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METU NORTHERN CYPRUS CAMPUS CHM 351 1 H 1.0079 3 Li 6,941 11 Na 22.990 19 K 39.098 37 Rb 85.468 55 Cs 132.91 87 Fr (223) Physical Chemistry Final Solutions 15 January 2016 Name, Surname: ID: 4 Be 9.0122 12 Mg 24.305 20 Ca 40.078 38 Sr 87.62 56 Ba 137.33 88 Ra (226) 21 Sc 44.956 39 Y 88.906 57 La* 138.91 89 Ac** (227) *Lanthanides ** Actinides 22 Ti 47.867 40 Zr 91.224 72 Hf 178.49 104 Rf (261) Time Allowed: 180 min Signature: 6 C 12.011 14 Si 28.086 32 Ge 72.64 50 Sn 118.71 82 Pb 207.2 114 Uuq (289) 7 N 14.007 15 P 30.974 33 As 74.922 51 Sb 121.76 83 Bi 208.98 115 UUp (288) 8 O 15.999 16 S 32.065 34 Se 78.96 52 Te 127.60 84 Po (209) 116 Uuh (291) 9 F 18.998 17 Cl 35.453 35 Br 79.904 53 I 126.90 85 At (210) 2 He 4.0026 10 Ne 20.180 18 Ar 39.948 36 Kr 83.798 54 Xe 131.29 86 Rn (222) 118 Uuo (294) 23 V 50.942 41 Nb 92.906 73 Ta 180.95 105 Db (262) 24 Cr 51.996 42 Mo 95.94 74 W 183.84 106 Sg (266) 25 Mn 54.938 43 Tc (98) 75 Re 186.21 107 Bh (264) 26 Fe 55.845 44 Ru 101.07 76 Os 190.23 108 Hs (277) 27 Co 58.933 45 Rh 102.91 77 Ir 192.22 109 Mt (268) 28 Ni 58.693 46 Pd 106.42 78 Pt 195.08 110 Ds (281) 29 Cu 63.546 47 Ag 107.87 79 Au 196.97 111 Rg (272) 30 Zn 65.41 48 Cd 112.41 80 Hg 200.59 112 Uub (285) 5 B 10.811 13 Al 26.982 31 Ga 69.723 49 In 114.82 81 Tl 204.38 113 Uut (284) 58 Ce 140.12 59 Pr 140.91 60 Nd 144.24 61 Pm (145) 62 Sm 150.36 63 Eu 151.96 64 Gd 157.25 65 Tb 158.93 66 Dy 162.50 67 Ho 164.93 68 Er 167.26 69 Tm 168.93 70 Yb 173.04 71 Lu 174.97 90 Th 232.04 91 Pa 231.04 92 U 238.03 93 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (262) Points will be deducted for incorrect use of SF/DP, missing and/or wrong units. USE THE ABOVE PERIODIC TABLE FOR ATOMIC/MOLAR MASSES. Notes: If you wish to truncate significant figures/decimal points before the last step of a series of calculations, use at least one more than the required number. Obey the rules for significant figures when quoting the final answer. Use units in all steps of a series of calculations. This will help you get the correct units for the answer. 1. (a) (3 points) Why do molecules in a gas have different speeds at a given temperature? Molecules are in constant motion. Thus, they collide with one another continually. Translational kinetic energy is transferred at each collision. Therefore, after a time interval, each molecule will have a different translational kinetic energy, resulting in a distribution of molecular speeds. (b) (10 points) Use the Maxwell-Boltzmann equation for distribution of speeds to estimate the fraction of N2 molecules at 400 K that have speeds between 200 m s 1 and 210 m s 1 . 3/2 m mv 2 /2 kT 2 The fraction is given by f (v) 4 v dv . Since the speed range is small, e 2 kT we may safely take v 205 m s 1 and dv v 10 m s 1 . Therefore m f (v) 4 21 1 2 1.38065 10 J K 298.15 K 3/2 e m (205 m s 1 2 ) /21.380651021 J K 1 298.15 K (205 m s 1 ) 2 (10 m s 1 ) m is the mass of one single molecule of nitrogen in kg. Therefore m 28.014 103 kg mol1 4.6518 1026 kg molecule1 . 6.02214 1023 molecules mol1 Inserting this value into the equation and evaluating it gives f (v) 6.87 103 2. (a) (3 points) Why does the ionic strength of a solution increase with increasing charge numbers of the constituent ions? As the charge number of an ion increases, interaction with other ions increases geometrically. Higher interactions will introduce higher deviations from non-ideality. Therefore, the sum of these interactions will result in a higher ionic strength. (b) (10 points) The solubility product of silver chloride in water at 25°C is 1.5 1010 . Assuming that the Debye-Hückel limiting law applies, calculate the mean activity coefficient of the electrolytes in a solution that is 0.05 M in KNO3 and 0.02 M in KCl. The mean activity coefficient is given by log Az | z | I 1/2 . This equation involves ionic strength, I. So we have to calculate it first. I is given by (with b o 1 mol dm 3 ) I 1 1 zi2 (bi / b o ) (0.05 12 ) (0.05 12 ) (0.02 12 ) (0.02 12 ) 0.07 2 i 2 The mean activity coefficient of the electrolytes can now be calculated. log A | z z | I 1/2 [0.5086 |1 (1) (0.07)1/2 ] 0.13456 0.734 2 3. (5 points) Fill in the blanks (no partial credits) Diffusion is the migration of matter down a concentration gradient. Electrical conductivity is migration of electric charge down an electrical potential gradient. Viscosity is the migration of linear momentum down a velocity gradient. Thermal conductivity is the migration of energy down a temperature gradient. 