CHM 152 Practice Final

Transcription

1 CM 152 Practice Final 1. Of the fllwing, the ne that wuld have the greatest entrpy (if cmpared at the same temperature) is, [a] 2 O (s) [b] 2 O (l) [c] 2 O (g) [d] All wuld have the same entrpy at the same temperature. [e] Mre infrmatin wuld be needed t make a selectin. 2. A prcess will be nnspntaneus at all temperatures if [a] the enthalpy change is and the entrpy change is +. [b] the enthalpy change is + and the entrpy change is +. [c] the enthalpy change is and the entrpy change is. [d] the enthalpy change is + and the entrpy change is. [e] all reactins will be spntaneus at sme temperature. 3. Given the fllwing standard state free energies f frmatin, G f (kj/ml) 2 O (l) NO 3 (l) 79.9 NO (g) 86.7 NO 2 (g) 51.8 Calculate G fr the reactin: 3NO 2 (g) + 2 O(l) 2NO 3 (l) + NO(g) [a] G = 8.7 kj [b] G = 192 kj [c] G = 414 kj [d] G = 192 kj [e] G = 155 kj 4. Cnsider a reactin fr which = 13.2 kj and S = 125 J/K. Calculate G rxn fr this reactin at a temperature f 95.0 K. [a] 25.1 kj [b] 13.3 kj [c] 13.1 kj [d] 11.9 kj [e] 1.33 kj 5. Cnsider the reactin, 2SO 2 (g) + O 2 g) 2SO 3 (g) G fr this reactin is 142 kj. Calculate the G fr this reactin at 25 C when the partial pressures f the gases are as fllws: P(SO 2 ) = atm, P(O 2 ) = atm, and P(SO 3 ) = atm. [a] 130 kj [b] 4.8 kj [c] 142 kj [d] 147 kj [e] 154 kj 6. The activatin energy f a reactin may be lwered by, [a] increasing the cncentratin f the reactants [b] increasing the temperature [c] adding a catalyst [d] answers b and c are bth crrect [e] answers a, b, and c are all crrect.

2 Cnsider the fllwing reactin and data fr questins 7 and 8. Exp# [NO] [ 2 ] Rate (ml/l s) M 0.20 M M 0.30 M M 0.20 M NO N O 7. If 2 were disappearing at the rate f ml/l s, N 2 wuld be appearing at the rate f, [a] ml/l s [b] 0.16 ml/l s [c] ml/l s [d] 0.24 ml/l s [e] The rate f N 2 appearance can nt be determined unless the temperature is knwn. 8. The rate law fr the reactin is, [a] rate = k[no] 2 [ 2 ] 2 [d] rate = k[no] 2 [ 2 ] [b] rate = k[no][ 2 ] 2 [e] nne f these are crrect. [c] rate = k[no][ 2 ] 9. The reactin 2NO 2NO + 2 is fund t bey the rate law, rate = k[no] 2. Of the fllwing three pstulated mechanisms, the ne(s) pssible is/are, 1) NO NO + (slw) 2) 2NO NO 2 + NO (slw) + NO NO 2 (fast) NO 2 NO + 2 (fast) NO 2 + NO 2NO + 2 (fast) 3) NO NO + (slw) NO + NO + 2 (fast) [a] 1 [b] 2 [c] 3 [d] 1 and 3 [e] 2 and At a certain temperature K c = fr the reactin: 2 (g) + F 2 (g) 2F(g) If 1.0 ml 2 and 1.0 ml F 2 are placed in a reactin vessel and allwed t react, then at equilibrium... [a] [F] will be much larger than [ 2 ] and [F 2 ]. [b] [F] will be much less than [ 2 ] and [F 2 ]. [c] [F] will be nearly equal t [ 2 ] and [F 2 ]. [d] the system will cntain nly 2 and F 2. [e] the system will cntain nly F. 11. The fllwing reactin cmes t equilibrium: 2 2 (g) + S 2 (g) 2 2 S(g) If the vlume f the reactin vessel is reduced, at cnstant temperature, the cncentratin f [a] 2 wuld increase. [b] 2 wuld decrease. [c] S 2 wuld increase. [d] 2 S wuld decrease. [e] 2, S 2 and 2 S wuld nt change. 2

