Solution W2009 NYB Final exam

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1 Solution W009 NYB Final exam Question Consider one liter of solution with [H SO mol/l. Then : Mass of solution: 000 ml x.3 g/ml 30 g solution mass of H SO 4 : 3.75 mol /L x g/mol 368 g mass of water : 30 g g 86 g water mole of water : 86g x ( mol/8.0 g) 47.9 mol water a) mass percent H SO 4 : 368 g / 30 g x 00% 9.9% b) molality H SO 4 : 3.75 mol / 0.86 kg solvent 4.35 molal c) mole fraction of H SO 4 : 3.75 mol H SO 4 /(3.75 mol mol) d) number of mole of H SO 4 required:.5l x 0.0 mol/l 0.5 mol H SO 4 volume of 3.75M corresponding to 0.5 mol: 0.5 mol x ( L/3.75mol) L or 40. ml Question NaCl g/mol, osmotic pressure ( ) 7.97 atm, T 37.0 C Concentration of NaCl: (0.4 g)x( mol/58.44 g) / L 0.66 mol/l Since i M R T then i i Π MRT 7.97atm (0.66 mol/l)( L.atm/K.mol)( )K.89 Question 3 From k A exp(- E a RT ) we can get ln k E a k R " % ' # T T & since k is used to find the rate, therefore M/t k and k can be replaced by M/t. Since M is the same in both cases the equation become: ln t t " % " ' x E a % ' à ln t ln t + # T T& # R & " % " ' x E a % ' # T T & # R & ln t ln t 8.6 min " 368.6K % " ' x 453x03 J/ mol% # 373.K& # 8.3 J/mol.K ' &

2 Question 4 a. Reaction rate k[clo n [OH - m If [OH - and k are kept constant between two experiments (example: exp and exp ) then Rate Rate n [ClO - n.30x0 & 0.00 # -!" n [ClO 5.75x0 % n therefore n for [ClO (the units cancelled out, therefore, they are not written here to make the numbers more apparent) Since n, then m can be found by using the combination of the experiments and 3 Rate [ClO [OH - m -.30x0 & 0.00 # Rate3 -!" [ClO 3[OH - m 3.30x0 % 0.00 & 0.00 #!" % m 4 4 m therefore m for [OH - Finally the rate law is: rate k[clo [OH - b. Any experiment can be used to determine the rate constant: experiment : rate k[clo [OH - then k rate [ClO [OH - k.30x0 [0.00 mol.l - - mol.l.s [0.00 mol.l - k 30. L mol - s - c. The overall order of this reaction is n+m 3 Question 5 a. If 85% have reacted, then the NOBr concentration is: 5/00 x M 0.0 M Since it is a second order reaction, then [NOBr [NOBr 0 k t + the equation becomes: t [NOBr [NOBr 0 k t [NOBr [NOBr k 0 [0.0 mol.l mol.l [0.080 mol.l - s s. b. The half-life for a second order reaction is: t / k [NOBr 0 t / 0.80 M - s - [0.080 M 0 t / 6 s c. True: ii. False: i, iii, iv, v, vi.

3 Question 6 a. Pressure of HI at equilibrium can be obtained from: K p ( P HI ) P H S Since K p uses values in atm, the initial pressure of H S has to be converted: 0. kpa x atm/0.3 kpa 9.97x0 - atm H S(g) + I (s) HI(g) + S(s) I 9.97x0 - atm C -x -x +x +x E 9.97x0 - x atm - +x atm - ( x) Therefore:.34x0-5 where x is the pressure of HI at equilibrium x0 x Since K p is small, then we can assume that: 9.97x0 - >> x. In this case, the equation becomes.34x0-5 x 9.97x0-4x x.34x x0-4 (the approximation was right since 5.79x0-4 /9.97x0 - x00% 0.58%) In this case, the HI pressure at equilibrium is: x P HI.6x0-3 atm ( M ) HIRT b. since PV nrt à P MRT then K p [ M HI K p [ M xrt HS K p therefore K RT à K M HS RT ( L.atm.K.34x0 - mol -5 )x( )K 4.90x0-7 c. the reaction is inverted and the amount of material is doubled. Therefore: ' K p & % K p #!! " à ' K p & %.34x0-5 #! " 5.57x0 9 Question 7 a. True: i, ii, iii, iv, v. False: vi. b. Left: iii, iv. no change: i, ii, vi. right: v.

4 Question 8 a) percent dissociation [dissociated salt or acid [initial concentration X 00% [dissociated percent dissociation x [initial 00% 8.0% x [.58 00% 0.3 M b) ph -log[h + à ph - log (0.3) 0.89 HClO (aq) H + (aq) + ClO - (aq) I C -x +x +x E 0.58 x +x +x K a [H+ [ClO - [HClO [0.3[0.3 [ x0 - c) Calculate first the concentration of H + or ClO - in solution: K a (x ) x if x << 4.5 then x (4.5)x(.x0- ) 0. The percent dissociation is 0. x00% 4.9% therefore the approximation was right 4.5 Question 9 a) b) CH 3 NH H O - ClO 4 strongest base weakest base NH 4 Cl KNO 3 KF Most acidic most basic c) KCN Concentration mol 0.54 g x 65. g 0.5 L 6.3x0 - M K b K w K a 0-4.6x x0 CN - (aq) + H- OH(l) HCN(aq) + OH - I 6.3x C -x - +x +x E 6.3x0 - x - x x (aq) K b (x ) 6.3x0 - - x if x << 6x0 - then x (6.3x0 - )x(.6x0-5 ).0x0-3 [OH -.

