(i.e., equilibrium is established) leads to: K = k 1

Transcription

1 CHEMISTRY 104 Help Sheet #8 Chapter 12 Equilibrium Do the topics appropriate for your lecture (Resource page) Prepared by Dr. Tony Jacob Nuggets: Equilibrium Constant Expression; Size of K; K c versus K p ; How K changes as Reaction Changes; Reaction Quotient (Q); Le Chatelier's Principle EQUILIBRIUM CONSTANT EXPRESSION and EQUILIBRIUM CONSTANT (K) Equilibrium occurs when the concentrations of products and reactants are no longer changing, i.e., they are constant (see diagram below) At equilibrium both products and reactants exist Equilibrium is dynamic which means both the forward and reverse reactions are occurring but there is no net change in chemical concentrations Equilibrium constant expressions are written with products multiplied together and raised to the power of their stoichiometric coefficients divided by the reactants multiplied together raised to their stoichiometric coefficients [C] c For example: aa(g) + bb(g) cc(g) K c = [A] a [B] b when the chemical concentrations are in molarity K c is used The K is the equilibrium constant Pure solids, liquids, and solvents are not included in an equilibrium constant expression as their concentrations are constant. K is unitless (it is divided by a reference concentration or pressure which is not shown) From kinetics with rate forward = rate reverse (i.e., equilibrium is established) leads to: K = k 1 /k -1 = [products] coefficient /[reactants] coefficient K is a function of T {expressed as K(T)} since the rate constants k 1 and k -1 are dependent on T SIZE OF K If K is large (e.g., 1 x 10 5 = [products] [reactants] = 105 ) the products are strongly favored (see left diagram below) 1 If K is small (e.g., 1 x 10-5 = [products] [reactants] = 1 ) the reactants are strongly favored (see right diagram below) 10 5 If K is 1 neither the reactants nor products dominate and both are present in significant quantities

2 K c versus K p : K c uses concentration values (M); K p uses pressure values (usually atm); often K (no subscript) implies K c ; K p = K c (RT) Δn Δn is the change in #gaseous moles: Δn = n gas products n gas reactants ; T in K; R = Latm/molK; Example 1: If the K p for this reaction: C(s) + O 2 (g) 2CO(g) is 1.5 x 10-3 what is the K c at 25.0 C? Answer 1: K p = K c (RT) Δn ; Δn = 2 1 = 1 (Note: C(s) not included because it s a solid); (1.5 x 10-3 )/[(0.0821)(298)] 1 = 6.1 x 10-5 HOW K CHANGES AS THE REACTION CHANGES 1. Reverse a reaction: K 1 / K (for A B K = [B]/[A] ; for B A K rev = [A]/[B] ; K rev = 1/K) 2. Multiply a reaction by a constant c: K K c (for A B K = [B]/[A] ; for 3A 3B K multiply = [A] 3 /[B] 3 = ([A]/[B]) 3 = K 3 ) 3. Add Reactions: Reaction 1 + Reaction 2 = Reaction 3: K Rxn3 = K Rxn1 x K Rxn2 (for A B K 1 = [B]/[A] ; for B C K 2 = [C]/[B] ; add rxns: A + B B + C A C K add = [C]/[A]; while K 1 x K 2 = ([B]/[A]) x ([C]/[B]) = ([B][C]/[B][A] = ([C]/[A]) = K 3 ) Example 2: If K c = for the reaction: N 2 (g) + O 2 (g) 2NO(g), what is the K c for the reaction: 1 / 2 NO(g) 1 /4 N 2 (g) + 1 / 4 O 2 (g)? Answer 2: The new reaction is the prior reaction: 1) reversed, and 2) multiplied by 1 / 4. The K c for the new reaction = K c (new) = ( 1 / K c )1/4 = ( 1 / ) 1/4 = (1.325) 1/4 = 1.07 EQUILIBRIUM CALCULATIONS (typical problems) 1. Given concentrations or moles at equilibrium find K eq 2. Given concentrations or moles not at equilibrium find K eq 3. Given K eq find equilibrium concentrations If algebraic expression is too difficult to solve easily, then: a. try taking square root of both sides; if not possible then b. make an approximation (presented in Ch 14; if used here, one method to determine if approximation is valid is if 100 x K eq < [ ] o is true approximation is generally valid); if not true: not valid c. use quadratic (some instructors skip quadratic equation problems) Example 3: (given [ ] at equilibrium find K eq ; problem type #1 above) If a 2.0L flask at 25 C contains 0.086g H 2, 0.963g I 2, and 12.2g HI at equilibrium, what is K eq? H 2 (g) + I 2 (g) 2HI(g) Answer 3: determine concentrations (M): [HI] = [12.2g HI x (1mol HI/127.9g HI)]/(2.0L) = M; [H 2 ] = [0.086g H 2 x (1mol H 2 /2.016g H 2 )]/(2.0L) = M; [I 2 ] = [0.963g I 2 x (1mol I 2 /253.8g I 2 )]/(2.0L) = M; Write K eq expression: K eq = [HI]2 [H2 ][I2 ] ; plug in values: K eq = (0.0477) 2 (0.0213)( ) = 56.2 Example 4: (given [ ] not at equilibrium find K eq ; problem type #2 above) If a 5.0L container initially has 0.50mol NO(g) and 0.30mol O 2 (g), and once equilibrium has been established, 0.10mol NO 2 (g) exist, what is K eq? 2NO(g) + O 2 (g) Answer 4: Step 1. Set up ICE Table with mol values (can also use M); initial mol NO 2 (g) = 0. I C E 2NO(g) + O 2 (g) 2NO 2 (g) NO 2 (g) Step 2. Complete ICE Table. Note C row is stoichiometric: 2 : 1 : 2 I C E 2NO(g) + O 2 (g) 2NO 2 (g) Step 3. Write K eq ; convert E row values to M by dividing by volume (L); plug in E row M values; and solve for K eq. K eq = [NO 2] 2 (0.10 / 5) 2 (0.02) 2 [NO] 2 = [O 2 ] (0.40 / 5) 2 = (0.25 / 5) (0.08) 2 =1.25 (0.05)

3 Example 5: (given K eq find [ ]; problem type #3 above) If the initial concentrations of H 2 (g) = I 2 (g) = 0.20M, and the K eq for the reaction below at 35 C is 144, what are the final concentrations of all the chemicals? H 2 (g) + I 2 (g) Answer 5: Step 1. Set up ICE Table with M values; initial concentration of HI(g) = 0. H 2 (g) + I 2 (g) 2HI(g) I HI(g) Step 2. Complete ICE Table. Note C row is stoichiometric: 1 : 1 : 2 H 2 (g) + I 2 (g) 2HI(g) I Step 3. Write K eq ; plug in K value and E row values; solve for x. K eq = [HI]2 [H 2 ][I 2 ] ; 144 = (2x) 2 (0.20 x)(0.20 x) = (2x) 2 (0.20 x) 2 C C -x -x +2x (2x) = E E 0.20-x 0.20-x 2x (0.20 x) 2 ; 12 = 2x 0.20 x ; x = 0.17 Step 4. x does not answer the question! Complete the question: [H 2 ] = [I 2 ] = 0.20-x = = 0.03M; [HI] = 2x = 2(0.17) = 0.34M REACTION QUOTIENT Q a snapshot in time of the reaction progress; Q expression is the same as a K eq expression but Q may or may not be a system at equilibrium; Q can be used to determine if system is at equilibrium, and if not, which way the reaction will shift to reach equilibrium [C] c aa + bb cc Q = [A] a [B] b Q > K system will shift left (to reactant side) to reach equilibrium (when reaction shifts L reactants and products Q and approaches K) Q < K system will shift right (to product side) to reach equilibrium (when reaction shifts R reactants and products Q and approaches K) Q = K at equilibrium (no shift) Example 6: If K c is 7.2 x 10-4 for the equilibrium below and the concentrations of PCl 5 is 0.50M, PCl 3 is 0.15M, and Cl 2 is 0.025M, is the system at equilibrium? If not, would this reaction proceed to the right or left to reach equilibrium? PCl (g) PCl (g) + Cl (g) Answer 6: Determine Q. Q = [PCl 3][Cl2 ] [PCl 5 ] = (0.15)(0.025) = 7.5x10 3 Since Q K the reaction is not at equilibrium. Since Q > K, the reaction will shift left. (0.50) Le CHÂTELIER S PRINCIPLE: a system at equilibrium when perturbed will shift in the direction to oppose the stress on the system Factors affecting equilibrium: 1. Concentration concentrations of chemicals in the equilibrium expression (not pure liquids or solids) are changed by adding reactants or products Example: if [ ] (what you do to the system) [ ] (system responds to lower [ ]) shift to side that [ ] of that chemical (value of K no change) 2. Pressure change the pressure by changing the volume of a system