8 ( g ) + O 2 ( g ) CO 2 ( g ) + H 2 O( g ) Q c = [CO 2 ] 3 [H 2 O] 4 [C 3 H 8 ][O 2 ] 5

10. PLAN: Writing the Reaction Quotient for an Overall Reaction PROBLEM:Understanding reactions involving N 2 and O 2, the most abundant gases in air, is essential for solving problems dealing with atmospheric pollution. Here is a reaction sequence between N 2 and O 2 to form nitrogen dioxide, a toxic pollutant that contributes to photochemical smog. (a) Show that the Q c for the overall reaction sequence is the same as the product of the Q c s of the individual reactions. (1) N 2 ( g ) + O 2 ( g ) 2NO( g ) K c1 = 4.3 x 10 -25 (2) 2NO( g ) + O 2 ( g ) 2NO 2 ( g ) K c2 = 6.4 x 10 9 (b) Calculate the K c for the overall reaction. Write the sum of the overall reactions; write the Q c. Write the Q c s for the individual reactions and then multiply the expressions. We are given the K c s for the individual reactions, so we multiply those values.

11. SOLUTION: Writing the Reaction Quotient for an Overall Reaction N 2 ( g ) + 2O 2 ( g ) 2NO 2 ( g ) Q c = [NO 2 ] 2 [N 2 ][O 2 ] 2 Q c1 = [NO] 2 [N 2 ][O 2 ] Q c2 =[NO 2 ] 2 [NO] 2 [O 2 ] [NO] 2 [N 2 ][O 2 ] Q c1 x Q c2 = [NO 2 ] 2 [NO] 2 [O 2 ] = [NO 2 ] 2 [N 2 ] [O 2 ] 2 (2) 2NO( g ) + O 2 ( g ) 2NO 2 ( g ) (1) N 2 ( g ) + O 2 ( g ) 2NO( g ) (a) (b)K c =K c1 x K c2 = (4.3 x 10 -25 ) x (6.4 x 10 9 ) = 2.8 x 10 -15 Sum of the reaction (1) and (2.) Overall reaction

12. SOLUTION: Determining the Equilibrium Constant for an Equation Multiplied by a Common Factor the equilibrium constant, K c, is 2.4x10 -3 at 1000K. If we change the coefficients of the equation, which we’ll call the reference (ref) equation, what are the values of K c for the following balanced equations? PROBLEM:For the ammonia formation reaction N 2 ( g ) + 3H 2 ( g ) 2NH 3 ( g ) (a) 1/3 N 2 ( g ) + H 2 ( g ) 2/3 NH 3 ( g )(b) NH 3 ( g ) 1/2 N 2 ( g ) + 3/2 H 2 ( g ) PLAN:Compare each equation to the reference. Keep in mind that changing the coefficients will be reflected in a power change in K c and a reversal of the equation will show up as an inversion of K c. (a) The reference equation is multiplied by 1/3, so K c(ref) will be to the 1/3 power. K c = [K c(ref) ] 1/3 = (2.4x10 -3 ) 1/3 = 0.13 (b) The reference equation is reversed and halved, so K c(ref) is to the -1/2 power. K c = [K c(ref) ] -1/2 = (2.4x10 -3 ) -1/2 = 20.

13. Summary When the forward reaction rate and reverse reaction rate are the same, the system reaches equilibrium. The equilibrium is dynamic and there is no change in concentration of product and reactants. Equilibrium constant K is the ratio of product concentration to the reactants concentration when the equilibrium is established. Reaction quotient (Q) is the particular ratio of product concentration to the reactants concentration at any time. K = Q, the equilibrium establishes. Q > K, reverse direction favors. For both K and Q, the balanced reaction equation is essential.

