Solutions to Equilibrium Practice Problems

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1 Solutios to Equilibrium Practice Problems Chem09 Fial Booklet Problem 1. Solutio: PO 4 10 eq The expressio for K 3 5 P O 4 eq eq PO 4 10 iit 1 M I (a) Q 1 3, the reactio proceeds to the right. 5 5 P O 1 M 1 M 4 iit iit PO 4 10 iit 5.0 M I (b) Q 3.7 3, the reactio proceeds to the left. 5 5 P O 8.0 M 0.70 M 4 iit iit 1 of 18

2 Problem. Solutio: Chem09 Fial Booklet To determie the fial cocetratios, the first thig eeded are the iitial reactat cocetratios ad a expressio for the reactio coefficiet Q. [A (g) ] = 1.00 mol 0.50 L = 4.00 M, [AB (g)] = [B (g) ] =.00 mol = 8.00 M, 0.50 L ad Q ABiit A B (8.00).0 (4.00) (8.00) iit iit The directio of the reactio eeds to be determied. To do this, we compare Q vs K. Sice Q =.0 > K = 0.5, the reactio proceeds to the left. To determie what fial cocetratios will be from iitial cocetratios, a hady tool the Iitial, Chage, Equilibrium (ICE) table ca be used. All compouds ivolved i the reactio are icluded i a ICE table as follows, with the species that will be cosumed o the left side, ad the species that will be produced o the right side. Sice the reactio proceeds to the left, A ad B will be formed ad AB will be cosumed. Cocetratio (M) AB (g) A (g) B (g) Iitial 8.00 M 4.00 M 8.00 M Chage -x +x +x Equilibrium 8.00 x x x As the reactio proceeds, x moles of A (g) ad B (g) are formed as x moles of AB (g) are cosumed. Make sure to cosider stoichiometric coefficiets appropriately. The equilibrium values are simply the sum of the iitial + chage cocetratios. Substitute the equilibrium cocetratios ito K, ad solve for x (remember that K is defied for the reactio i the way that it was iitially described): K ABeq A B (8.00 x) 0.5 (4.00 x) (8.00 x) eq eq so, of 18

3 (8.00 x) 0.5(4.00 x)(8.00 x) 64 3x 4x 0.5(3 1 x x ) 64 3x 4x 16 6x 0.5x 3.5x 38x 48 0 Chem09 Fial Booklet To solve for x, the quadratic formula must be used, 38 ( 38) 4(3.5)(48) b b 4ac x 1.46 or 9.39, a (3.5) Sice x 9.39 = > 8.00 (which would leave the equilibrium cocetratio of AB at equilibrium egative), the the x = 1.46 So, fial cocetratios are: [A] = = 5.46 M [B] = = 9.46 M [AB] = 8.00 (1.46) = 5.08 M To check, substitute these cocetratios ito the Equilibrium costat expressio, ABeq A B (5.08) K 0.5, matches up. (5.46) (9.46) eq eq 3 of 18

4 Chem09 Fial Booklet Problem 3. Aswer: K c 4 [ NO] [ H ] [ NO ] Solutio: Pure solids ad pure liquids ad solvets are ot icluded i the expressio. Products go over Reactats, ad coefficiets i the equatio are writte as superscripts. Problem 4. Solutio: K c = [CO (g) ] Remember: Solids ad liquids are ot icluded i the expressio. Problem 5. Solutio: Equatio is equal the double ad reverse of equatio1 therefore K p = K p1 = ( ) = 40.6 Problem 6. Aswer: Equilibrium Costat = K 1 1/ = 1 Solutio: The secod equatio if reversed (1/K) ad halved (K1/). Combiig these two gives 1/K1/. K 1 Problem 7. Aswer: The reactio equatio is: The relatioship betwee K c ad K p is: CO (g) + O (g) CO (g) K c = K p = K c (RT) gas I this case there are 3 moles of gas i the reactats ad moles of gas i the products, so = -1 So solvig for K p : K p = K c (RT) gas = ( ) [(0.081 L atm mol K ) (98 K)] 1 = of 18