4. (15 points) Invertase (an enzyme) catalyses the hydrolysis of sucrose (table sugar). In a series of experiments, the following data was obtained. Use the data to determine by a graphical method the Michaelis constant of the reaction. Show your work on your graph paper and insert the result below. (Hint: Make a suitable table and do a double-reciprocal plot) [Sucrose]o (M) 0.029 0.059 0.088 0.117 0.175 Rate (M s-1) 0.182 0.266 0.310 0.330 0.362 K M 0.0392 M 5. (a) (2 points) What is effusion? Effusion is the escape of a gas in a container through a small hole into vacuum. (b) (3 points) When does an ion reach its drift speed at a given temperature in an electrical field? When the drag (viscous force due to diffusion) equals the electrical force acting upon the ion. (c) (2 points) What is chemisorption? It is the attachment of an adsorbate on a substrate by forming chemical bonds with units on the surface of the substrate. (d) (4 points) Consider the equation j ideal RT ln , which gives the chemical potential of an j ion in a solution as the sum of two terms. The last term accounts for the nonideal behaviour of ion j. Why do we need to make this correction? Explain in a few sentences. All molecules in a solution interact. Ions interact appreciable and supersede nearly all other interactions. These, inevitably, will produce nonidealities, since the interactions will no longer be directly proportional to the number of ions. Hence a correction is needed to account for the sum of nonidealities. 3 6. (8 points) Consider the cell Co(s) | Co2 (aq,1.0 m) || Ni 2 (aq, 0.1m) | Ni(s) where the cell reaction is Co(s) Ni 2 (aq) Ni(s) Co2 (aq) . What is the equilibrium constant of the reaction at 25°C? Is the reaction spontaneous? Since the cell reaction is Co(s) Ni 2 (aq) Ni(s) Co2 (aq) , the equilibrium constant is [Co2 ]eq RT o given by K . At equilibrium, Ecell 0 , and therefore Ecell ln K . 2 F [Ni ]eq o From the stoichiometry of the reaction, 2 . We have to determine Ecell from tables after splitting the overall reaction into two half reactions. They are reduction Ni 2 (aq) 2e Ni(s) E o 0.257 V oxidation Co(s) Co2 (aq) 2e E o 0.280 V o [(0.257 V) (0.280 V)] 0.023 V , Therefore, Thus, Ecell 8.3144 J mol1 K 1 298.15 K 0.023 V ln K 2 96485 C mol1 Thus, ln K 1.790 . Therefore, K 5.99 o 0 V , the reaction is spontaneous. Since Ecell 7. Decomposition of gaseous N2O5 is a unimolecular first order reaction N2O5 (g) 2NO2 (g) + 1/2 O2 (g) with a rate constant of k 4.8 104 s 1 at 25C. Because all the components are gases, the reaction kinetics is conveniently investigated by the measurements of the pressure changes in the reaction chamber. Assuming that the reaction takes place in a closed chamber (constant volume) at 25°C and the initial pressure of N2O5 is 100 kPa, (a) (6 points) what percentage of N2O5 decomposes after 60 minutes? Since the reaction is first order, the relevant integrated rate law is [A] ln kt [A]o We can use partial pressures to replace molar concentrations, Thus ln p kt , where p partial pressure of N 2O5 at time t. po 4 Thus, when t 60 min p 60 s (4.8 104 s 1 3600 s) 1.728 . Therefore, 3600 s , ln 1 min po p ln 1.728. Thus, p 17.76 kPa. 100 Let X percentage of N2O5 that decomposes after 60 minutes. It can be determined from po p (100 17.76) kPa 100 100 82.24% 82% (2 SF) po 100 kPa (b) (4 points) what is the total pressure in the chamber after 60 minutes? X pT pN 2O5 pNO2 pO2 pN2O5 17.76 kPa (above) pNO2 2 82.24kPa =164.48 kPa pO2 1 82.24 kPa 41.12kPa 2 Therefore, pT 223.36 kPa 223 kPa 8. (10 points) Carbon monoxide is produced during photolysis of acetone when it is irradiated with radiation at 313 nm with a quantum yield of unity. After 20 min of radiation at this wavelength, 18.4 cm3 of carbon monoxide is produced at a pressure of 1008 Pa and 22C. Calculate the number of photons absorbed by acetone. Since, the primary quantum yield is unity, the number of absorbed photons is equal to the number of carbon monoxide molecules formed by photolysis. Let N number of carbon monoxide molecules. If the gas is assumed to behave ideally under the given conditions, then N PVN A RT 106 m3 6.02214 1023 molecules mol1 1 cm3 8.3144 J mol1 K 1 (273.15 22) K 1008 Pa 18.4 cm3 4.71 1018 molecules. Therefore, the number of photons is 4.55 1018 photons, since 1 . 9. (15 points) A certain sample of activated charcoal has a surface area of 1000 m2 g 1 . What amount of ammonia gas, measured at 25C and 1 bar will be adsorbed at full coverage on the surface of 45 g of charcoal? Take the diameter of a molecule of ammonia as 3.0 1010 m, and assume that four adjacent ammonia molecules are at the four corners of a square (as illustrated below). 5 First find the total surface, S, offered by 45 g of charcoal. It is given by S 45g 1000 m2 g 1 45000 m2 . It is evident from the diagram that the rectangle contains one ammonia molecule. Therefore, area covered by one ammonia molecule, A, is given by A (3.0 1010 m)2 9.0 1020 m2 Total number of ammonia molecules, N, is given by S 45000 m 2 5.00 1023 molecules . So the mass of ammonia, mNH 3 , can now be A 9.0 1020 m2 calculated from N mNH3 5.00 1023 molecules 17.0306 g mol1 14.1 g 6.022 1023 molecules mol1 6