14. Manipulating equilibrium : For example: the reaction HA  H + + A - K = It is very common to have two or more simultaneous equilibria in a system. HA  H + + A -(1) K1 H + + C  HC +(2) K1 The sum of the reaction HA + C  A - + HC + K=? K=K 1 K 2 =  = [H + ][A - ] [HA] [A - ][H + ] [HC + ] [H + ][C][HA] [HA][C] [A - ][HC + ] Take home message

15. Example: Reaction 1 H 2 O (l)  H +( aq) + OH - (aq) K 1 =K w =[H + ][OH - ]=1.0  10 -14 Reaction 2 NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) K 2 =1.8  10 -5 Find the equilibrium constant of NH 4 + (aq)  NH 3 (aq) + H + (aq) Solution: 1. Convert the direction of reaction 2. reaction 3 : NH 4 + (aq) + OH - (aq)  NH 3 (aq) + H 2 O (l) K 3 =1/K 1 2. Combine reaction 1 and 3 to obtain the sum reaction: NH 4 + (aq)  NH 3 (aq) + H + (aq) 3. Multiply the K 1 by K 3. K = K 1  K 3 = K 1  1/K 2 =5.6  10 -10 Take home message

16. Suppose that a system at equilibrium is subjected to a change. Left 1A + 2B  3C + 1DRight Initial [A i ] [B i ] [C i ] [D i ] Change -x -2x +3x +x Final [A i ]-x [B i ]-2x [C i ]+3x [D i ]+x Q = reaction quotient If Q > K, the reaction proceeds to the left (reverse reaction favors). (decrease the numerator and increase the denominator) Q = K, the equilibrium established. Q < K, the reaction proceeds to the right (forward reaction favors). ([C i ]+3x)([D i ]+x) ([A i ]-x)([B i ]-2x) CH 13 equilibrium Take home message

17. Example: BrO 3 - + 2 Cr 3+ + 4 H 2 O  Br - + Cr 2 O 7 2- + 8H + Bromate Chromium (III) Dichromate 0.043 M 0.0030 M 1.00 M 0.10 M 5.00 M The reaction reaches the equilibrium, K=1.00  10 11 at 25  C Adding 0.10 M Cr 2 O 7 2-, in what direction the reaction proceeds? Q = [Br - ][Cr 2 O 7 2- ][H + ] 8 [BrO 3- ][Cr 3+ ] 2 [H 2 O] 4 Note: concentration for H 2 O is considered as 1 Q=2.00  10 11 > K, the reaction must go to the left, reverse direction favors. Take home message

18. SOLUTION: Calculating K c from Concentration Data PROBLEM:In order to study hydrogen halide decomposition, a researcher fills an evacuated 2.00-L flask with 0.200mol of HI gas and allows the reaction to proceed at 453 0 C. 2HI(g) H 2 (g) + I 2 (g) At equilibrium, [HI] = 0.078M. Calculate K c. PLAN:First find the molar concentration of the starting material and then find the amount of each component, reactants and products, at equilibrium. [HI] = 0.200 mol 2.00 L = 0.100 M Let x be the amount of [H 2 ] at equilibrium. Then x will also be the concentration of [I 2 ] and the change in of [HI] will be the original concentration minus the stoichiometric amount that reacted, 2x, or 0.078M. Determination of K K c = [C] c [D] d [A] a [B] b

19. Calculating K c from Concentration Data continued concentration (M)2HI(g) H 2 (g) + I 2 (g) I nitial C hange E quilibrium 0.100 -2x+ x 00 0.100 - 2xxx [HI] = 0.078 = 0.100 - 2x ; x = 0.011M K c = [H 2 ] [I 2 ] [HI] 2 = [0.011][0.011] (0.078) 2 = 0.020

20. Expressing Equilibria with Pressure Terms K c and K p PV = nRT P = n V RT Q p = P  M so for 2NO(g) + O 2 (g) 2NO 2 (g) p(NO 2 ) 2 p(NO) 2 x p(O 2 ) Q c = [NO 2 ] 2 [NO] 2 x [O 2 ] K p = K c (RT)  n(gas) P = MRT (RT) 2 [NO 2 ] 2 (RT) 2 (RT) x [NO] 2 [O 2 ] (RT) 2 M(NO 2 ) 2 (RT) 2 M(NO) 2 x RTM(O 2 ) = =  n = moles of reactant - moles of product R = 8.314 J  mol -1  K -1 R= 0.082 l  atm  mol -1  K -1

