This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 37
10.76) u = 3RT = 3RT 1/ 2 At constant T, u = (1/ )1/2 a) The larger the molar mass, the slower the avg. speed (at constant T). SF6 < HBr < Cl2 < H2S < CO (g/mol) 146.06 80.91 Dec 70.91 34.08 , Inc speed (u) 28.01 b) Calculate the rms speeds of CO and Cl2 at 300 K. Use eqn above and make sure R, T and are in SI units (8.314 J/molCK or kgCm2/s2CmolCK, Kelvin, kg/mol, respectively) to give SI units for u (m/s) 3 (8.314 kgCm2/s2CmolCK)(300 K) 1/2 uCO = () = 516.85 m/s = 516 m/s 28.01 x 103 kg/mol 3 (8.314 kgCm2/s2CmolCK)(300 K) 1/2 uCl = () = 324.84 m/s = 325 m/s 2 70.91 x 103 kg/mol As expected, the lighter CO molecules move as the greater rms speed. 10.79) 38
10.79) (cont.) 10.80) 39
10.80) (cont.) 10.81) a) Gases behave nonideally at HIGH Pressure and LOW Temp (Real gases behave ideally at low P and high T) b) Real gases behave nonideally because they have: Finite Volumes: real gas particles occupy some of the volume of the container so the volume of free space for the molecules to move in is less than the container (causes Pmeasured > Pideal)  Positive deviations from ideality Intermolecular Attractive Forces: Real gases have AF and attract each other  causes the measured pressure to be less than ideal (Pm < Pi)  Negative deviations from ideality c) For an ideal gas, PV/RT = n, the number of moles of gas particles, which should be a constant for all pressure, volume and temperature conditions (for 1 mole of gas PV/RT = 1 or PV/nRT = 1 is equivalent). If this ratio changes with increasing pressure the gas is not behaving ideally (i.e. if PV/nRT = 1 the gas is behaving ideally, if not then it is exhibiting nonideal behavior). If PV/nRT > 1 the gas is exhibiting a positive deviation from ideality (due to the volume of the molecules). If PV/nRT < 1 the gas is exhibiting a negative deviation from ideality (due to the IAF between molecules). 40
10.84) The van der Waals equation is given below: n2a ( P +  ) ( V  nb ) = nRT V2 P= nRT n2a   V  nb V2
causes (+) deviation Pm > Pi causes (!) deviation Pm < Pi Pm = measured (real) P; Pi = ideal P or due to vol due to IAF of molecules a: corrects for attractive forces between particles  real gas particles have AF  ideal gas particles have NO AF (a=0) b: corrects for finite molecular volume of particles  real gases occupy some vol of the container (we are interested in free vol)  ideal gas part. Have NO vol (b=0) Both tend to inc. with inc. in Mol. Wt. And structural complexity. The "a" constant depends on AF between particles and these will be discussed further in Chapter 11. Vcont  nb < Vcont so nRT/(Vnb) > nRT/Videal n2a/V2 accounts for fact that AF cause particles to hit walls w. less force than ideal gases, ^ smaller P then expected for an ideal gas (negative deviation from ideality) 41
10.85) Under these conditions this gas exhibits a negative deviation from ideality (Pmeasured will be closer to PVDW than Pideal & PVDW < Pideal) The intermolecular attractive forces reduce the effective number of particles and the real pressure. This is reasonable for 1 mole of gas at relatively low temp. and pressure. The molecules hit the wall with less force than in ideal case. 10.87) 10.92) 42
10.95) It is simplest to calculate the partial pressure of each gas as it expands into the total volume, then sum the partial pressures. Use Boyle's Law to calculate the new pressures of each gas in the mixture after the expansion. The final vol is: V2 = 1.0 L + 1.0 L + 0.5 L = 2.5 L 10.96) Use mole fraction to calculate moles of O2 in air (nO = PO nair).