Example: if P (what you do to the system) P (system responds to lower P) gas moles (to lower P reduce gas moles) shift to side with fewer moles gas (only consider gaseous moles not liquid or solid chemicals) (value of K no change) 3. Temperature heat or cool a system; endothermic reaction (Δ r H > 0) write heat on reactant side since heat is absorbed; exothermic reaction (Δ r H < 0) write heat on product side since heat is released Example: T (what you do to the system) T (system responds to lower T) to consume heat, T K is a function of T {written K(T)} so reaction shifts AND the value of K changes How K changes: If change in T results in the reaction shifting: to the product side (shift R) K to reactant side (shift L) K 4. Add Solvent this can shift a reaction by changing the chemical concentrations as the solvent is added (value of K no change)

4 FACTORS NOT AFFECTING EQUILIBRIUM 1. Add a catalyst (reaction speeds up but no shift in equilibrium) 2. Add a pure liquid or a solid to the system (since these don't appear in equilibrium expression, they don't affect the equilibrium unless they change concentrations; see Factors affecting equilibrium, #3) Quadratic equation: from ax 2 + bx + c = 0 x = b ± b2 4ac 2a (skip if not covered) 1. Write the K c expression for the following: a. H 2 (g) + I 2 (g) 2HI(g) d. C(s) + O 2 (g) CO 2 (g) b. 2NO(g) + O 2 (g) 2NO 2 (g) e. CS 2 (g) + 4H 2 (g) CH 4 (g) + 2H 2 S(g) c. 2NO 2 (g) 2NO(g) + O 2 (g) f. 4H 2 (g) + 4I 2 (g) 8HI(g) 2. a. How do we know when equilibrium is achieved? b. What can you say about the relative rates of the forward and reverse reactions at equilibrium? c. When a reaction has not reached equilibrium and is still proceeding to the left, what can you say about the relative rates of the forward and reverse reactions? 3. In which of the following reaction(s) will the products be most favored? (Note: The K values below aren t accurate!) Rxn I: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) K = 1.0 x Rxn II: PCl 3 (g) + Cl 2 (g) PCl 5 (g) K = 49 Rxn III: AgCl(s) Ag + (aq) + Cl - (aq) K = 2 x Calculate the equilibrium constant, K, for the reaction below in terms of K 1, K 2, and K 3 2C(s) + H 2 (g) C 2 H 2 (g) K = K given the individual equilibrium constants for the reactions below Rxn I C 2 H 2 (g) O 2 (g) 2CO 2 (g) + H 2 O(l) Κ = K 1 Rxn II C(s) + O 2 (g) CO 2 (g) Κ = K 2 Rxn III H 2 (g) O 2 (g) H 2 O(l) Κ = K 3 5. Calculate the equilibrium constant, K, for the reaction below in terms of K 1, K 2, and K 3 N 2 H 4 (l) + CH 4 O(l) CH 2 O(g) + N 2 (g) + 3H 2 (g) K = K given the individual reaction standard enthalpy changes Rxn I N 2 H 4 (l) + H 2 (g) 2NH 3 (g) Κ = K 1 Rxn II N 2 (g) + 3H 2 (g) 2NH 3 (g) Κ = K 2 Rxn III CH 2 O(g) + H 2 (g) CH 4 O(l) Κ = K 3 6. Why are solids and pure liquids not included in an equilibrium reaction? 7. Given the equilibrium concentrations for the reaction below, [PCl 5 ] = M, [PCl 3 ] = M, [Cl 2 ] = M, what is the K value? PCl 5 (g) PCl 3 (g) + Cl 2 (g) 8. The value of K p is 3.02 x 10-5 for the reaction at 25 C. What is the value of K c for this reaction at 25 C? CS 2 (g) + 4H 2 (g) CH 4 (g) + 2H 2 S(g)

5 9. A mixture at equilibrium in a 0.45L flask contains 2.8g N 2, 0.063g H 2, and 0.85g NH 3. What is the value of K for this reaction? N 2 (g) + 3H 2 (g) 2NH 3 (g) 10. Initially, a 250ml flask contains 0.063mol COBr 2 and 0.25mol CO. When the reaction has reached equilibrium, the final number of moles of CO is COBr 2 (g) CO(g) + Br 2 (g) a. Calculate the final concentrations of COBr 2, CO, and Br 2. b. Calculate K c. 11. Initially, a flask contains 1M N 2 and 1M O 2. When the reaction has reached equilibrium, the final concentration of N 2 is 0.9M. N 2 (g) + O 2 (g) 2NO(g) a. Calculate the