Chm 1316 Honors Freshman Chemistry II
Spring 1999
FINAL EXAM     Due 8 am, 5 May 1999

Work any 10 problems!     Open book & periodic table; Closed all else.

 

  1. 16.0 g of a colorless liquid containing only C, H, and O are boiled and the resultant vapor at 200°C and 1 atm occupies 10.0 L. Combustion of this vapor in excess O2 results in 22.7 g of CO2 and 13.9 g of H2O. Write and balance this combustion reaction.

    To write and balance the reaction equation, we have to first know what the molecular formula of the CxHyOz is. So this is a deja vu empirical/molecular formula problem masquerading as an ideal gas one.

    n(CxHyOz) = PV/RT = 1 atm (10.0 L) / [ 0.08206 atm L/mol K (473 K) ] = 0.258 moles which makes the molecular weight 16.0 g/0.258 mol = 62.1 g/mol

    n(C) = 22.7 g CO2 × 1 mol CO2/44.01 g CO2 = 0.516 mol C
    n(H) = 13.9 g H2O × 1 mol H2O/18.02 g H2O × 2 mol H/1 mol H2O = 1.543 mol H

    wt(C+H) = 12.01 g/mol × 0.516 mol + 1.008 g/mol × 1.543 mol = 7.75 g
    wt(O) = wt(CxHyOz) - 7.75 g = 16.0 - 7.75 - 8.25 g O

    n(O) = 8.25 g / 16.00 g/mol = 0.515 mol O

    CxHyOz = C0.516H1.543O0.515 = CH3O for an empirical formula. But it would have an empirical formula weight of 12.01+3(1.008)+16.00 = 31.05 g/mol, half the known molecular weight. So the molecular formula must be C2H6O2, probably ethanediol, HOCH2CH2OH, but we can't be sure of that since there's more than one isomer of that combination of elements.

    C2H6O2(liq) + (5/2) O2(g) ---> 2 CO2(g) + 3 H2O(liq)

     

  2. In the 2 H2(g) + O2(g) ---> 2 H2O(liq) fuel cell, both electrical and pressure-volume work are involved. Give both at standard conditions and with the proper sign (i.e., "work on the system") and in the same units (J/mol of reaction).

    welec = DG° = 2 DGf°[H2O(liq)] = 2(-237 kJ/mol) = - 474 kJ/mol, the negative sign implying the battery does work.

    wPV = - PDV = - RT Dngas = - 8.314 J/mol K (298 K) (-3) = + 7.43 kJ/mol, the plus sign implying that the atmosphere does work on the system (by compressing the gases flat).


     

  3. That fuel cell is run in 1 M base solution at both anode and cathode for a standard electromotive force, E°, of 1.23 V (pp. 843 and A26). But the reverse reaction, water electrolysis, is run with 1M acid at the anode and 1 M base at the cathode for a standard (applied) voltage of - 2.06 V (p. 851). Except for the reversal, those reactions are the same! Where does the extra (1.23-2.07=) - 0.83 V go in the electrolysis and by what mechanism?

    Looking at the reaction equation on page 851,

    6 H2O ---> 2 H2 + O2 + 4 (H+ + OH - )

    we see that the electrolysis produces not only the expected gaseous products but also a (briefly) separated quartet of (H+ , OH - ) pairs. The latter's recombination to form water gives up four Heats of Neutralization (which you can calculate from Appendix 4 if you are of a mind to) which heats up the solution, wasting exactly the difference in the energies which give rise to that extra voltage.


     

    1. What voltage sacrifice is made in the fuel cell (#2) by using air which is, after all, only 0.21 atm O2(g)?

      DE = (0.0592 V / n) log(PO2) = (¼)(0.0592 V) log(0.21) = - 0.01 V = - 10 mV

       

    2. By what voltage would we improve E° if we ran the fuel cell (#2) at 35 atm each of H2(g) and O2(g)?

      DE = (0.0592 V / n) log(PO2 × P2H2) = (¼)(0.0592 V) log(353) = + 0.069 V = + 69 mV


     
  5. How many moles of ice (at 0°C) would it take to condense only one mole of steam (at 100°C) to equilibrium at 0°C? [ DHfus(0°C) = 6.02 kJ/mol, DHvap(100°C) = 40.7 kJ/mol, CP(0° to 100°) = 4.197 J/g°C (on average) ]

    DHvap / DHfus = 40.7 kJ/mol steam / 6.02 kJ/mol ice = 6.76 mol ice/mol steam.

    But that leaves the steam at 100°C. We have to cool it to 0°C.

