Chm 1316 Honors Freshman Chemistry II
Spring 1999
Exam 3     Due 9 am, 5 April 1999

Work any 5 problems!     Open book & periodic table; Closed all else.

 

  1. Rika didn't care for bomb calorimetry since it didn't make enough noise. So imagine that nitroglycerin, C3H5N3O9(liq), is detonated instead. It doesn't need any extra oxygen! In fact, if you balance its detonation reaction, you'll find that while almost all of the nitrogen ends up as N2(g), an extra ½ mole of O atoms steal ½ mole of N atoms to make ½NO one of the reaction products! With that in mind, the reaction will balance:

    C3H5N3O9(liq) ---> 3CO2(g) + (5/2)H2O(liq) + (5/4)N2(g) + ½NO(g)

    Data from a bomb shows DH°detonation as - 1.4796 MJ/mol. What's nitroglycerin's DHf°? (Nitroglycerin earned Nobel the money now given away as his prize.)

    3CO2(g) + (5/2)H2O(liq) + (5/4)N2(g) + ½NO(g) ---> C3H5N3O9(liq)
    DH1 = + 1,479.6 kJ/mol
    3 C(gr) + 3 O2(g) ---> 3 CO2(g)     DH2 = 3DHf°[CO2(g)]
    (5/2)H2(g) + (5/4)O2(g) ---> (5/2)H2O(liq)     DH3 = (5/2)DHf°[H2O(l)]
    ¼N2(g) + ¼O2(g) ---> ½NO(g)     DH4 = ½DHf°[NO(g)]

    If we now add all of those reactions (and their enthalpies) together, the reactants of reaction #1 will be cancelled out by the products of reactions 2-4, leaving the formation reaction for nitroglycerin:

    3 C(gr) + (5/2) H2(g) + (3/2) N2(g) + (9/2) O2(g)---> C3H5N3O9(liq) DHf°

    DHf° = 1,479.6 + 3(-393.5) + (5/2)(-286) + ½(90) = - 370.9 kJ/mol

  2. 1 mole of an ideal gas with CV = 12.47 J/mol°C starts at 1 atm and 298 K.
    1. With the volume held fixed, it is brought in contact with a hot plate held at 596 K (double the original temperature). When the gas has heated to this new T, how much heat has flowed from the hot plate?

      DE = CVDT = qV = 12.47 J/mol°C (298°C) = 3.716 kJ/mol

       

    2. Still in contact with the plate and held at 596 K, the gas is now allowed to expand against a fixed 1 atm external pressure. Since DT=0, then DE=0 for an ideal gas. How much new heat flows in which direction between the plate and the gas by the time the gas reaches equilibrium?

      DE = q + w = 0

      q = - w = + PextDV

      Since V doubled,
      DV = Vinitial = RTinitial/Pinitial
      = 0.08206 atm L/mol K (298 K) / 1 atm = 24.45 L/mol

      qnew = + 24.45 atm L (8.314 J/mol K)/(0.08206 atm L/mol K) = 2.478 kJ/mol flowing into the system.

  3. Predict and justify the sign of DS for the following reactions.

    1. CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(liq)

      Dngas = - 2 so DS < 0

    2. 2 CrO3(s) + H2O(liq) ---> 2 H+(aq) + Cr2O72-(aq)

      A solid has dissolved. DS > 0

    3. H2C=CH2(g) + H2O(liq) ---> CH3CH2OH(liq)

      Dngas = - 1 so DS < 0

    4. 2 Au(CN)2-(aq) + Zn(s) ---> 2 Au(s) + Zn(CN)42-(aq)

      Two moles of ions become solid but only one mole of solid went back in solution. DS < 0

    5. C12H22O11(s) + 11 H2SO4(conc) ---> 12 C(s) + 11 H2SO4·[11 H2O(liq)]
      where the water is bound up tightly with the sulfuric acid.

      A large, complex molecule decomposed. DS > 0

  4. You are working in the modern standard of pressure as 1 bar; so the thermochemical quantities you seek are DGfq, but for one gaseous compound, all you can find is DGf° which presumes the standard is 1 atmosphere instead of 1 bar. Suddenly you realize that you can correct the old value without doing any experiment! Assuming the gas to be ideal (good bet), what's the difference (in J/mol) between DGfq and DGf° at 25°C?