5 Problem 8. Solutio: Chem09 Fial Booklet N (g) + C H (g) HCN (g) i: c: +x +x -x e: x x 1.00-x Q=1>K so the rx goes to the left (1.00 x) (1.00 x) K c (1.00 x) (1.00 x) K c 1 x K K c c Problem 9. Solutio: COCl (g) CO (g) +Cl (g) iitial: chage: -x +x +x equil: 0.04-x x x x Kc = 0.04 x x + Kcx 0.04Kc = 0 Kc Kc (4)(0.04)( Kc) x of 18

6 Problem 10. Solutio: Kc [ CO ][ H ] [ CO][ H O] 5.10 Chem09 Fial Booklet CO + H O CO +H I 0.1M C -x -x +x +x E 0.1-x 0.1-x 0.1+x 0.1+x (0.1 x) x x Kc 5.10 (0.1 x) x x x x x x x 1.x x 0.034M Therefore, at equilibrium [H] = x = M = 0.134M Problem 11. Solutio: Oly C is true A icrease i volume will shift the equilibrium to the right thus causig a icrease i the total moles of CO at equilibrium. Problem 1. Solutio: a) shift to the left b) o effect c) shift to the right d) shift to the right Problem 13. Solutio: The reactio will shift left formig Ni(CO) 4(g) to reach equilibrium Problem 14. Aswer: C Solutio: Q = [NO] [Cl ] [NOCl] = (1.) (0.56) (1.3) = Q=K, therefore we are already at equilibrium. 6 of 18

7 Chem09 Fial Booklet Solutios to Acids ad Bases Practice Problems Problem 15. Solutio: Equal volumes of 0.1 M NaF ad 0.1 M HF Addig a acid ad the salt of its cojugate base ca form a buffer, this is the case i (b). Problem 16. Solutio: 1 ad 3. A buffer ormally cosists of a weak acid ad its cojugate base i roughly equal amouts. The ratio should be o greater tha 0.1 to 10 of weak base to weak acid. I 1, the ratio of acid to base is 3:1. I 3 the ratio is 4:1. I, there is a greater amout of strog base tha acid, so all of the acetic acid is cosumed so it is ot a buffer. 4 is ot a buffer solutio sice all of the weak acid is cosumed with strog base, 5 is ot a buffer because sodium acetate is a weak base, so you have a weak base with a strog base i solutio. Problem 17. Solutio: B This is a buffer solutio, therefore: [ A ] ph pka [ HA] eq eq log log( ) log 1.74 Problem 18. Solutio: This is a buffer solutio, therefore: [ A ] eq ph pka log [ HA ] eq [ CH COONa] = = 4.75 log [ CH3COONa ] Therefore log =0 ad [ CH3COONa ] =1 [CH3COONa] = 0.10M # moles CH3COONa = 0.10 moles Problem 19. Solutio: For this equilibrium [H+] = [I-] = 10-8 = 1 x 10-8 Ka = 1 x 10-6 = [ H ][ I [ HI] ] = -8 (1x 10 ) [ HI] [HI] = 1 x 10-10, therefore [HI]/[I- ] = 1 x 10-10/1 x 10-8 =0.01 = 1/100 7 of 18

8 Problem 0. Solutio: ph = 8.91 Chem09 Fial Booklet ph = pka + = = 8.91 CN log HCN 1.0 log.0 Problem 1. Solutio: C This is a buffer system as there are equal amouts of cojugate acid ad base, but the additio of H+ ca chage the ph slightly. The origial ph of the buffer is approximately equal to pka. Whe H+ ios are added from the strog acid HCl, A- is coverted ito HA. Therefore the additio of 0.01 moles H+ produces 0.01 moles HA ad cosumes 0.01 moles A-. HA= = 0.51 mol A- = = 0.49 mol [H+]=Ka x HA/A- = (1.8 x 10-4) (0.51/0.49) = x 10-4 ph = -log (1.873 x 10-4) = 3.77 Problem. Solutio: C There are more moles of carboate buffer i solutio c tha ay of the other solutios Problem 3. Solutio: E The solutio is a buffer with ph above 7. A buffer is resistat to both additio of strog acid ad strog base ad the cocetratio of the hydroium io is ot more tha the hydroxide io (ph >7). Problem 4. Solutio: E HCN + H O CN + H 3 O + I 0.5 y 0 C E 0.5 y Ka = 6. x = [(1x10-7)( Y+1x10-7)]/ 0.5 Y=0.003M x 1L = 0.003moles Mass = g/molNaCN x moles = 0.15g 8 of 18