final concentrations of O 2 and NO. b. Calculate K c. 12. If the equilibrium constant for the reaction below is , what are the final concentrations of all chemicals if the initial concentration of HI was 1.50M and the initial concentrations of H 2 and I 2 were M? 2HI(g) H 2 (g) + I 2 (g) 13. If the equilibrium constant for the reaction below is 4.50 x 10-5, what are the final concentrations of all chemicals if the initial concentration of O 2 and N M? O 2 (g) + N 2 (g) 2NO(g) 14. Initially 1.50 moles of A are added to a 3.00 L reaction vessel and the system is allowed to come to equilibrium. At equilibrium, the amount of A has decreased by 35.0%. 2A 2B + C a. What is the concentration of A, B, and C at equilibrium? b. What is the value of K c? 15. K is 0.75 for the reaction A(g) B(g) + C(g). If a 2.0 L flask contains 5.0mol A, how many mol of A, B, and C will exist once equilibrium is achieved? (Hint: Requires the quadratic equation; skip if not required.) 16. For each of the following questions, determine if the system is at equilibrium. If the system is not in equilibrium, which way will the reaction shift to reach equilibrium (left or right)? a. For the following reaction, K = CO(g) + 2H 2 (g) CH 3 OH(g) The initial concentrations are: [CO] = M, [H 2 ] = M, [CH 3 OH] = 1.33 M b. For the following reaction, K = 4.8 x CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) The starting concentrations are: [CO] = M, [H 2 ] = 0.60 M, [CH 4 ] = 1.2 M, and [H 2 O] = 0.30 M. c. For the following reaction, K = NO 2 (g) N 2 O 4 (g) The starting concentrations of [NO 2 ] and [N 2 O 4 ] are 0.50 M. 17. What factors will affect equilibrium and cause a reaction to potentially shift?

6 18. Given the reaction below, determine which way the exothermic reaction will shift when the various changes are applied to the system that is at equilibrium (left, right, or not shift) and specify what happens to K eq (increase, decrease, no change). C(s) + O 2 (g) 2CO(g) a. oxygen is added b. carbon monoxide is added c. the volume of the container is reduced d. a catalyst is added e. carbon is added f. the reaction is placed in an ice bath and cooled 19. Given the reaction below, determine which way the endothermic reaction will shift when the various changes are applied to the system that is at equilibrium (left, right, or not shift) and specify what happens to K eq (increase, decrease, no change). N 2 (g) + O 2 (g) 2NO(g) a. oxygen is added b. nitrogen monoxide is added c. the volume of the container is increased d. a catalyst is added e. nitrogen is removed f. the reaction is placed in an ice bath and cooled 20. Using the reaction below, answer the questions given. CS 2 (l) + 2N 2 O(g) CO 2 (g) + 2S(s) + 2N 2 (g) a. When N 2 O is added what happens to the amount of CO 2? b. When N 2 is added what happens to the amount of CS 2? c. When a catalyst is added what happens to the amount of N 2? d. When N 2 O is removed what happens to the concentration of N 2? e. When N 2 is removed what happens to the concentration of CO 2? f. When CS 2 is added what happens to the amount of N 2 O? g. When CO 2 is removed what happens to the concentration of CS 2? (Hint: This is a tricky one!) h. If this reaction is endothermic, what happens to the amount of S when the system is heated? ANSWERS 1. a. K = [HI] 2 [H 2 ][I 2 ] [NO 2 ] 2 b. K = [NO] 2 [O 2 ] c. K = [NO]2 [O 2 ] [NO 2 ] 2 d. K = [CO 2 ] e. K= [CH 4 ][H 2 S]2 [O 2 ] [CS 2 ][H 2 ] 4 f. K = [HI] 8 [H 2 ] 4 [I 2 ] 4 2. a. When the concentrations of the reactants and products no longer change. b. The forward and reverse rates are equivalent at equilibrium. c. The reverse rate is greater than the forward rate. 3. Rxn I 4. K = (K 2 2 )(K 3 )/(K 1 ) {multiply Rxn II by 2 (K 2 2 ); no change to Rxn III; reverse Rxn I (1/K1 ); add rxns and multiply new K s; K = (K 2 2 )(K3 )(1/K 1 )} 5. K = K 1 /[(K 2 )(K 3 )] {no change to Rxn I; reverse Rxn III (1/K 3 ); reverse Rxn II (1/K 2 ); add rxns and multiply K s; K = (K 1 )(1/K 2 )(1/K 3 ) = K 1 /[(K 2 )(K 3 )]}