3 12. Fr the equilibrium: 2N 2 O(g) + O 2 (g) 4NO(g), = +198 kj. Which f the fllwing changes will increase the amunt f N 2 O when the system reaches its new equilibrium state? [a] The temperature f the system is lwered. [b] The vlume f the cntainer is increased. [c] Sme O 2 is added. [d] e(g) is added withut changing the vlume. [e] Sme NO is remved. 13. Fr the reactin: 2 (g) + CO 2 (g) 2 O(g) + CO(g), the equilibrium cnstant, K p, is 4.40 at 2000 K. Calculate G fr this reactin. [a] 73.2 kj [b] 24.6 kj [c] 243 J [d] 243 J [e] 24.6 kj What is the K b value fr the carbnate in, CO 3? [a] K b = [d] K b = ( )( ) [b] K b = [c] K 11 b = [e] Nne f these are crrect Which f the fllwing wuld increase the K a fr acetic acid? [a] Decrease the p f the slutin. [b] Add sme sdium acetate. [c] Add sme sdium hydrxide. [d] Add sme water. [e] Nne f the abve, the K a is temperature dependent nly. 16. Which ne f the fllwing salts will frm a basic slutin n disslving in water? [a] Na [b] KO 4 [c] N 4 NO 3 [d] NaCN [e] Nne f these slutins are basic 17. Which is the cnjugate base f 2 PO 4? [a] 3 PO 4 [b] PO 4 [c] PO 4 2 [d] PO 4 3 [e] 2 PO The p f a slutin is The [O ] is: [a] [O ] = M [b] [O ] = M [c] [O ] = 5.50 M [d] [O ] = M [e] Nne f these is crrect. 19. What is the p f a M NO 3 slutin? [a] p = [b] p = 1.82 [c] p = 2.59 [d] p = [e] This cannt be answered withut mre infrmatin. 3

4 20. What is the p f a 0.10 M slutin f C 3 COONa? [a] p = 4.74 [b] p = 5.13 [c] p = 8.87 [d] p = 9.25 [e] Nne f these 21. Which f the fllwing is the mst acidic slutin? [a] 0.10 M C 3 COO and 0.10 M C 3 COONa [b] 0.10 M C 3 COO [c] 0.10 M NO 2 and 0.10 M NaNO 2 [d] 0.10 M NO 2 [e] 0.10 M C 3 COONa 22. Which f the fllwing mixtures is suitable fr making a buffer slutin with an ptimum p f ? [a] C 3 COONa/C 3 COO [b] N 3 /N 4 [c] NaO/O [d] NaNO 2 /NO 2 [e] Na/ 23. Which f the fllwing 0.10 M slutins cannt act as a gd buffer? [a] F & NaF [b] C 3 COO & C 3 COONa [c] NaCO 3 & Na 2 CO 3 [d] pure 2 O [e] All f these slutins are gd buffers. 24. What is the p f a slutin that is 0.41 M O and M NaO? [a] p = 0.39 [b] p = 3.94 [c] p = 6.58 [d] p = 7.49 [e] p = Fr which titratin will the p be basic at the equivalence pint? [a] with NaO [b] with N 3 [c] O with NaO [d] All f these titratins will be neutral at the equivalence pint. [e] A predictin cannt be made withut additinal infrmatin. 26. The p at the equivalence pint f an acid-base titratin may differ frm 7.00 because f [a] the indicatr used. [c] the initial p f the titrate. [e] the hydrlysis f the salt frmed. [b] the aut-inizatin f water. [d] the cncentratin f the titrant. 27. Chse the statement that describes the rle f B 3 and N 3 in the reactin: B + N B [a] B 3 is a Lewis base and N 3 is a Lewis acid. [b] Bth N 3 and B 3 are Lewis acids. [c] Bth N 3 and B 3 are Lewis bases. [d] B 3 is a Lewis acid and N 3 is a Lewis base. N 4

5 28. What is the p f the slutin resulting frm the additin f 10.0 ml f 0.10 M NaO t 50.0 ml f 0.10 M slutin? [a] p = 1.00 [b] p = 1.08 [c] p = 1.18 [d] p = 7.00 [e] Insufficient data t answer. 29. The mlar slubility f MgCO 3 is ml/l. What is K sp fr this cmpund? [a] [b] [c] [d] [e] The K sp f Ag 2 CrO 4 is The mlar slubility f Ag 2 CrO 4 in 0.10 M AgNO 3 slutin is: [a] M [b] M [c] M [d] M [e] M 31. Fr Pb 2, K sp = Will a precipitate f Pb 2 frm when 0.50 L f M Pb(NO 3 ) 2 is added t 0.50 L f M Na and why? [a] Yes, Q > K sp [b] N, Q < K sp [c] N, Q = K sp [d] Yes, Q < K sp [e] Mre infrmatin is needed t slve the prblem. Fr questins 32 and 33 cnsider the fllwing redx reactin, Zn + NO 3 acidic Zn 2+ + NO 32. In the balanced equatin in acidic slutin, the cefficient f 2 O is, [a] 1 2 O [b] 2 2 O [c] 3 2 O [d] 4 2 O [e] 5 2 O 33. The reducing agent is, [a] Zn [b] NO 3 [c] NO [d] + [e] 2 O 34. Referring t the Standard Reductin Ptential Table n this exam, the strngest xidizing agent listed belw is [a] Mn 2+ [b] Mn [c] O 3 [d] 2 [e] Ag Of the fllwing reactins, the spntaneus ne wuld be, [a] Mn 2+ + Fe Mn + Fe 2+ [b] 3Ag + NO Ag + + NO + 22 O [c] 3Sn Cr O 3Sn 2+ + Cr 2 O [d] all are spntaneus. [e] nne are spntaneus. 36. In a galvanic cell, the reactin ccurring at the cathde is [a] reductin [b] xidatin [c] hydrlysis [d] ismerizatin [e] acid-base 5

