CHM 3400 – Fundamentals of Physical Chemistry
First Hour Exam
February 8, 2017
There are five problems on the exam. Do all of the problems. Show your work
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R = 0.08206 Latm/moleK
NA = 6.022 x 1023
R = 0.08314 Lbar/moleK
1 Latm = 101.3 J
R = 8.314 J/moleK
1 atm = 1.013 bar = 1.013 x 105 N/m2
1 atm = 760 torr
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1. (20 points) A gas mixture contains two gases, neon (Ne, M = 20.17 g/mol) and argon (Ar, M = 39.95
g/mol). The density of the gas mixture, at T = 300.0 K, is  = 0.562 g/L, and the partial pressure of neon in
the gas mixture is p(Ne) = 328. torr. You can assume ideal gas behavior for the gases in the mixture.
a) Find the mass of neon contained in 1.000 L of the above gas mixture.
b) Find p(Ar), the partial pressure of argon in the above gas mixture. Give your final answer in
units of torr.
2. (20 points) In an experiment, 4296. J of heat is added to a sample of a substance (not an ideal gas). The
addition of heat is carried out reversibly and at a constant external pressure p ex = 24.10 atm. The
temperature of the substance increases from an initial value Ti = 22.8 C to a final value T f = 37.9 C.
During the process, the volume occupied by the substance increases by 9.8 mL. Based on the above
information, find the following
a) Cp, the average value for the constant pressure heat capacity for the substance over the
temperature range of the process carried out above.
b) q, w, U, and H for the process.
3. (24 points) Ammonia (NH3, M = 17.03 g/mol) is a trace gas in the Earth’s atmosphere, and is also found
in the atmospheres of other planets and moons.
a) What is the c, the rms average speed of an ammonia molecule, at T = 500.0 C?
b) One formula for modeling the temperature dependence of the constant pressure molar heat
capacity of a gas has been developed by researchers at NASA. They fit experimental data to a polynomial
expression in terms of temperature. For ammonia (NH3) the equation they use is as follows:
Cp,m(NH3(g)) = a + bT + cT2
(3.1)
For ammonia, the values for the fitting parameters are a = 24.619 J/molK, b = 3.75 x 10-2 J/molK2, and c
= - 0.138 x 10-5 J/molK3. Note that T in eq 3.1 is the absolute temperature (in Kelvin).
Find the values for q, w, U, and H when the temperature of 1.000 mole of ammonia gas
changes from an initial value Ti = 300.0 K to a final value T f = 800.0 K by a reversible heating of the gas at
a constant pressure p = 5.00 torr. You may assume that for the conditions of pressure and temperature in
the problem ammonia obeys the ideal gas law.
4. (20 points) Pyrogallol (1,2,3-trihydroxybenzene, C6H3(OH)3(s)) is used in some hair dyes and also in
the analysis of the oxygen content of gas samples.
a) Give the correctly balanced formation and combustion reaction for pyrogallol.
b) Based on the data below, find the value for Hf(C6H3(OH)3(s)), the enthalpy of formation for
pyrogallol.
Combustion data
Formation data
Hc(C6H3(OH)3(s)) = - 2672.3 kJ/mol
Hf(CO2(g)) = - 393.51 kJ/mol
Hf(H2O()) = - 285.83 kJ/mol
5. (16 points) Hess’ law (based on the fact that enthalpy is a state function) makes it possible to find the
enthalpy change for processes that are difficult (or impossible) to carry out in the laboratory.
One example of the use of Hess’ is to find the enthalpy change when a specific bond is broken in a
specific molecule. Consider the following process
CH3(g)  CH2(g) + H(g)
(5.1)
Because CH3(g) is highly reactive, it is difficult to measure Hrxn for reaction 5.1 directly.
Using the information given below, find the value for Hrxn for reaction 5.1
H2(g)  2H(g)
CH4(g)  CH3(g) + H(g)
C2H4(g)  2 CH2(g)
2 CH4(g)  C2H4(g) + 2 H2(g)
Hrxn = + 435.9 kJ/mol
Hrxn = + 438.5 kJ/mol
Hrxn = + 730. kJ/mol
Hrxn = + 201.9 kJ/mol
Solutions.