7 6. Solids and pure liquids are not included in an equilibrium expression because their concentrations are constant and only chemicals whose concentrations can change are included. 7. K = 7.21 x 10-4 {K = [PCl 3 ][Cl 2 ]/[PCl 5 ] = (0.0101)( )/(0.0350) = } 8. K = 1.81 x 10-2 {K p = K c (RT) Δn ; T = 298, Δn = 3-5 = -2, K c = K p /[(0.0821)(298)] -2 = (3.02 x 10-5 )(24.47) 2 = 1.81 x 10-2 } 9. K = 170 {K = [NH 3 ] 2 /([N 2 ][H 2 ] 3 ); 2.8g N 2 x (1mol N 2 /28.02g N 2 ) = 0.010mol N 2 /0.45L = 0.222M; 0.063g H 2 x (1mol H 2 /2.016g H 2 ) = mol H 2 /0.45L = M; 0.85g NH 3 x (1mol NH 3 /17.03g NH 3 ) = mol NH 3 /0.45L = 0.111M; K = (0.111) 2 /[(0.222)(0.0694) 3 ] = 166} 10. a. [COBr 2 ] = 0.21M; [Br 2 ] = 0.040M; [CO] = 1.0M {write ICE Table: I row: 0.063, 0.250, 0; C row: , , ; E row: 0.053, , 0.010; convert to M by dividing by 0.25L; E row: 0.21, 1.04, 0.040} b. K = 0.19 {K = [CO][Br 2 ]/[COBr 2 ] = (1.0)(0.040)/(0.21) = 0.190} 11. a. [O 2 ] = 0.9M; [NO] = 0.2M {write ICE Table: I row: 1, 1, 0; C row: -0.1, -0.1, +0.2; E row: 0.9, 0.9, 0.2} b {K = [NO] 2 /[N 2 ][O 2 ] = (0.2) 2 /[(0.9)(0.9)] = } 12. [HI] = 1.21M; [H 2 ] = [I 2 ] = 0.149M {write ICE Table: I Row: 1.50, , ; C Row: -2x, +x, +x; E Row: x, x, x; K = [H 2 ][I 2 ]/[HI] = ( x)( x)/(1.50-2x) 2 ; = ( x) 2 /(1.50-2x) 2 ; take sqrt of both sides: = ( x)/(1.50-2x); x = x; 1.245x = ; x = ; [HI] = x = (0.1436) = 1.213M; [H 2 ] = [I 2 ] = x = = M} 13. [NO] = 5.01 x 10-3 M; [O 2 ] = [N 2 ] = 0.747M {write ICE Table, I Row: 0.750, 0.750, 0; C Row: -x, -x, +2x; E Row: x, x, 2x; K = [NO] 2 /[O 2 ][N 2 ] 4.50 x 10-5 = (2x) 2 /(0.750-x)(0.750-x); 4.50 x 10-5 = (2x) 2 /(0.750-x) 2 ; take sqrt of both sides: 6.71 x 10-3 = (2x)/(0.750-x); 5.03 x x 10-3 x = 2x; x = 5.03 x 10-3 ; x = x 10-3 ; [O 2 ] = x = x 10-3 = 0.747M; [N 2 ] = x = x 10-3 = 0.747M; [NO] = 2x = 2(2.507 x 10-3 ) = } 14. a. [A] = 0.325M; [B] = 0.175M, [C] = M {write ICE Table, I Row: 1.50, 0, 0; C Row: -2x, +2x, +x; E Row: x, 2x, x; from the information provided that A decreases by 35.0%, then 2x = (0.35)(1.50) = 0.525mol; at equilibrium (E Row): A = = 0.975mol; B = 2x = 0.525mol; C = x = 0.525/2 = mol; [A] = 0.975mol/3.00L = 0.325M; [B] = 0.525mol/3.00L = 0.175M; [C] = mol/3.00L = M} b. K = 2.54 x 10-2 {K = [B] 2 [C]/[A] 2 [(0.175) 2 (0.0875)]/[(0.325) 2 ] = } mol A; 2.1 mol B; 2.1mol C {write ICE Table, I row: 2.5, 0, 0; C row: -x, +x, +x; E row: 2.5-x, x, x; K = [B][C]/[A]; 0.75 = (x)(x)/(2.5-x); can t solve easily; can t take sqrt use quadratic; cross multiply: x 2 = -0.75x + (2.5)(0.75); x 2 = -0.75x ; x x = 0; solve for x; a = 1; b = 0.75; c = ; x = 0.75± (0.75)2 4(1)( 1.875) 0.75± ± ± ; x = ; x = ; x = ; x = ; 2(1) x = 1.045; [A] = 2.5-x = 1.455M; mol = M x L = (1.455)(2.0) = 2.91mol A; [B] = [C] = x = 1.045M; mol = M x L = (1.045)(2) = 2.09mol B and 2.09 mol C} 16. a. Q > K; not at equilibrium; shifts left {Q = [CH 3 OH]/([CO][H 2 ] 2 ) = (1.33)/[( )( ) 2 = 1.93 x 10 8 ; Q > K (1.93 x 10 8 > 10.5) shifts left} b. Q < K; not at equilibrium; shifts right {Q = [CH 4 ][H 2 O]/([CO][H 2 ] 3 ) = (1.2)(0.30)/[(0.020)(0.60) 3 ] = 83.3; Q < K (83.3 > 4.8 x 10 3 ) shifts right} c. Q < K, not at equilibrium; shifts right {Q = [N 2 O 4 ]/[NO 2 ] 2 = (0.50)/(0.50) 2 = 2.0; Q < K (2.0 < 8) shifts right} 17. concentration, pressure, and temperature (sometimes adding the solvent will also cause a reaction to shift though this is often not included in the description of this topic)