    CP = 4.197 J/g°C × 18.02 g/mol H2O = 75.63 J/mol°C
    CPDT / DHfus = 75.63 J/mol°C (100°C) / 6020 J/mol ice = 1.256 mol ice/mol "steam"

    Total moles of ice = 6.76 + 1.26 = 8.02 moles of ice to condense 1 mol steam to 0°C.


     

  6. In a classic demonstration experiment (not done in lecture), a wooden toothpick is dropped into still water and observed to be stationary. Next a similar toothpick, one end of which has been soaped is dropped into water. It moves!

     

    1. Which way? (Do the experiment, if you like; it's neat.)
       
      The toothpick starts moving in the direction of its non-soapy end.

       
    2. Why?
       
      The soap destroys water's surface tension and so reduces its grip on the soapy end.

       
    3. Why does it finally come to rest?
       
      Eventually, the soap spreads, reducing the tension all over the water's surface uniformly. No direction is then favored.

     
  7. Catalysts work by influencing the "activated complex" of the reaction by their presence. They might even predissociate some reactants, lowering Eact. But heterogenous catalysts have an added advantage of forcing the reactants onto a surface, lowering the effective volume in which reactants are seeking one another.

    Imagine an overall second order elementary gas phase reaction A + B at 1 atm total pressure at 25°C. Imagine that a solid catalyst captures multilayers of both molecules at liquid-like concentrations, say 10 M each. By what factor is the reaction rate improved through this "concentration" effect, ignoring any other kind of surface catalytic activity?
     
    Rate = k2 [A] [B] = k [A]2 since [A] = [B]

    Ratecondensed / Rategaseous = [A]2condensed / [A]2gaseous

    [A]gas = nA / V = PA / RT = ½ atm / [0.08206 atm L/mol K (298 K)] = 0.0204 mol/L

    Ratecondensed / Rategaseous = (10 M)2 / (0.0204 M)2 = 2.4×105 (!)


     

  8. "An Energy Solution" on p. 508 describes a layered salt pond that generates a (perhaps amazing) 107°C temperature at its bottom just by solar heating. Read that article.

    It maintains that the dense salt solution at the bottom of the tank cannot boil at 107°C since its boiling point is "much higher than 100°C." With the tools at our command, we can test that prediction.

    1. What is the normal boiling point elevation of a 27% (by mass) salt solution? (By normal, we mean if the pressure is 1 atm.)
       
      In 1 kg of solution there are 270 g NaCl and 730 g H2O. We'd have to scale the water's weight up by (1000/730)=1.37 to make a kg of solvent, and the solution that had that much solvent would, perforce, contain 1.37×270 = 370 g NaCl.

      molality NaCl = 370 g NaCl ( 1 mol NaCl / 58.5 g NaCl ) = 6.3 m

      Assuming it dissociates 100% (not bloody likely), there'd be 12.6 m ions in solution, and the boiling point elevation would be:

      DT = Kb mions = 0.51°C kg/mol (12.6 mol/kg) = 6.4°C

      Actually it's less since the vant Hoff factor for NaCl isn't as much as 2 at such concentrations. Even at 1% of that concentration, i = 1.9 for NaCl (Table 11.6). So boiling point elevation alone cannot account for unboiled salt water at 107°C.


       
    2. Since that doesn't reach the observed 107°C, the rest of the boiling point elevation must be due to the increased pressure of the (salt) water on the pool bottom. Fill in the missing (pink) entries in the following table to find that pressure, then calculate the new boiling point at the total pressure on the bottom of the pool. (Estimate the density by assuming that 1000 kg of solvent still occupies 1 L of volume...but carries the extra mass of the solute as well.) [ r(Hg) = 13.6 g/cc and DHvap = 40.7 kJ/mol (deja vu) ]

      Layer Thickness (m) % by wt NaCl Est. Density (g/cc) DP (atm)
      Convective 0.5 m 2%    
      Non-convective 1.5 m 15%    
      Bottom 1.0 m 27%    
      Extra pressure due to weight of all the water (atm):  

       
      We've already seen in part (a) that 1000 kg of bottom water (I like the sound of that) carries an additional 370 g of salt. So if it still occupies only a liter (unlikely), the density of bottom water would be 1370 g/L or 1.37 g/mL, 37% heavier than pure water.

      The pressure that the bottom layer exerts on the pond floor is that which an equivalent height of mercury would exert scaled by the difference in their densities.