2 2 43
10.97) Ammonia, NH3(g), and hydrogen chloride, HCl(g), react to form solid ammonium chloride, NH4Cl(s), according to the following equation, NH3 (g) + HCl (g) v NH4Cl (s) Two 2.00L flasks at 25C are connected by a stopcock, as below: The stopcock is opened and the gases react until one is completely consumed. a) Which gas remains in the system (i.e. which is the limiting reactant)? This is a stoichiometry problem so we need to convert to moles. 1 mol NH3 nNH = 5.00 g NH3  = 0.29359 mol NH3 3 17.03 g NH3 1 mol HCl nHCl = 5.00 g HCl  = 013713 mol HCl 36.46 g HCl Since the reactants react in a 1:1 mole ratio, the HCl is the limiting reactant and all 013713 moles of it reacts with 013713 moles of NH3 leaving 0.15646 moles of NH3 as the only gas present. b) What is the final pressure of the system after the reaction is complete (assuming the volume of the NH4Cl(s) can be ignored)? Total volume is 4.00 L. n = 0.15646 mol; V = 4.00 L; T = 25C + 273.15 = 298.15 K nRT (0.15646 mol) (0.08206 L Catm/mol CK) (298.15 K) P =  = V 4.00 L P = 0.95701 atm = 0.957 atm 44
10.99) Calculate the average molar mass of the mixture of O2 and Kr. Then relate the average molar mass to the molar masses of O2 and Kr using mole fractions . Remember mole fractions add to 1. Also, the mass of 1 mole of any substance is its molar mass. 45
10.103) 46
10.103) (cont.) 47
10.104) (Enough moles of gas so they occupy a fairly significant volume compared to volume of the container.) 48
10.105) Cyclopropane is composed of 85.7% C and 14.3% H, by mass. a) What is the molecular formula of cyclopropane if 1.56 g occupies a volume of 1.00 L at 0.984 atm and 50C? Remember the molecular formula (MF) is an integer multiple of its empirical formula (EF) and the molecular weight (MW) is the same multiple of the empirical formula weight (EFW). Also, the molecular weight (amu) is the same value as the molar mass (g/mol). So if you know the EF you know the EFW and if you can then determine the MW you can find the MF. MF = (EF)n MW n = EFW MW = n (EFW) MW / m mass =  = n moles Have mass ==> need moles Given mass % comp ==> empirical form Given P, V, T ==> moles Give mass & moles ==> (molar mass) & MW 1) Get Empirical Formula & Empirical Formula Weight (EFW) a) Convert mass % to moles (assume 100 g cyclopropane) 1 mol C ? mol C = 85.7 g C  = 7.1357mol C 12.01 g C 1 mol H ? mol H = 14.3 g H  = 14.186 mol H 1.008 g H 49
10.105) (cont.) b) Get mole ratio  divide by smallest number of moles 7.1357mol C C :  = 1.00 = 1 (mol C/mol C) 7.1357mol C 14.186 mol H H :  = 1.988 = 2 (mol H/mol C) 7.1357mol H The empirical formula of cyclopropane is CH2. EFW for CH2 = 12.01 + 2 (1.008) = 14.026 amu 2) Get Molecular Weight (MW) and Molecular Formula Have mass (1.56 g). Need moles. Use the IGL to get moles. PV (0.984 atm) (1.00 L) n =  =  = 3.710 x 102 mol RT (0.08206 L Catm/mol CK) (323.15 K) m 1.56 g =  =  = 42.04 g/mol = 42.0 g/mol (amu) n 3.710 x 102 mol MW 42.0 amu n =  =  = 3.00 = 3 EFW 14.0 amu MF = (EF)n = (CH2)3 = C3H6 50
10.105) (cont.) b) Which gas, Ar or C3H6, deviates more from idealgas behavior. AWAr = 39.95 amu MWC2H6 = 42.0 amu The following are the van der Waals constants for Ar and C3H6: Ar: CH6: a = 1.34 L2atm2/mol2 a = 2.25 L2atm2/mol2 b = 0.0322 L/mol b = 0.0428 L/mol Would expect C3H6, a larger and more complex molecule than CH4, to have larger a and b values than CH4. ...
View
Full
Document
 Winter '08
 Zellmer

Click to edit the document details