8 18. a. right; no change in K {when O 2 (g) is added reaction shifts to the right; K no change since no change in T} b. left; no change in K {when CO(g) is added reaction shift to the left; K no change since no change in T} c. left; no change in K {V reduced P increased need to reduce gas moles; 1 gas mol on the left; 2 gas mol on the right; shift to the left to reduce gas mol; K no change since no change in T} d. no shift; no change in K {no shift a catalyst speeds up the reaction but does not change the equilibrium position; K no change since no change in T} e. no shift; no change in K {when C(s) is added no shift since solids are not part of the K expression; K no change since no change in T} f. right; K increases {write reaction and include heat for exothermic reaction: C(s) + O 2 (g) 2CO(g) + heat; when reaction is cooled system will shift to generate heat which is to the right; when a reaction shifts right and this is caused by a change in T, K will increase (more products and fewer reactants)} 19. a. right; no change in K {when O 2 (g) is added reaction shifts to the right; K no change since no change in T} b. left; no change in K {when NO(g) is added reaction shifts to the left; K no change since no change in T} c. no shift (equal #gas moles); no change in K {V increased P decreased need to increase gas moles; 2 gas mol on the left; 2 gas mol on the right; same gas mol on each side no shift since the shift does not increase gas mol; K no change since no change in T} d. no shift; no change in K {no shift a catalyst speeds up the reaction but does not change the equilibrium position; K no change since no change in T} e. left; no change in K {when N 2 (g) is removed reaction shifts to the left; K no change since no change in T} f. left; K decreases {write reaction and include heat for endothermic reaction: N 2 (g) + O 2 (g) + heat 2NO(g); when reaction is cooled the system will shift to generate heat which is to the left since heat is then a product; when a reaction shifts left and this is caused by a change in T, K will decrease (less products and more reactants)} 20. a. amount of CO 2 increases {when N 2 O is added, since it is part of the K expression, this causes the reaction to shift to the right and increase the amount of CO 2 } b. amount of CS 2 increases {when N 2 is added, since it is part of the K expression, this causes the reaction to shift to the left and increase the amount of CS 2 } c. no change {when a catalyst is added, it speeds up the reaction but does not change the equilibrium position of the chemicals} d. concentration of N 2 decreases {when N 2 O is removed, since it is part of the K expression, this causes the reaction to shift to the left and decrease the amount of N 2 ; as the amount of N 2 decreases its concentration also decreases} e. concentration of CO 2 increases {when N 2 is removed, since it is part of the K expression, this causes the reaction to shift to the right and increase the amount of CO 2 ; as the amount of CO 2 increases its concentration also increases} f. no change {when CS 2 is added, since it is not part of the K expression, this does not cause the reaction to shift and the amount of N 2 O does not change} g. no change {when CO 2 is removed, since it is part of the K expression, this causes the reaction to shift to the right and will decrease the amount of CS 2 ; however, this does not change the concentration of CS 2 because it is a liquid and the concentrations of liquids and solids are constant (i.e., the concentration of a liquid is just its density which doesn t change unless T changes)} h. amount of S increases {when heat is added to an endothermic reaction, this causes the reaction to shift to the right and will increase the amount of S}