21. Converting Between K c and K p PROBLEM: A chemical engineer injects limestone (CaCO 3 ) into the hot flue gas of a coal-burning power plant for form lime (CaO), which scrubs SO 2 from the gas and forms gypsum. Find K c for the following reaction, if CO 2 pressure is in atmospheres. CaCO 3 (s) CaO(s) + CO 2 (g) K p = 2.1x10 -4 (at 1000K) PLAN: We know K p and can calculate K c after finding  n gas. R = 0.0821 L*atm/mol*K. SOLUTION:Dn gas = 0 - 1 since there is only a gaseous product and no gaseous reactants. K p = K c (RT)  n K c = K p /(RT)  n = (2.1x10 -4 )(0.0821 x 1000) -1 = 2.6x10 -6

22. SOLUTION: Determining Equilibrium Concentrations from K c PROBLEM:In a study concerning the conversion of methane to other fuels, a chemical engineer mixes gaseous CH 4 and H 2 O in a 0.32-L flask at 1200 K. At equilibrium the flask contains 0.26mol of CO, 0.091mol of H 2, and 0.041mol of CH 4. What is the [H 2 O] at equilibrium? K c = 0.26 for the equation CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) PLAN:Use the balanced equation to write the reaction quotient and then substitute values for each component. CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) [CH 4 ] eq = 0.041mol 0.32 L = 0.13 M[CO] eq = 0.26mol 0.32 L = 0.81 M [H 2 ] eq = 0.091mol 0.32 L = 0.28 M Q c = [CO][H 2 ] 3 [CH 4 ][H 2 O] [H 2 O] = [CO][H 2 ] 3 [CH 4 ] K c = 0.53 M = (0.81)(0.28) 3 (0.13)(0.26) Take-home message

23. Determining Equilibrium Concentrations from Initial Concentrations and K c PLAN: PROBLEM:Fuel engineers use the extent of the change from CO and H 2 O to CO 2 and H 2 to regulate the proportions of synthetic fuel mixtures. If 0.250 mol of CO and 0.250 mol of H 2 O are placed in a 125-mL flask at 900K, what is the composition of the equilibrium mixture? At this temperature, K c is 1.56 for the equation CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) We have to find the concentrations of all species at equilibrium and then substitute into a K c expression. CO(g) + H 2 O(g) CO 2 (g) + H 2 (g)concentration I initial C change E equilibrium 2.00 00 SOLUTION:All concentrations must be recalculated as M, so [CO] = 0.250/0.125L -x-x -x+x 2.00-x xx

24. Determining Equilibrium Concentrations from Initial Concentrations and K c continued Q c = K c = [CO][H 2 ] [CO 2 ][H 2 O] = (x)(x)(x)(x) (2.00-x) = (x)2(x)2 (2.00-x) 2 = +/- 1.25 = x 2.00-x x = 1.11M 2.00 - x = 0.89M [CO 2 ] = [H 2 O] = 0.89M [CO] = [H 2 ] = 1.11M

25. Calculating Equilibrium Concentration with Simplifying Assumptions PROBLEM:Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It decomposes by the reaction COCl 2 (g) CO(g) + Cl 2 (g) K c = 8.3x10 -4 (at 360 0 C) Calculate [CO], [Cl 2 ], and [COCl 2 ] when the following amounts of phosgene decompose and reach equilibrium in a 10.0-L flask. (a) 5.00 mol COCl 2 (b) 0.100 mol COCl 2 PLAN:After finding the concentration of starting material, write the expressions for the equilibrium concentrations. When solving for the remaining amount of reactant, see if you can make an assumption about the initial and final concentrations which could simplifying the calculating by ignoring the solution to a quadratic equation. SOLUTION:(a) 5.00 mol/10.0 L = 0.500M (b) 0.100 mol/10.0 L = 0.0100M Let x = [CO] eq = [Cl 2 ] eq and 0.500-x and 0.0100-x = [COCl 2 ] eq, respectively, for (a) and (b).

