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Chemistry 2 week 3 notes

by: Ariel Kamen

Chemistry 2 week 3 notes chem 132

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Ariel Kamen

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general chemistry lecture 2
Sarah T. Stokes
Class Notes
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This 5 page Class Notes was uploaded by Ariel Kamen on Friday February 19, 2016. The Class Notes belongs to chem 132 at Towson University taught by Sarah T. Stokes in Winter 2016. Since its upload, it has received 26 views. For similar materials see general chemistry lecture 2 in Chemistry at Towson University.


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Date Created: 02/19/16
 2/16/16                                         Volatile Solute and  Raoult   s La  P A = XAP(degrees) B  P B= X A P(degrees) B P T= P A + PB B.P. elevation: (delta)Th = i(Kb)m      (delta)b= Tb —  T(degrees) b * (delta)Th — change in b.p. * K b— b.p. elevation constant * m  — molality * Tb — new b.p. * b— pure b.p. T(degrees) F.P. depression: (delta)T f = i(f)m        (deltaf= T(degrees) f— T f i van’t Hoff factor —> number of particles Non­electrolytes: CH3CH2OH (l) —> CH3CH2OH (aq) i = 1 C6H12O6 (s) —> C6H12O6 (aq) i = 1 i = 2       NaCl (s) —> Na+ (aq) + Cl— (aq) i = 4       Na3PO4 (s) —> 3Na+ (aq) + PO4 ^ ­3 (aq) i = 3       BaCl2 (s) —> Ba ^ 2+ (aq) + 2Cl— (aq) Electrolytes (strong acids): HCl      hydrochloric acid HBr      hydrobromic acid HI H2SO4 HNO3 HClO4 Non­electrolytes (non­metals): weak acids sugars ­ ­covalent compounds ­alcohol Practice 13.7: Find b.p. and f.p. of a solution with 478 g of ethylene glycol (C2H6O2) and 3202 g of H2O (delta)T b= i(Kb)m                  b = 0.52 (degrees) C / m H2O (for solvent)                                                     m = mol solute / hg solvent                    i = 1 478 g EG(1 mol EG) / 62.07 g EG = 7.70 mol EG 3202 g H2O(1 hg H2O) / 1000 g =  3.202 hg H2O   (divide both and get mol/ hg) = 2.40 m = (1)(0.52 C/ m)(2.40 m)   (delta)Tb = 1.2 (degrees) C 100 (degrees) C + 1.2 (degrees) C = T b = 101.2 (degrees) C b.p. of solution K f= 1.86 (degrees) C    (delta)Tf= (1)(1.86 C / m)(2.40 m) = 4.5 (degrees) C T  = 0 ­ (4.5 C) = —4.5 (degrees) C f.p. of solution Osmotic Pressure ­ semipermeable membrane  —> solvent molecules pass ; solute molecules DO NOT PASS osmosis: solvent moves from high solvent concentration (low solute concentration) to low  solvent concentration (high solute concentration) *They run opposite of each other   P 13.9: 202 mL of benzene contains 2.47 g of organic polymer (i = 1) osmitc pressure of 8.63  mmHg at 21 (degrees) C *Find molar mass of polymer 8.63 mm Hg(1 atm) / 760 mm Hg = 0.0114 atm        21(degrees) C + 273 = 294K 0.0114 atm = (1)M(0.0824 L * atm/molK)(294 K) 0.0114 atm / 0.0824 L * atm/molK(294K) = M = 4.72 x 10^ ­4 mol solute polymer / L soln x 0.202 L soln = 9.54 x 10^ ­5 mol polymer MM = g solute polymer / mol soly polymer —> 2.47 g / 9.54 x 10^ ­5 = 2/18/16           Practice 13.8 0.85g organic compound in 100.0g Kf = 5.12 (degrees) C/m benzene. F.p. is 5.16 degrees C.  Find molality and molar mass of solute.       Tf = 5.5 degrees C (5.5 — 5.16) = (1)(5.12)m     0.34 / 5.12 = 0.066 m 0.066 mol O.C. / 1 Kg benzene x 0.1000 Kg = 0.0066 mol O.C. 0.85g O.C. / 0.0066 mol O.C. = 128 g/mol Table 13.3 Theoretical i Experimental i Sucrose 1 1.0 HCl 2 1.9 NaCl 2 1.9 MgSO4 2 1.3 MgCl2 3 2.7 FeCl3 4 3.4 Practice 13.10 F.p. depression for a 0.100 m MgSO4 soln is 0.225 degrees C, find van’t Hoff factor      Kf = 1.86 degrees C/m 0.225 = i(1.86)(0.100)      1.21 = i END OF MATERIAL FOR EXAM 1 Kinetics how quickly will a reaction proceed (pressure that reaction does happen) ­ * thermodynamics (chpt 18)  — does the reaction happen (short or long??) * t = time      T = temperature • reactants  —> products •  —> B A  • rate= ­delta [A]/ delta t = rate= + delta [B]/ delta t • 2A  —> B—> rate= ­1/2 delta [A]/ delta t = + delta [B]/ delta t • aA + bB  —> cC + Dd •   rate = ­1/a delta [A]/ delta t =  ­1/ b delta [B]/ delta t = 1/ C  delta [C]/ delta t = 1/ d delta[D]/  delta t A —> B 4PH3 (g) —> P4 (g) + 6H2 (g)  rate equation            r = ­1/4 delta[PH3]/ delta t = delta [P4]/ delta t = 1/6 delta [H2]/ delta t rate expression                                ­H2 is formed at rate of 0.078 M/s           Find  rate at  which PH3  is  consumed ­1/ 4  delta[PH3]/  delta t = 1/6 (0.078 M/s) ­1/4x =  1/6(0.078) = x = ­0.052  M/s       delta [PH3]/ delta t = ­0.052 M/s rate of [ClO2) rate of [F2] rate of [F2][ClO2] rate = K [F2][ClO2] K = rate at constant    M/S = (1 / M x S) M^2      M/S = M/S 1.2 x 10^­3 M/S = K(0.10 M)(0.010 M)            1.2 1 / M x S = K 1.21 M^­1 S^­1 aA + bB —> cC + dD rate laws: rate = K[A]^x[B]^y rate = K[F2]^1 [ClO2]^1


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