Chemistry C2407xGeorge FlynnOxtoby 4.10Oxtoby 4.42Oxtoby 4.48Oxtoby 4.52Oxtoby 7.2Oxtoby 13.44CCoolluummbbiiaa UUnniivveerrssiittyy in the City of New York New York, N.Y.10027 Department of Chemistry 3109 Havemeyer Hall 212-854-4162 Chemistry C2407x George Flynn Problem Set 1 Solutions to Even Numbered and Other Problems Oxtoby, 4.6 a) Since 76 cm difference in Hg levels is 1 atm, P = [(9.50) / 76.0] × 1 atm = 0.125 atm. Using SI units and Pascals: P = ρhg ρ = 13.60 g/cm3 × (106 cm3/ 1 m3)(1 kg/ 1000 g) ρ = 13.6 × 103 kg/m3 h = 9.50 cm = 0.0950 m g = 980.7 cm/s2 = 9.807 m/s2 P = (13.6 × 103 kg/m3)(0.095 m)(9.807 m/s2) = 1.267 × 104 Pascals 1 atm = 1.01325 × 105 Pascals P = (1.27 × 104 Pa) × (1 atm/ 1.01325 × 105 Pa) P = 0.125 atm. b) Pressure is same for both gauges so P = ρ1h1g = ρ2h2g or ρ1h1 = ρ2h2 h1 / h2 = ρ2 / ρ1 ρ1 = 13.6 × 103 kg/m3 ρ2 = 1.045 × 103 kg/m3 h1 = 9.5 cm = .095 m h2 = h1(ρ1)/ρ2 = (.095 m)(13.6 / 1.045) h2 = 1.24 meters Oxtoby 4.10 1 atm = 760 torr = 1.01325 × 105 Pa (5× 10-10 torr) × (1 atm/ 760 torr) = 6.58 × 10-13 atm (6.58 × 10-13 atm) × (1.01325 × 105 Pa/ 1 atm) = 6.67 × 10-8 Pa Oxtoby 4.42Crms = (3 kT/m)1/2 = (3 RT/M)1/2 M = atomic weight of Na atoms. M = 23 gm/mole = 0.023 kg/mole. SI units here are kg and Joules for R. R = 8.314 Joules/mole-deg T = 0.00024 deg Kelvin Crms = ((3)(8.314 J/mole-deg)(0.00024 deg) / (0.023 kg/mole))1/2 Crms = (0.260 J/kg)1/2 = 0.510 m/s (J = kg(m/s)2 → J/kg = (m/s)2 Oxtoby 4.48 In class we derived the (slightly oversimplified) formula for the number of collisions with an area A: I = #/sec = (1/6)(N/V)AC where we took C = Crms = 3RT/M A more rigorous treatment gives #/sec = (1/4)(N/V)AC where C is the average speed = 8RT/πM The 1/4 replacing the 1/6 comes from our oversimplified assumption that all atoms fly on a trajectory ⊥ to the wall. We will calculate the value of A using both formulas. In either case we need to know the number of atoms/sec escaping from the vessel A useful formula is PV = nRT = (N/N0)RT N = N0(PV/RT) At start P = 0.99 atm, at end (1 hr) P = 0.989 atm. V= 200mL, T = 298 K, R 82 mL-atm/mole-deg N1 = N0(P1V/RT) N2 = N0(P2V/RT) ∆N = (N1 – N2) = [(P1 – P2)V/RT] N0 ∆N = [(0.001 atm)(200 mL)/(82 mL-atm/mole-deg) × (298 deg)}N0 ∆N = (8.185 × 10-6 mole)N0, N0 = 6.023 × 1023 / mole ∆N = (8.185 × 10-6)(6.023 × 1023) = 4.93 × 1018 1 hour = 60 × 60 sec = 3600 sec I = ∆N/t = 4.93 × 1018 / 3600 = 1.37 × 1015 sec-1 More rigorous formula A = 4 (∆N /t ) (V / N) / c V/N = V/ {N0 (PV / RT)} = RT / N0P V/N = (82 cm3-atm / mole-deg)(298 deg) / (6.023 × 1023 / mole) × (0.99 atm) V/N = 4.1 × 10-20 cm3 / molecule c = 8RT/πM = (8 (8.314)(298)/π(0.002))1/2 m/s c = 1776 m/s = 1.78 × 105 cm/s A = 4(∆N/t)(V/N)/ c = 4(1.37 × 1015 sec-1)(4.1 × 10-20 cm3)/(1.78 × 105 cm/s) A = 1.26 × 10-9 cm2 = πr2 r = 2.0 × 10-5 cm = 2.0 × 10-7 m Crms = 3RT/M = [(3)(8.314)(298) / (0.002)]1/2 = 1.927 × 105 cm/sA = (6)(∆N/t)(V/N)/Crms = 1.74 × 10-9 cm2 r = 2.35 × 10-5 cm = 2.35 × 10-7 m (this is the less rigorous value) Oxtoby 4.52 (Rate Ef He) / (Rate Ef H2) = 3 Experimental observatioin (RHe / R) = NH2He CHe / N H2CH2CHe / C = H23RT/MHe/3RT/MH2 = (M / MH2He)1/2 RHe / R = (NH2He / N) (M / MH2H2He)1/2 = (NHe / N) (2/4)H21/2 (NHe / N) = H24/ 2 (RHe / R) = H22 (3) (NHe / N) = 4.24 H2XH2 = n / (n + nH2H2He) = 1 / (1 + nHe / n) H2nHe / n = (NH2He / N0 ) / (N / NH20 ) = NHe / N H2[n = # moles, N = # molecules, N0 = 6.023 × 1023] nHe / n = NH2He / N = 4.24 H2XH2 = 1/(1 + 4.24) = 0.191 Oxtoby 7.2 P = 0.98 atm Vi = 150 mL = 0.15 liter Vf = 0.80 liter w = -P∆V = -(0.98 atm) (0.8 – 0.15 liter) w = -0.637 l-atm R = 8.314 Joules / mole-deg = 0.082 l-atm / mole-deg 1 = (8.314 / 0.082) Joules / l-atm w = (-0.637 l-atm) (8.314 / 0.082 Joules/l-atm) w = -64.6 Joules An alternate approach is to express P, V in SI units: P = (0.98 atm)(101325 Pascals/atm) P = 99299 Pa ∆V = 0.65 liter = 0.65 × 103 cm3 102 cm = 1 m → 106 cm3 = 1 m3 ∆V = 0.65 × 103 cm3 / (106 cm3 / m3) = 0.65 × 10-3 m3 w = -P∆V = (-99299 Pa)(0.65 × 10-3 m3) w = -64.5 JoulesAssigned Problem Calculate the collision frequency for: a) a sample of oxygen at 1.00 atm. Pressure and 250 C b) a molecule of hydrogen in a region of interstellar space where the number density is 1.0x1010 molecules per cubic meter and the temperature is 30 K. [Take the diameter of oxygen to be 2.92x10-10 meter and that of hydrogen to be 2.34x10-10 meter.] From class for collisions of one molecule of A with all other A’s, the collision rate is z: z = 2 (NA/V)π ρ 2 cavge cavge = (8kT / π mA)1/2=(8RT / π MA)1/2 R = gas constant and MA = Molecular weight ρ= σA + σA = 2σA = d (diameter of A) PV = nRT = (NA/N0)RT (NA/V) = PN0/RT Part a: (NA/V) = (1.00)(6.023×1023)/(0.082)(298) = 2.46×1022 molecules per liter (NA/V) = 2.46×1022 molecules per liter × 103 liter/ m3 = 2.46×1025 molecules/m3 cavge =(8RT / π MA)1/2=[8 (8.314)(298)/(π)(.032)]1/2 = 444.0 m/sec. z = 2 [2.46×1025 molecules/m3 ][(π)(2.92 × 10 –10 m)2][444.0 m/sec] z = (1.414) [2.46×1025 molecules/m3 ][(3.1416)(8.53 × 10-20 m2)] [444.0m/s] z = 4.14×109 collisions/sec Part b: (NA/V) = 1.00×1010 molecules/ m3 cavge =(8RT / π MA)1/2=[8 (8.314)(30)/(π)(.002)]1/2 = 563.5 m/sec. z = 2 [1.00×1010 molecules/ m3][(π)(2.34 × 10 –10 m)2][563.5 m/sec] z = (1.414) [1.00×1010 molecules/ m3][(3.1416)(5.48 × 10-20 m2)] [563.5 m/s] z = 1.37×10-6 collisions/sec (Or, 1 collision every 728,996 sec = 1 collision every 8.44 days)Oxtoby 13.44 From class for collisions of molecule A with itself, the total collision rate is ZAA : ZAA = (1/2) π σAA2 <urel> (NA / V)2 urel = (8kT / π µ)1/2 µ = mA mA / (mA + mA) = mA / 2 urel = 2 (8kT / π mA)1/2 σAA = σA + σA = 2σA = d (diameter of A) Using the Arrhenius model, the reaction rate R is R = ZAAe−EA/RT …