26. Calculating Equilibrium Concentration with Simplifying Assumptions continued K c = [CO][Cl 2 ] [COCl 2 ] (0.500 - x) = 4.8x10 -2 we can drop x in the denominator assume x is << 0.500 so that we can drop x in the denominator 8.3x10 -4 = (x) (0.500) 4.15x10 -4 = x 2 x ≈ 2 x 10 -2 CHECK: 0.020/0.500 = 0.04 or 4% percent error (b)K c = 8.3x10 -4 = (x) (0.010 - x) Dropping the -x will give a value for x = 2.9x10 -3 M. (0.010 - x) ≈ 0.0071M CHECK: 0.0029/0.010 = 0.29 or 29% percent error quadratic formula Using the quadratic formula produces x = 2.5x10 -3 and 0.0100-x = 7.5x10 -3 M (a)K c = 8.3x10 -4 = (x) (0.500-x)

27. Reaction direction and the relative sizes of Q and K. Equilibrium: no net change Reactants products Reaction Progress reactants products Reaction Progress Applications of Equilibrium constant Q c = [C] c [D] d [A] a [B] b

28. SOLUTION: Predicting Reaction Direction and Calculating Equilibrium Concentrations PROBLEM:The research and development unit of a chemical company is studying the reaction of CH 4 and H 2 S, two components of natural gas. CH 4 (g) + 2H 2 S(g) CS 2 (g) + 4H 2 (g) In one experiment, 1.00mol of CH 4, 1.00mol of CS 2, 2.00mol of H 2 S, and 2.00mol of H 2 are mixed in a 250-mL vessel at 960 0 C. At this temperature, K c = 0.036 (a) In which direction will the reaction proceed to reach equilibrium? (b) If [CH 4 ] = 5.56 M at equilibrium, what are the equilibrium concentrations of the other substances? PLAN:Find the initial molar concentrations of all components and use these to calculate a Q c. Compare Q c to K c, determine in which direction the reaction will progress, and draw up expressions for equilibrium concentrations. [CH 4 ] initial = 1.00mol/0.25 L = 4.0M [H 2 S] initial = 2.00mol/0.25 L = 8.0M[H 2 ] initial = 2.00mol/0.25 L = 8.0M [CS 2 ] initial = 1.00mol/0.25 L = 4.0M

29. Predicting Reaction Direction and Calculating Equilibrium Concentrations continued Q c = [CS 2 ][H 2 ] 4 [CH 4 ][H 2 S] 2 = [4.0][8.0] 4 [4.0][8.0] 2 = 64 A Q c of 64 is >> than K c = 0.036 The reaction will progress to the left. CH 4 (g) + 2H 2 S(g) CS 2 (g) + 4H 2 (g)concentrations initial change equilibrium 4.08.04.08.0 + x+ 2x- 4x 4.0 + x8.0 + 2x - x 4.0 - x8.0 - 4x At equilibrium [CH 4 ] = 5.56M, so 5.56 = 4.0 + x and x = 1.56M Therefore -[H 2 S] = 8.0 + 2x = 11.12M [CS 2 ] = 4.0 - x = 2.44M [H 2 ] = 8.0 - 4x = 1.76M

30. Steps in solving equilibrium problems PRELIMINARY SETTING UP WORKING ON A REACTION TABLE SOLVING FOR X AND EQUILIBRIUM QUANTITIES 1.Write the balanced equation. 2.Write the reaction quotient, Q. 3.Convert all of the amounts into the correct units (M or atm). 4.When reaction direction is not known compare Q with K. 5.Construct a reaction table. Check the sign of x, the change in the quantity. 6.Substitute the quantities into Q. 7.To simplify the math, assume that x is negligible. 8.[A] ini - x = [A] eq = [A] ini 9.Solve for x. 10.Find the equilibrium quantities Check to see that calculated values give the known K. Check that assumption is justified (<5% error). If not, solve quadratic equation for x.

31. Calculating Variations on Q and K aA + bB cC + dD Q c = [C] c [D] d [A] a [B] b cC + dD aA + bB Q ’ = 1/Q c aA + bB cC + dD n Q c ’ = (Q c ) n For a sequence of equilibria, K overall = K 1  K 2  K 3  … 1.A reaction quotient for a forward reaction is the reciprocal of the reverse reaction quotient. 2.If all the coefficients of the balanced equation are multiplied by a factor, the factor becomes the exponent of the reaction quotients. Take-home message

32. Steps in solving equilibrium problems PRELIMINARY SETTING UP WORKING ON A REACTION TABLE SOLVING FOR X AND EQUILIBRIUM QUANTITIES 1.Write the balanced equation. 2.Write the reaction quotient, Q. 3.Convert all of the amounts into the correct units (M or atm). 4.When reaction direction is not known compare Q with K. 5.Construct a reaction table. Check the sign of x, the change in the quantity. 6.Substitute the quantities into Q. 7.To simplify the math, assume that x is negligible. 8.[A] ini - x = [A] eq = [A] ini 9.Solve for x. 10.Find the equilibrium quantities Check to see that calculated values give the known K. Check that assumption is justified (<5% error). If not, solve quadratic equation for x.