9 Chem09 Fial Booklet 9 of 18

10 Chem09 Fial Booklet Problem 5. Solutio: E A buffer ormally cosists of a weak acid ad its cojugate base i roughly equal amouts. The ratio should be o greater tha 0.1 to 10 of weak base to weak acid. I 1, the ratio of acid to base is 3:1. I 3 the ratio is 4:1. I, there is a greater amout of strog base tha acid. Problem 6. Solutio: C or E I order to determie which solutios are able to act as buffer solutios, determie what ios will be foud i the solutio ad whether those ios are acidic, basic or spectator ios. If acidic ad/or basic ios are foud, calculate the amout of ios that are preset. For solutio A) all ios preset (H+, NO3-, ad Na+) are spectator ios. No buffer abilities possible. For solutio B) all ios preset (Na+, OH- ad Cl-) are spectator ios. No buffer abilities possible. For solutio C) This is a 1:1 ratio of cojugate acid to its cojugate base. This IS a buffer. For solutio D) the HCl completely dissociates to form H+ ad Cl- ios. NH3 ca form a equilibrium where NH3 + H+ NH4+. However, HCl ad its dissociated ios are preset i 0.01mol amouts the same as the amout of NH3. Therefore, all the NH3 is used up i reactig with the HCl, so o bufferig abilities are possible. For solutio E) the NaOH completely dissociates to form Na+ ad OH- ios. The 0.004mol of OH- will react with the 0.01mol of HF to form 0.004mol of F- ad have 0.006mol left of HF sice the OH- is the limitig reaget. This meas we have HA ad A- both preset i a ratio of 3:. This is a buffer. Problem 7. Solutio: C B + HO BH+ + OH- Calculate from the give ph, the cocetratio of OH- ios that dissociate form from the reactio of the base with water. poh = 14 ph = 14 (8.88) = 5.1 [OH-] = 10-pOH = = x 10-6M Assume 1 L of solutio, therefore the [B] = 0.40mol/L ad [BH+] = 0.50mol/L. K = [BH +][OH ] [B] (0.50)( x 10 6) = = 4.74 x of 18

11 Problem 8. Solutio: D Chem09 Fial Booklet Because equal volumes of the acid ad weak base are beig mixed, all cocetratios (M) ca be treated as moles (mol). HNO3 is a strog acid so it completely dissociates HNO3 H+ + NO NH3 will react with the H+ released by the HNO3 to form NH4+. Iitially there is 0.3M or 0.3moles of NH3. Upo additio of 0.1moles H+ (from the HNO3), 0.1mol of NH3 will react to form 0.1mol of NH4+, leavig 0.mol NH3 ureacted. Therefore, NH3 + H+ NH4+ Usig the followig equatio, solve for [H+]. [H+] = Ka x [HA] [A-] Kw = Ka x Kb so that Ka = Kw = 1.0 x = 5.56 x Kb 1.8 x 10-5 [H+] = 5.56 x x [0.1] =.78 x 10-10M [0.] ph = -log[h+] = -log[.78 x 10-10] = of 18

12 Problem 9. Solutio: B Chem09 Fial Booklet Write out the equatios that are occurrig i the solutio described above. Because both solutes are beig added to 1L of water, all cocetratios (M) ca be treated as moles (mol). CH5COONa is a soluble salt so it completely dissociates CH5COONa CH5COO- + Na HCl is a strog acid so it completely dissociates HCl H+ + Cl CH5COO- will react will all the H+ to form CH5COOH. Iitially there is 0.1 mole of CH5COO-. Whe 0.01mol of H+ is added, the CH5COO- reacts leavig 0.09mol. There is also 0.1mol of CH5COOH to start, but whe the CH5COO- reacts with the H+, it forms 0.01mol more CH5COOH so that the total amout of CH5COOH is 0.11mol. CH5COOH CH5COO- + H+ Before HCl is added After HCl is added Calculate [H+] usig the followig equatio. ( = moles) [H+] = Ka x HA = 1.41 x 10-5 x (0.11) = 1.7 x 10-5M A- (0.09) Calculate ph from the [H+]. ph = -log[h+] = -log[1.7 x 10-5] = of 18