1)
a) pV = nRT, and so for the neon gas in the mixture
(n/V) = p = (328. torr) (1 atm/760 torr)
= 0.01753 mol Ne/L
RT (0.08206 Latm/molK)(300.0 K)
and so
(Ne) = 0.01753 mol Ne
L
20.17 g = 0.3536 g Ne/L
mol
Therefore, 1.000 L of the gas mixture contains 0.3536 g of neon.
b)  = (Ne) + (Ar), and so (Ar) =  - (Ne)
and so
(Ar) = 0.562 g/L – 0.3536 g/L = 0.2084 g Ar/L
We can convert this to the number of moles of argon per liter of gas mixture
n/V = (0.2084 g Ar/L) (1 mol Ar/39.95 g) = 0.0522 mol Ar/L
From the ideal gas law
p = (n/V)RT = (0.0522 mol/L) (0.08206 Latm/molK) (300.0 K)
= 0.1284 atm (760 torr/1 atm) = 97.6 torr
2)
a) Cp = dqp  qp =
4296 J
284.5 J/C = 284.5 J/K
dT
T (37.9 – 22.8) C
b) The process is constant pressure, and so q = H = 4296. J
The process is reversible and at constant pressure, and so
w = - if pex dV = - if p dV = - p if dV = - p V
= - (24.10 atm) (9.8 x 10-3 L) = - 0.236 Latm (101.3 J/Latm) = - 23.9 J
Finally, from the first law, U = q + w = 4296. J + (- 23.9 J) = 4272. J
3)
a) crms = (3RT/M)1/2 = [ 3 (8.314 J/molK) (773.2 K)/(17.03 x 10-3 kg) ]1/2 = 1064. m/s
b) The process is the constant pressure reversible heating of a gas, and so
q = H = if n Cp,m dT = n if (a + bT + cT2) dT
= n { a (Tf – Ti) + (b/2) (Tf2 – Ti2) + (c/3) (Tf3 – Ti3) }
= (1.000 mol) { (24.619 J/molK) (800.0 K – 300.0 K)
+ (1/2)(3.75 x 10-2 J/molK2) [ (800.0 K)2 – (300.0 K)2 ]
+ (1/3) (- 0.138 x 10-5 J/molK3) [ (800.0 K)3 – (300.0 K)3 ] }
= (1.000 mol) { (12309.5 J/mol) + 10312.5 J/mol) + (- 223.1 J/mol) } = 22399. J
Because the gas is ideal, Cp,m – CV,m = R, or CV,m = Cp,m – R
and so
U = if n CV,m dt = n if CV,m dV = n if (Cp,m – R) dT = H – nR (Tf – Ti)
= 22399. J – (1.000 mol)(8.314 J/molK)(800.0 K – 300.0 K) = 18242. J
Finally, from the first law U = q + w, and so w = U – q = (18242. J) – (22399. J) = - 4157 J
4)
a)
formation
6 C(s) + 3 H2(g) + 3/2 O2(g)  C6H3(OH)3(s)
combustion
C6H3(OH)3(s) + 6 O2(g)  6 CO2(g) + 3 H2O()
b) Using the balanced combustion reaction above and the general expression for the enthalpy
change for a chemical reaction, we may say
Hc(C6H3(OH)3(s)) = [ 6 Hf(CO2(g)) + 3Hf(H2O()) ] – [ Hf(C6H3(OH)3(s)) ]
or
Hf(C6H3(OH)3(s)) = [ 6 Hf(CO2(g)) + 3Hf(H2O()) ] – [ Hc(C6H3(OH)3(s)) ]
= [ 6 (- 393.51 kJ/mol) + 3 (- 285.83 kJ/mol) ] – (- 2672.3 kJ/mol) = - 546 3 kJ/mol
5) The reaction we are interested in can be found as follows
CH3(g)  CH2(g) + H(g)
Hrxn
(1)
½ C2H4(g)  CH2(g)
H1 = (1/2) (730. kJ/mol)
(2)
CH3(g) + H(g)  CH4(g)
H2 = (-1) (438.5 kJ/mol)
(3)
CH4(g)  ½ C2H4(g) + H2(g)
H3 = (1/2) (201.9 kJ/mol)
(4)
H2(g)  2 H(g)
H4 = 435.9 kJ/mol
Since the four reactions above, when combined, give us reaction 5.1, then from Hess’ law
Hrxn = H1 + H2 + H3 +H4
= (365. kJ/mol) + (- 438.5 kJ/mol) + (100.95 kJ/mol) + (435.9 kJ/mol) = 463.4 kJ/mol