      DP3 = (1 atm/0.76 m Hg)(1.0 m)(1.37 g/mL H2O / 13.6 g/mL Hg) = 0.133 atm

      The intermediate layer is only 150 g salt to 850 g water; so 1000 g water contains 176 g of salt in a liter (ho ho ho) making the density 1.176 g/mL. It's pressure differential is:

      DP2 = (1/0.76)(1.5)(1.176/13.6) = 0.17 atm

      And the top layer similarly: 20 g salt in 980 g water means 1020 g/L or 1.020 g/mL hence

      DP1 = (1/0.76)(0.5)(1.020/13.6) = 0.05 atm

      and the table becomes:

      Layer Thickness (m) % by wt NaCl Est. Density (g/cc) DP (atm)
      Convective 0.5 m 2% 1.020 0.05
      Non-convective 1.5 m 15% 1.176 0.17
      Bottom 1.0 m 27% 1.37 0.13
      Extra pressure due to weight of all the water (atm): 0.35

      Now we relate pressure to boiling point temperature by the Clausius-Clapeyron equation:

      P = 1 atm e-(DHvap/R) [(1/T) - (1/Tnbp)] where "nbp" is the normal boiling point.

      We know we want P = 1.35 atm so we have to rearrange C-C to find the corresponding T:

      T = 1 / [(1/Tnbp) - (R/DHvap)ln(P in atm)]
      T = 1 / [(1/373 K) - ln(1.35)(8.314 J/mol K)/40,700 J/mol] = 382 K

      DT = 382 - 373 = 9°C


       
    3. Assuming both effects simply add, how hot could the pond water get at the bottom before boiling? (107°C doesn't seem so silly now.)
       
      Tbottom = 100°C + 6°C + 9°C = 115°C

     
  9. 14C is an unstable radioisotope of 12C that has an extra 2 neutrons. It decays in a first-order process (by converting one of the extra neutrons into a proton and an electron, emitting the latter as a b- particle, and becoming a stable 14N) with a half-life of 5700 years. Anthropologists use the ratio of 14C to 12C in wood to tell when the tree died (and stopped taking up the background 14C that surrounds us from cosmic ray sources). If they can measure the concentration of 14C accurately down to 1% of its initial value, how far back can they date wooden artifacts?

    k1 = ln(2) / t½ = 0.693 / 5700 yr = 1.22×10-4 yr-1

    [14C] = [14C]0 e-k1t
    0.01 = e-1.22×10-4 t (in yr)
    ln(0.01) = - 1.22×10-4 yr-1 t
    t = - 4.61 / ( - 1.22×10-4 yr-1 ) = 38,000 yr


     

  10. The sign of the entropy of any chemical reaction that yields a non-zero Dngas is easy to guess: DS takes the sign of Dngas because of the massive change in positional entropy.

    The situation when Dngas = 0 is less obvious. What sort of differences between reactants and products can be exploited to guess the sign of DS then?

    We can look for complexity changes. If few large molecules become several small ones, DS increases.

    If there are the same number of molecules in reactants and products but some reactant molecules are solids while products are all in solution, DS increases.

    If all are in the same phase and there are the same number of reactant and product molecules but products represent more mixed atoms (e.g., H2 and F2 reacting to make 2 HF), DS increases.


     

  11. You've been asked to increase the rate tenfold (above its 25°C value) for a reaction with an activation energy Eact = 20 kJ/mol. What temperature do you suggest?

    k1 = A e-Eact/RT1
    k2 = A e-Eact/RT2
    k2/k1 = e(-Eact/R)[(1/T2) - (1/T1)]

    ln(k2/k1) = (-Eact/R) [ (1/T2) - (1/T1) ]

    T2 = 1 / [ (1/T1) - (R/Eact)ln(k2/k1) ]
    T2 = 1 / [ (1/298 K) - ln(10) (8.314 J/mol K) / (20000 J/mol) ] = 417 K
        = 144°C


     

  12. While DGf° for PbSO4(s) and SO42 -(aq) are given in Appendix 4, DGf° for Pb2+(aq) is not. However, Appendix 5.4 shows Ksp for PbSO4 to be 1.3×10-8. Can we use that to obtain the missing lead(II) ion formation Gibbs Free Energy? What would it be?

    Not a problem:     PbSO4(s) <---> Pb2+(aq) + SO42 - (aq)

    DGsp° = - RT ln Ksp = - 8.314 J/mol K (298 K) ln (1.3×10-8) = 45.0 kJ/mol

    But DGsp° = DGf°[Pb2+(aq)] + DGf°[SO42 - (aq)] - DGf°[PbSO4(s)] or

    DGf°[Pb2+(aq)] = DGsp° - DGf°[SO42 - (aq)] + DGf°[PbSO4(s)]
          = 45.0 - (-745) + (-813) = - 23 kJ/mol

Comments to Chris Parr Return to CHM 1316 HomePage Last modified 5 May 1999.