33. LeChatelier’s Principle – allow us to predict the reaction direction When a chemical system at equilibrium is subjected to a stress, the system will return to equilibrium by shifting to reduce the stress. What does the “disturbance of a system stand for? The change in concentration (adding or removing gas) The change in pressure (compression or expandion) The change in temperature Net reaction is referred to the as a shift in the equilibrium position of the system to the left or right. Effect of changes in conditions on an equilibrium system

34. The Effect of Added Cl 2 on the PCl 3 -Cl 2 -PCl 5 System Concentration (M)PCl 3 (g) + Cl 2 (g) PCl 5 (g) Disturbance New initial Change New equilibrium *Experimentally determined value. Original equilibrium0.2000.1250.600 +0.075 0.200 0.600 -x-x-x-x+x+x 0.200 - x 0.200 + x (0.637)* If the concentration increases, the system reacts to consume some of it. If the concentration decreases, the system reacts to produce some of it. Adding or removing gas

35. The effect of added Cl 2 on the PCl 3 -Cl 2 -PCl 5 system. PCl 3 ( g ) + Cl 2 ( g ) PCl 5 ( g ) the concentration of Cl 2 increases, the system reacts to consume some of it. Therefore, the reaction shifts to the right. In general, when the reactant is added, the equilibrium position shifts to the right. When the product is moved, the reaction shifts to the right.

36. SOLUTION: Predicting the Effect of a Change in Concentration on the Equilibrium Position PROBLEM:To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O 2 ; 2H 2 S(g) + O 2 (g) 2S(s) + 2H 2 O(g) What happens to (a) [H 2 O] if O 2 is added? (b) [H 2 S] if O 2 is added? (c) [O 2 ] if H 2 S is removed? (d) [H 2 S] if sulfur is added? PLAN:Write an expression for Q and compare it to K when the system is disturbed to see in which direction the reaction will progress. Q = [H 2 O] 2 [H 2 S] 2 [O 2 ] (a) When O 2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H 2 O] increases.

37. Predicting the Effect of a Change in Concentration on the Equilibrium Position continued (b) When O 2 is added, Q decreases and the reaction progresses to the right to come back to K. So [H 2 S] decreases. Q = [H 2 O] 2 [H 2 S] 2 [O 2 ] (c) When H 2 S is removed, Q increases and the reaction progresses to the left to come back to K. So [O 2 ] increases. (d) Sulfur is not part of the Q (K) expression because it is a solid. Therefore, as long as some sulfur is present the reaction is unaffected. [H 2 S] is unchanged. 2H 2 S(g) + O 2 (g) 2S(s) + 2H 2 O(g)

38. The effect of pressure (volume) on an equilibrium system. + lower P (higher V) more moles of gas higher P (lower V) fewer moles of gas The pressure change significantly influences on the equilibrium system with gaseous component. Changes in pressure: Change conc of gas Add inert gas Change volume of vessel

39. SOLUTION: Predicting the Effect of a Change in Volume (Pressure) on the Equilibrium Position PROBLEM:How would you change the volume of each of the following reactions to increase the yield of the products. (a) CaCO 3 (s) CaO(s) + CO 2 (g) (b) S(s) + 3F 2 (g) SF 6 (g) (c) Cl 2 (g) + I 2 (g) 2ICl(g) PLAN:When gases are present a change in volume will affect the concentration of the gas. If the volume decreases (pressure increases), the reaction will shift to fewer moles of gas and vice versa. (a) CO 2 is the only gas present. To increase its yield, we should increase the volume (decrease the pressure). (b) There are more moles of gaseous reactants than products, so we should decrease the volume (increase the pressure) to shift the reaction to the right. (c) There are an equal number of moles of gases on both sides of the reaction, therefore a change in volume will have no effect.

