Chemistry 192 Final Exam Spring 2018 Solutions

Transcription

1 Chemistry 192 Final Exam Spring 2018 Solutions R = J mol 1 K 1 R=.0821 L atm mol 1 K 1 R= L bar mol 1 K 1 N A = molecules mol 1 T = t F = C mol 1 c = m s 1 1. kg = g First order: [A] = [A] 0 e kt Second order: 1 [A] = 1 + kt [A] 0 [H 3 O + ][OH ] = Each question is worth 25 points. 1

2 1. The pk a of iodoacetic acid (CH 2 ICOOH) is pk a = Calculate the ph of a M iodoacetic acid solution. Approximations do not work for this problem. CH 2 ICOOH (aq) + H 2 O (l) CH 2 ICOO (aq) + H 3O + (aq) K a = = [CH 2 ICOOH] [CH 2 ICOO ] [H 3 O + ] initial M 0 M 0 M change y M y M y M equilibrium ( y) M y M y M = [CH 2ICOO ][H 3 O + ] [CH 2 ICOOH] = y y y y = 0 y = ± [( ) 2 + 4( )] 1/2 2 = [H 3 O + ] = M ph = log 10 ( ) =

3 2. The base dissociation constant of n-butylamine [CH 3 (CH 2 ) 3 NH 2 ] is When an L sample of n-butylamine of unknown concentration is titrated with M HCl, the equivalence point is attained after the addition of 0.27 L of the strong acid. Calculate the concentration of the original n-butylamine solution and the ph of the final solution at the equivalence point. Approximations work for this problem. n CH3 (CH 2 ) 3 NH 2 = n H3 O + = ( mol L 1 )(0.27 L) = mol [CH 3 (CH 2 ) 3 NH 2 ] initial = mol = 0.14 M L [CH 3 (CH 2 ) 3 NH + 3 ] = mol L L = M CH 3 (CH 2 ) 3 NH + 3(aq) + H 2O (l) CH 3 (CH 2 ) 3 NH 2(aq) + H 3 O + (aq) K a = K w K b = = [CH 3 (CH 2 ) 3 NH + 3 ] [CH 3 (CH 2 ) 3 NH 2 ] [H 3 O + ] initial M 0 M 0 M change y M y M y M equilibrium ( y ) M y M y M = y y y y = [H 3 O + ] = ph =

4 3. A buffer solution is made by combining 0.25 moles of methylamine (CH 3 NH 2, having pk b = 3.38) and 0.45 moles of CH 3 NH + 3 with water to produce L of solution. The buffer is then mixed with L of M sodium hydroxide. Calculate the ph of the original buffer solution, and the ph of the solution formed by mixing the buffer with the aqueous sodium hydroxide. Approximations work for this problem. CH 3 NH + 3(aq) + H 2O (l) CH 3 NH 2(aq) + H 3 O + (aq) Original buffer Mixed pk a = pk b = ph = pk a + log 10 [CH 3 NH 2 ] [CH 3 NH + 3 ] = log = n OH = (0.100 mol L 1 )(0.500 L) = mol ph = log =

5 4. The ph of a saturated solution of the sparingly soluble base Ca(OH) 2 is at 298K and at 370K. Calculate r,m H, r,m G and r,m S at 298 K for the reaction Ca(OH) 2(s) Ca 2+ (aq) + 2OH (aq). At 298 K [H 3 O + ] = M = M [OH ] = = M [Ca 2+ ] = 1 2 [OH ] = M K sp (298) = [Ca 2+ ][OH ] 2 = At 370 K [H 3 O + ] = M = M [OH ] = [Ca 2+ ] = 1 2 [OH ] = M = M K sp (370) = [Ca 2+ ][OH ] 2 = ln K sp(t 2 ) K sp (T 1 ) = r,mh ( 1 1 ) R T1 T ln = r,m H ( J mol 1 K K 1 ) r,m H = J mol K r,m G = RT ln K sp = ( J mol 1 K 1 )(298 K) ln( ) = J mol 1 r,m S = r,mh r,m G T = J mol 1 K 1 5

