INSERT: READ THIS FIRST!
Show your work and approach in all problems involving a calculation or derivation in
order to receive credit.
1) Useful constants and conversion factors
R = 8.31451 J/K-mole = 0.08205 L-atm/K-mole
kB = 1.3807 x 10-23 J/K-molecule
NA = 6.022 x 1023 molecules/mole
h = 6.626 x 10-34 J-s
atomic weights of isotopes (g/mole): H, 1.008; D, 2.014; 12C, 12.000; 13C, 13.003
atomic weights (g/mole): H, 1.0; C, 12.0
2) Decomposition of diimide at 298 K
Absorbance at 350 nm versus time
1.2
1
A
0.8
0.6
0.4
0.2
0
0
500
1000
time (sec)
1500
NAME:_________SOLUTIONS_________ 4th. midterm exam, Chem 1b, 2007
1) (40 points) Diimide, the trans isomer of HN=NH, decomposes to form H2 and N2:
N2H2 H2 + N2. Diimide absorbs UV light at 350 nm. A small amount of diimide is
created in a flask through the rapid decomposition of a precursor molecule. Its
concentration was monitored at 298 K by measuring its absorbance at 350 nm as a
function of time. The results are shown in the graph on the insert.
a) Determine the rate law for the decomposition and the rate constant at 298 K.
The plot of absorbance shows an exponential decay with a constant halflife of ca.
460 s so the kinetics are first order, i.e. d[HN=NH]/dt = -k[HN=NH].
k = ln(2)/ = 0.693/(460 s) = 0.0015 1/s.
b) The rate constant at 400 K is 0.0091 s-1. Using this result and your answer
from (a), determine the activation energy for the reaction. [If you are unable to
answer part (a), assume that k(298 K) = 0.0035 s-1.]
Determine the slope, m of a plot of ln(k) versus 1/T(K).
Ea = - Rm = -(8.314 J/K)ln[(0.0091 1/s)/(0.0015 1/s)]/[(1/400 K) - (1/298 K)]
= 7.6 kJ/mole
c) Two possible mechanisms, A and B, for the reaction are
A-1) trans HN=NH  cis HN=NH (slow step)
A-2) cis HN=NH  N2 + H2
B) trans HN=NH  N2 + H2
i) Are both mechanisms consistent with the empirical rate law? Explain.
Yes. The kinetics for mechanism A with two steps are determined by the
slow step which is the first, unimolecular step so first order kinetics are
expected. Step B has one step and is also unimolecular so first order
kinetics are expected as well.
ii) Are both mechanisms consistent with the entropy of activation for the
decomposition, -260 J/K-mole? Explain.
No. In meachnism B, the diimide breaks up directly to form two
molecules so a positive entropy of activation is expected. In mechanism
A, the trans isomer is converted first to the cis isomer. A linear, more
symmetric transition state with a slightly negative entropy of activation is
expected. However, the entropy of activation is large and negative.
Neither mechanism is accept.
2) (15 points) Calculate the ratio of diffusion rates for each of the following pairs of
isotopomers: (12CH4 , 13CH4) and (H2, HD). Why are substances enriched in carbon-13
more expensive than those enriched in deuterium?
From kinetic gas theory <KE> = 0.5M<v2> = 3RT/2. Therefore, for a pair of molecules
M and M’, isotopomers in this case, R = [<v2>/<v’2>]0.5 = [M’/M]0.5
For the pair (12CH4/13CH4) R =[16/17]0.5 = 0.97, a value very close to one so separation
will require several stages of purification with a resulting increase in cost.
For the pair (H2/HD), R = [2/3]0.5 = 0.82, a value not as close to one so separation will be
easier.
3) (17 points) A solution is prepared by mixing 25 g of n-hexane, C6H14, and 15 g of
n-octane, C8H18.
a) Is the solution ideal? Discuss.
Yes. Both compounds are non-polar hydrocarbons with similar molecular
weights. The octane-hexane interactions will be very close to octane-octane and
hexane-hexane interactions. We can use Raoult’s Law in part (b).
b) Calculate at 25C the partial pressure of hexane in equilibrium with the
solution. The partial pressure of pure n-hexane is 0.19 atm 25C.
We require the mole fraction of hexane in the solution, XH.
25 g of H corresponds to (25 g)/(86 g/mole) = 0.29 mole
15 g of O corresponds to (15 g)/(114 g/mole) = 0.13 mole
XH = (0.29 mole)/(0.29 mole + 0.13 mole) = 0.69
pH = XHpH(pure) = (0.69)(0.19 atm) = 0.13 atm.
4) (28 points) Chemists use phase transitions to calibrate thermometers. A laboratory in
Denver, Colorado at an elevation of 5000 feet orders a sample of benzene and uses the
uncorrected CRC Handbook values of its melting and boiling point.
a) Explain why the laboratory takes great care to purify the benzene.
Impurities will raise the boiling point and cannot be present if an accurate
measurement is needed.
b) Which measurement, that of the melting point or of the boiling point, will be
more suitable for calibrating their thermometers? Why?
Use the melting point. Since the molar volume of gases is much greater than that
of liquids and solids, the boiling point is very sensitive to atmospheric pressure.
The value measured in Denver will differ significantly from the CRC Handbook
or sea level value. In contrast, solids and liquids have comparable molar volumes
so the value is fairly insensitive to modest changes in pressure.
c) Solid benzene is more dense than liquid benzene. In a p (vertical axis) versus T
(horizontal axis) phase diagram, what is the sign of the slope of the boundary
between the liquid and solid phases? Provide the rationale for your answer.
We have an illustration of Le Chatelier’s principle. At the normal melting point,
the solid and liquid phases are in equilibrium. Now increase the pressure. The
equilibrium will shift in a manner to relieve the stress. That is, towards the denser
phase with the smaller molar volume. The sample will freeze. In order to reach
the melting point, one has to raise T. The slope, (p/T) = (+/+), is positive.
d) What is the maximum number of phases that can co-exist in a mixture of
benzene and ethanol? Provide a rationale for your answer.
The Gibbs Phase Rule states that f = c - p +2. For a maximum value of p, f =0 or
p(max) = c +2. For 2 components, c =2 so p(max) = 4.
e) At 25.00C, the standard enthalpy and entropy of vaporization of liquid
benzene are 33.8 kJ/mole and 95.8 J/K-mole, respectively. Calculate the vapor
pressure of liquid benzene at 30.0C.