13 Chem09 Fial Booklet Solutios to Kietics Practice Problems Problem 30. Solutio: v k[ I ] [ S O ] v k I S O m m [ ] [ 8 ] m k(0.080) (0.040) m k(0.040) (0.040) v v 1 3 m m m k(0.080) (0.040) m k(0.080) (0.00) 1 rate k[ I ] [ S O ] k rate [ I ][ S O ] k M s (0.080) (0.040) Problem 31. Solutio: Rate = k[h O] x [CH 3 Cl] y x Rate1 k(0.0100) x Rate k(0.000) x =.5, therefore x = ad the reactio is secod-order i H O x Rate1 k(0.0100) x Rate k(0.000) ( y 3 ) = 3.69 ( 3 )y =, therefore y = 1 ad the reactio is first-order i CH 3 3Cl Rate = k[h O] [CH 3 Cl] 1 13 of 18

14 Problem 3. Solutio Chem09 Fial Booklet [ A] 0 l[ A] t l[ A] kt, l kt 1/ [ A ] t1/ [ A] 0 for t1/ [ A] t 1/ [ A] 0 [ A] 0 l l l kt1/ [ A] t [ A] 1/ 0 l t1/ k 0 1/ 1/ Problem 33. Solutio: If 0% decomposes, the 80% of the sample remais. l[ A] l[ A] kt rearrage 0 l 0.8[A] o k(50s) [A] o k sec 1 t 1/ l k 155 sec Problem 34. Solutio: (a) v v k[ CH NNCH ] k[ CH3NNCH 3] ( ) (.0510 ) Therefore, the reactio is the first order. Rate = k [CH 3 NNCH 3(g) ], therefore k = (b) l[a] t = l[a] 0 kt M / s = 5.46 x 10-5 s-1 M l[ ] = l[ ] ( ) t -5.7 = ( )t ( )t =.30 t = s = 11.7 hours (c) [A]10% = = M Rate = k[a] = ( s-1)( M) = M/s 14 of 18

15 Chem09 Fial Booklet 1 1 Problem 35. Solutio: Because this reactio is secod order kt if 1/[A] was plotted vs. t [ A] [ A] o the the y-itercept would be 1/[A]0 ad the slope would be k 1/[A] slope = +k 1/[A] Problem 36. Aswer t Problem 37. Solutio: Sice Hrx > 0, the Hproducts > Hreactats 15 of 18

16 Thus, Ea rev = = 5 kj/mol (draw a eergy diagram to see this better) Chem09 Fial Booklet Problem 38. Solutio: Let T1 = 96 C = 369 K, ad T = 5 C = 98 K. Usig the Arrheius Equatio l ( k 1 k ) = E a R ( 1 T 1 T 1 ) x 10 s E 1 1 a l 1. x 10 s JK mol 369 K 98 K ad solve for Ea, Ea = 1.8 kj/mol Problem 39. Solutio: from Arrheius equatio sice vk, the l k 1 E 1 R T T k a 1 l k l v E a 1 1 k 1 v 1 R T T 1 l(3) E a E a J / mol 77.4 kj / mol Problem 40. Solutio: Sice we have a elemetary process, the rate of the reactio is 1 1 This reactio is secod order therefore kt [ A] [ A] 1 0.1M 1 0.M 5 M-1 = k(35. mi) k(35. mi) k = 1.4 x 10-1 M-1mi-1 o rate k[a], 16 of 18

17 Chem09 Fial Booklet Problem 41. The steps are elemetary, ad the first step is the rate determiig step. So the overall rate law oly depeds o the first step: rate = k[o3][no] 17 of 18

18 Chem03 Fial Booklet 18 of 18

Chem 209 Final Booklet Solutions

Chem 209 Final Booklet Solutions Chem09 Final Booklet Chem 09 Final Booklet Solutions 1 of 38 Solutions to Equilibrium Practice Problems Chem09 Final Booklet Problem 3. Solution: PO 4 10 eq The expression for K 3 5 P O 4 eq eq PO 4 10