    DG(P2) = DG(P1) + RT ln(P2/P1)
    DG(1 bar) = DG(1 atm) + RT ln(100 kPa/101.325 kPa)

    So the difference is
    8.314 J/mol K (298 K) ln(0.986923) = - 32.6 J/mol

    But does that solve the problem for gaseous compounds? No, some of the elements from which they are composed could also be gaseous. So we have to use Hess's Law:

    Elements(1 bar) ---> Compound(1 bar)       DGfq

    can be replaced with the following alternate path giving the same value for state function G, 

                         DGf°
    Elements(1 atm) -----> Compound(1 atm)
         D                       |
    DG1  |                       |  DG2
         |                       V
    Elements(1 bar)        Compound(1 bar)

    So

    DGfq = DG1 + DGf° + DG2
           = - ngas(elements) RT ln(1 bar/1 atm) + DGf°
              + (1 mol gaseous compound) RT ln(1 bar/1 atm)
    (The minus sign is due to using bar/atm instead of atm/bar for path 1.)

    Or, for the formation reaction as normally written,

    DGfq = DGf° - 32.6 J/mol Dngas

    Note that the correction would be zero if as many moles of compound gas were produced as moles of element gas were consumed, e.g., formation of CO2(g) from its elements. Also, since condensed phase G's vary insignificantly with such tiny pressure changes, we can use the above for formation of any compound, gaseous or not, as long as we employ Dngas in its usual sense.

    What would you do if it were a gaseous element instead of a compound?

    DGfq = 0.00 by definition for an element as did DGf°.

  5. You're in the lab and need an estimate of the room temperature (25°C) vapor pressure of benzene. You know its normal boiling point is 80°C; so that means the room temperature vapor pressure is less than 1 atm. But you don't know how much less. Remembering that DSvap is almost always close to 88 J/mol°C (called Trouton's Rule), it suddenly hits you how to calculate a good estimate. What value do you get (using only the values given here)?

    For the Clausius-Clapeyron equation, we'd have needed DHvap, but we have it if we know DSvap which we do!

    DHvap = TvapDSvap ~ (80+273 K)×88 J/mol K = + 31 kJ/mol

    p(25°C) = 1 atm × e-(DHvap/R)[(1/298 K)-(1/353 K)] = 0.142 atm = 108 torr

  6. One of the acid rain reactions,

    2 SO2(g) + O2(g) ---> 2 SO3(g)

    has an equilibrium constant that varies with temperature as:

    T 300 K500 K700 K
    K 4.0×1024 2.5×1010 3.0×104

    Estimate DH at 500 K. (You can use a spreadsheet if you like, but you don't have to.)

    OK, you probably whipped out your spreadsheet and did it up brown, but, just for jollies, let's do it without. First off, we need to recast the data into the form that is useful, given

    ln K = (-DH/R)(1/T) + DS/R

    Hence we want to plot (or evaluate) not K but ln(K) vs not T but 1/T.

    1/T 3.33×10-3 K-1 2.00×10-3 K-1 1.43×10-3 K-1
    ln(K) 56.65 23.94 10.31

    Now since 500 K is precisely mid-range, we'd probably get a good answer using the two endpoint data,

    e.g., slope = [ln(K700 K) - ln(K300 K)] / [(1/700 K)-(1/300 K)]

    but we'll get a much better answer by finding the slopes between points 1 and 2 (i.e., 400 K) and points 2 and 3 (i.e., 600 K) and averaging them. THAT will improve the estimate of the slope at 500 K because we're including more of the data correctly.

    slope400 = (23.94-56.65) / (2.00-3.33)×10-3 = + 2.46×10+4 K
    slope600 = (10.31-23.94) / (1.43-2.00)×10-3 = + 2.39×10+4 K
    slope500 = ½(2.46+2.39)×104 = 2.425×104 K

    DH(500 K) = - slope500×R = - 202 kJ/mol

    We're not surprised by the sign since the very large K implies an exothermic reaction. But notice how quickly K falls off with T? Dngas = -1 means that TDS < 0 becomes more important as T increases.

Comments to Chris Parr Return to CHM 1316 HomePage Last modified 7 April 1999.