40. The Effect of a Change in Temperature on an Equilibrium Only temperature changes can alter K. Consider heat as a product or a reactant. In an exothermic reaction, heat is a product (HT-LEXON). In an endothermic reaction, heat is a reactant. A temperature rise will increase K c for a system with a positive  H 0 rxn. A temperature rise will decrease K c for a system with a negative  H 0 rxn.

41. SOLUTION: Predicting the Effect of a Change in Temperature on the Equilibrium Position PROBLEM:How does an increase in temperature affect the concentration of the underlined substance and K c for the following reactions? (a) CaO(s) + H 2 O(l) Ca(OH) 2 (aq)  H 0 = -82kJ (b) CaCO 3 (s) CaO(s) + CO 2 (g)  H 0 = 178kJ (c) SO 2 (g) S(s) + O 2 (g)  H 0 = 297kJ PLAN:Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on K c. (a) CaO(s) + H 2 O(l) Ca(OH) 2 (aq) + heat An increase in temperature will shift the reaction to the left, decrease [Ca(OH) 2 ], and decrease K c. (b) CaCO 3 (s) + heat CaO(s) + CO 2 (g) The reaction will shift right resulting in an increase in [CO 2 ] and increase in K c. (c) SO 2 (g) + heat S(s) + O 2 (g) The reaction will shift right resulting in an decrease in [SO 2 ] and increase in K c.

42. The van’t Hoff Equation - The Effect of T on K ln K2K2 K1K1 = -  H 0 rxn R 1 T2T2 1 T1T1 - Temperature Dependence R = universal gas constant = 8.314 J/mol*K K 1 is the equilibrium constant at T 1 ln k2k2 k1k1 = - EaEa R 1 T2T2 1 T1T1 - ln P2P2 P1P1 = -  H vap R 1 T2T2 1 T1T1 - ln K2K2 K1K1 = -  H 0 rxn R 1 T2T2 1 T1T1 -

43. Summary 1.To solve the equilibrium problems, we use quantities (concentration or pressure) find K/Q, or use K/Q to find quantities. 2.We can simplify the calculation by assuming the change in concentration (x) is negligible. If the error is less than 5 %, the approximation is fine. 3.If the approximation is not justified, we use the quadratic formula to find x value. 4.La Chatelier principle states that a disturbed equilibrium undergoes the net reaction which will reduce the disturbance and re-establish a new equilibrium. 5.Change in concentration cause a net change away from the added component (remove the component). 6.An increase in pressure results in a net reaction to produce fewer molecules. 7.For an exothermic reaction, the temperature increase causes the lower K. (negative  H  < 0: HT-LEXON) shifting to the left. 8.Concentration and pressure will not alter the K, but the temperature.

44. Determining Equilibrium Parameters from Molecular Species For the reaction A(g) + B 2 (g) AB(g) + B(g)  H>0 the following molecular scenes depict different reaction mixtures. A = g, B = p) (a) If K c = 2 at the temperature of the reaction, which scene represents the mixture at equilibrium? (b)Will the reaction mixtures in the other two scenes proceed toward reactant or toward products to reach equilibrium? (c)For the mixture at equilibrium, how will a rise in temperature affect [Y 2 ]? SOLUTION:The equilibrium constant, K, is [AB][B] [A][B 2 ]. scene 1: Q c = (5)(3)/(1)(1) = 15 scene 2: Q c = (4)(2)/(2)(2) = 2.0scene 3: Q c = (3)(1)/(3)(3) = 0.33 Take-home message

45. Determining Equilibrium Parameters from Molecular Species Q c = 15Q c = 2.0Q c = 0.33 (b) In scene 1, Q c > K c, so the system will proceed to reactants to reach equilibrium. in scene 3, Q c < K c, so the system will proceed to products. (a) In scene 2, Q c = K c, so it represents the system at equilibrium. (c) If  H > 0, heat is a reactant (endothermic). A rise in temperature will favor products and [Y 2 ] will decrease as the system shifts to products.