6 37. Use data frm the Table f Standard Reductin Ptentials t calculate the equilibrium cnstant fr the fllwing reactin at 25 C. 2 (g) + 2Br (aq) 2 (aq) + Br 2 (l) [a] K = [b] K = [c] K = [d] K = [e] K = The radiactive decay f 87 Kr prduces a beta particle and, [a] an alpha particle [b] a psitrn [c] + [d] 87 Br [e] 87 Rb 39. A first rder reactin has a rate cnstant f k = s 1 at 25 C. The half-life f the reactin is: [a] 1660 s [b] 576 s [c] 289 s [d] s [e] s 40. A certain radiactive element has a half-life f 10.0 minutes. If 2.00 grams f this element were present initially, the amunt f the element remaining after 8.30 minutes has elapsed wuld be: [a] 1.00 grams [b] 1.13 grams [c] 1.32 grams [d] 1.08 grams [e] grams 6

7 Ptentially Useful Infrmatin G = Σn G f (prducts) Σn G f (reactants) Rate = [cnc] t = Σn f (prducts) Σn f (reactants) Rate = k[a] x [B] y S = ΣnS (prducts) ΣnS (reactants) ln[a] t = kt + ln[a] 0 S univ = S sys + S surr [A] t ln = kt [A] 0 G = T S [A] ln 0 = [A] t kt G = RTlnK G = G + RTlnQ t 1 = 2 k R = J/ml K = L atm/ml K 1 1 = kt + [A] [A] 1 K = C t 1 = k[a] PV = nrt mlarity( M ) mles f slute = Lfslutin A quadratic equatin f the frm ax 2 + bx + c = 0, has the slutins: x x = t 2 0 ln k 0 Ea 1 = + T R k1 Ea 1 1 ln = k2 R T2 T 1 b± [prducts] K c = K y p = K c ( T) n [reactants] K w = [ 3 O + ][O ] = at 25 C p + po = 14 p = lg[ 3 O + ] po = lg[o ] K w = K a K b [base] p = pk a + lg [acid] G = nf F = 96,500 J/V ml Ecell Ecell = Ecathde Eande Ecell = RT ln K nf E = E RT ln Q nf b 2a 2 4ac Ecell = Exi + Ered ln A V E cell = ln K at 25 C n V E = E ln Q at 25 n 7

8 Acid Inizatin Cnstants (25 C) Acid Frmula K a pk a ydrfluric F Nitrus NO Benzic C 6 5 COO Acetic C 3 COO Carbnic 2 CO Bicarbnate in CO ypchlrus O Ammnium in + N ydrcyanic CN Standard Reductin Ptentials alf-reactin E (Vlts) Mg 2+ (aq) + 2 e Mg (s) 2.37 Mn 2+ (aq) + 2 e Mn (s) 1.18 Zn 2+ (aq) + 2 e Zn (s) 0.76 Cr 3+ (aq) + 3 e Cr (s) 0.74 Fe 2+ (aq) + 2 e Fe (s) (aq) + 2 e 2 (g) 0.00 Sn 4+ (aq) + 2 e Sn 2+ (aq) Fe 3+ (aq) + e Fe 2+ (aq) Ag + (aq) + e Ag (s) NO 3 (aq) + 4 (aq) + 3 e NO (g) O (l) Br 2 (l) + 2 e 2 Br (aq) Cr 2 O 7 (aq) + 14 (aq) + 6 e 2 Cr (aq) O (l) (l) + 2 e 2 (aq) O 3 (aq) + 12 (aq) + 10 e 2 (g) O (l) Ce 4+ (aq) + e Ce 3+ (aq) Answer Key 1) c 2) d 3) a 4) e 5) a 6) c 7) c 8) d 9) b 10) a 11) b 12) a 13) b 14) b 15) e 16) d 17) c 18) a 19) b 20) c 21) d 22) b 23) d 24) c 25) c 26) e 27) d 28) c 29) d 30) e 31) b 32) d 33) a 34) c 35) b 36) a 37) b 38) e 39) c 40) b 8