5 ph 4 - (-log.0x0-3 ).00 (Check: (.0x0-3 /6.3x0 - )x00%.6% Therefore, the approximation was right) Question 0 a) The system is already at equilibrium. Since the ph 4.00 then [H +.0x0-4. Therefore: HNO (aq) H + (aq) + NO - Eq. 0.0 M.0x0-4 x (aq) K a [H+ [NO - [HNO à NO - K a x [HNO [H + 4.6x0-4 x 0.0 M.0x0-4 M 0.9 M The mass of KNO required is: (0.9 mol/l) x L x 85. g/mol 39 g Question a) After the addition of 5.0 ml Ba(OH), the concentration of the species are: [HCOOH (0.000 L) x (0.0 M) ( ) L M [Ba(OH) (0.050 L) x (5.0x0 - M) ( ) L 0.0 M HCOOH(aq) + Ba(OH) (aq) Ba(HCOO) (aq) + H O(l) Initial M 0.0 M 0 excess reaction - x 0.0 M -0.0 M +0.0 M + x 0.0 M product 0.05 M M excess After the reaction, the following equilibrium is present HCOOH(aq) H + (aq) + HCOO - (aq) I 0.05 M M C -x +x +x E 0.05 M - x x 0.04 M + x If x << 0.05 M then: [H + [HCOO - K a [HCOOH à [H + K a x [HCOOH [HCOO -.8x0-4 x 0.05 M 0.04 M 6.4x0-5 M ph -log(6.0x0-5 ) 4.9

6 [H + [HCOO - check: K a (6.4x0-5 )( x0-5 ) [HCOOH ( x0-5 ) therefore, the approximation was right..8x0-4

7 Question (Cont.) b) At equivalence point, n mole acid n mole base n mole acid (0.000 L) x (0.0 mol/l).0x0-3 mole acid.0x0-3 mole acid x base acid.0 x0-3 mole of base required The volume of Ba(OH) required is:.0x0-3 mole Ba(OH) x L 5.0x0 - mole 0.00 L of 0. ml c) At equivalence point, Ba + and HCOO - are in large amount. However, only HCOO - is responsible for the ph (conjugated base of a weak acid). Therefore: [HCOO -.0x0-3 mole L 5.0x0 - M K b K w K a x x0 HCOO - (aq) + H-OH(l) HCOOH(aq) + OH - I 5.0x C -x -x +x +x E 5.0x0 - x - +x +x (aq) K b (x ) 5.0x0 - - x if x << 5.0x0 - then x (5.6x0 - )x(5.0x0 - ).7x0-6 [OH -. [H + The approximation was right since x is less than 5% of 5x x x0 ph -log(6.0x0-9 ) 8.3 Question Two reactions are present NaI(aq) + Pb(NO 3 ) (aq) PbI (s) + NaNO 3 (aq) K sp.4x0-8 a) NaI(aq) + AgNO 3 (aq) PbI(s) + NaNO 3 (aq) K sp.5x0-6.4x0-8 [Pb + [I - [I -.4x x0-4 M 0.0.5x0-6 [Ag + [I - [I -.5x0-6.0x x0-3 M Ag + will be the first one to precipitate when [I - 7.5x0-3 M b) The second species will begin to precipitate when [I - 3.7x0-4 M. Therefore, the concentration of Ag + in solution when Pb + will begin to precipitate will be: K sp AgI [Ag + [I - à.5x0-6 [Ag + [3.7x0-4 à [Ag + 4.0x0-3 M

8 Question 3 a) i. A liquid that boils ΔS < 0 ΔS > 0 ii. Sugar that crystallized out from a supersaturated sugar solution iii. Iron rusts (formation of Fe O 3 from pure Fe and O ) iv. A-B(g) + C-D(s) A-B-C(g) + D(s) v. N O 4 (g) NO (g) H O vi. NaCl(s) Na + (aq) + Cl - (aq) ΔH sol +4.0 kj/mol b) ΔG ΔH - TΔS. At T boiling temperature, the system is at equilibrium. Therefore, ΔG 0 -ΔG o -ΔS o T T -58.5x0 3 J.mol 9.9 J.K K or C mol Question 4 ΔS mol x (40.5 J J ) - mol x 304 K mol K mol +77 J K or 0.77 kj K ΔH x (33.8 kj kj ) - mol x 9.67 mol mol kj since : ΔG ΔH - TΔS therefore: ΔG kj - (T)K x 0.77 kj K a) if T 5.0 C, then T ( )K T 98.K ΔG kj - (98.K) x 0.77 kj K à ΔG + 5. kj ΔG > 0 therefore not spontaneous b) if T 60.0 C, then T ( )K T 333.K ΔG kj - (333.K) x 0.77 kj K à ΔG -. kj ΔG < 0 therefore spontaneous

9 Question 5 Data for the Unknown Solute/Cyclohexane Solution Mass of empty test tube, stopper, beaker g Mass of test tube, stopper, beaker, & cyclohexane g Mass of test tube, stopper, beaker, & unknown solute/cyclohexane solution g Mass of cyclohexane g Mass of unknown solute g 0.4 Freezing Temperature of unknown solute/cyclohexane solution C 4.7 Molar mass of unknown solute g mol 88 Mass of cyclohexane: g Mass of the unknown: g ΔT f K f x m à m ΔT f K f ( ) C 0. C.mol mol kg mol solute m à kg solvent finally, the molar mass is: mol solute (m) x (kg solvent) mol solute (0.3 mol kg ) x (9.350x0-3 kg).9x0-3 mole molar mass mass 0.4 g n.9x g/mol mole

CHEM1109 Answers to Problem Sheet Isotonic solutions have the same osmotic pressure. The osmotic pressure, Π, is given by:

CHEM1109 Answers to Problem Sheet 5 1. Isotonic solutions have the same osmotic pressure. The osmotic pressure, Π, is given by: Π = MRT where M is the molarity of the solution. Hence, M = Π 5 (8.3 10 atm)