6 5. The gas-phase decomposition of benzene diazonium chloride C 6 H 5 N 2 Cl (g) C 6 H 5 Cl (g) + N 2(g) is first order. At 500. K pure benzene diazonium chloride is placed in a container of constant volume at an initial pressure of P 0 = 3.0 bar. After 5.0 hours the total pressure in the container is found to be P tot = 4.6 bar. Calculate the half life of benzene diazonium chloride at 500. K. n C6 H 5 N 2 Cl n C6 H 5 Cl n N2 initial n change αn 0 αn 0 αn 0 final n 0 (1 α) αn 0 αn 0 n tot = n 0 (1 + α) P tot = n totrt = n 0RT (1 + α) = P 0 (1 + α) V V 4.6 = 3.0(1 + α) α = 0.53 n 0 (1 α) = n 0 e kt k = 0.15 hr 1 t 1/2 = ln 2 k 0.47 = exp ( k(5.0 hr)) = 4.6 hr 6

7 6. The EMF of the electrochemical cell at 298K Ag (s) AgBr (s) Br (aq) (1.75 M) F (aq) ( M) F 2(g)(P ) Pt (s) is V. Given the standard half-cell potentials E F /F 2 /P t = V and E Ag/AgBr/Br = V, calculate the pressure of fluorine gas in the half cell on the right-hand side. Identify the cathode and the anode of the cell. Right : F 2(g) + 2e 2F (aq) Left : 2AgBr (s) + 2e 2Ag (s) + 2Br (aq) P F2 = 0.74 bar Overall : F 2(g) + 2Br (aq) + 2Ag (s) 2F (aq) + 2AgBr (s) Q P = [F ] 2 P F2 [Br ] 2 E = E R E L = V E = E RT nf ln Q P V = V ( J mol 1 K 1 )(298 K) 2(96485 C mol 1 ) ln P F2 The fluorine electrode is the cathode and the silver bromide electrode is the anode. 7

8 7. The formation equilibrium constant for the nickel ammonia complex, Ni(NH 3 ) 2+ 6 is K f = and the solubility product of solid nickel phosphate, Ni 3 (PO 4 ) 2 is K sp = A solution is made by adding moles of nickel ions into 0.50 L of 0.75 M aqueous ammonia. Calculate the concentration of free Ni 2+ ions in the solution. If moles of phosphate ions are then added to the solution, do the necessary calculations to determine if a precipitate of nickel phosphate forms. Approximations work for this problem. Ni(NH 3 ) 2+ 6 Ni 2+ (aq) + 6NH 3(aq) K = 1 K f = Assuming initially all nickel is in the form of the ammonia complex [Ni(NH 3 ) 2+ 6 ] = mol 0.50 L = M n NH3 = (0.75 mol L 1 )(0.50 L) 6(0.025 mol) = 0.23 mol [NH 3 ] = 0.23 mol 0.50 L = 0.45 M [Ni(NH 3 ) 2+ 6 ] [Ni 2+ ] [NH 3 ] initial M 0 M 0.45 M change y M y M 6y M equilibrium (0.050 y ) M y M y M = [Ni2+ ][NH 3 ] 6 [Ni(NH 3 ) 2+ 6 ] = y( y) y y(0.45) y = [Ni 2+ ] = M [PO 3 3 ] = mol 0.50 L = M Q = [Ni 2+ ] 3 [PO 3 4 ] 2 = > K sp Precipitate does form. 8

9 8. Determine the balanced nuclear reaction when Po decays by α emission to a daughter nucleus. Given the masses α= u, Po= u and the mass of the daughter = u, calculate the energy change for the decay reaction in kj mol Po 4 2 He Pb m = u u u = u E = ( m)c 2 = ( g mol 1 )(kg)(10 3 g) 1 ( m s 1 ) 2 = J mol 1 = kj mol 1 9

10 N(Score) Score Figure 1: High = 193, Median = 165, Mean =