CHM 3400 – Problem Set 4
Due date: Tuesday, September 23rd. The first hour exam will be on Thursday, September 25 th. It will cover material
from Chapters 1, 2, and 3 of Atkins, Chapter 4, sections 4.1 to 4.6, and handouts.
Do all of the following problems. Show your work.
"Scientists do not believe; they check." John Cornforth
1) Consider the gas phase chemical reaction
NH3(g) + 3 CH4(g)  N(CH3)3(g) + 3 H2(g)
(1.1)
a) Using the table of average bond enthalpies in Chapter 3 of Atkins, estimate the value for Hrxn for
reaction 1.1.
b) Based on your answer in a and the enthalpy of formation data in the Appendix of Atkins, estimate the
value for Hf(N(CH3)3(g)), the enthalpy of formation for gas phase trimethylamine.
2) The temperature and pressure for 4.000 moles of N2(g) are changed from initial values pi = 3.000 atm, Ti = 320.0
K to final values pf = 1.000 atm, Tf = 400.0 K by some unspecified process. Over this range of pressures and
temperatures nitrogen behaves as an ideal gas, and has a temperature independent constant pressure molar heat
capacity Cp,m = 29.125 J/molK. Find U, H, and S for the process.
3) 20.00 g of liquid water (H2O, M = 18.01 g/mol) at T = 80.0 C is added to 80.00 g of liquid water at T = 20.0 C
in an insulated container, and the resultant 100.00 g of water is allowed to come to equilibrium. Note that for water
Cp,m(H2O(l)) = 75.29 J/molK over the temperature range of the problem.
a) What is Tf, the final temperature of the 100.00 g of water?
b) What is S for the above process?
4) Consider the process where heat is slowly added to a 1.000 mole sample of bromine (Br 2) at a constant pressure p
= 1.000 atm. The initial and final temperature of the bromine are T i = 300.0 K and Tf = 400.0 K. The process can
be represented as
Br2(l, T = 300.0 K)  Br2(g, T= 400.0 K)
(4.1)
What is S for this process? Note that the normal boiling point of liquid bromine is Tvap = 332.4 K, and that the
enthalpy of vaporization for bromine is Hvap(Br2) = 29.45 kJ/mol. Also note that the constant pressure molar heat
capacities for liquid and gaseous bromine are Cp,m(Br2(l)) = 75.69 J/molK and Cp,m(Br2(g)) = 36.02 J/molK, and
that these values can be assumed to be constant over the range of temperatures in the problem.
Solutions.
1)
a) We may estimate the value for Hrxn for a gas phase reaction as
Hrxn   (bonds broken) -  (bonds formed)
where  (bonds broken) is the enthalpy change for breaking all of the chemical bonds in the reactant molecules, and
 (bonds formed) is the enthalpy change for forming the product molecules out of gas phase atoms. The
approximation sign is due to our use of average bond enthalpies in the calculation.
For the reaction in the problem
Bonds broken
3 N-H
12 C-H
3 (388)
12 (412)
Bonds formed
3 C-N
9 C-H
3 H-H
________
6108. kJ/mol
So
3 (305)
9 (412)
3 (436)
________
5931. kJ/mol
Hrxn  (6108. kJ/mol) - (5931. kJ/mol) = 177. kJ/mol
b) By Hess' law, we may also say that
Hrxn = (  Hf(products)) - (  Hf(reactants))
So
Hrxn = [ Hf(N(CH3)3(g)) ] - [Hf(NH3(g)) + 3 Hf(CH4(g)) ]
If we solve the above equation for Hf(N(CH3)3(g)), we get
Hf(N(CH3)3(g)) = Hrxn + [Hf(NH3(g)) + 3 Hf(CH4(g)) ]
= 177. kJ/mol + [ (- 46.11 kJ/mol) + 3 (- 74.81 kJ/mol) ]
= - 94. kJ/mol
where we have used our answer in part a for an estimated value for Hrxn, and the data in the back of the book for
the enthalpy of formation for NH3(g) and CH4(g) (note that Hf(H2(g)) = 0., and so this formation enthalpy does not
have to be included in the calculation.).
The literature value for the enthalpy of formation for gas phase trimethylamine (given in the CRC
Handbook) is Hr (N(CH3)3(g) = - 23.6 kJ/mol. so the value obtained indirectly using the average bond enthalpy
values is not very accurate.
2)
Since the gas is ideal, and Cp,m = 29.125 J/molK, then
CV,m = Cp,m - R = 29.125 J/molK - 8.314 J/molK = 20.811 J/molK
For an ideal gas,
U = if CV dTU = n if CV,m dT
H = if Cp dTU = n if Cp,m dT
Since both heat capacities are constant over the temperature range of the problem
U = n CV,m (Tf - Ti) = (4.000 mol)(20.811 J/molK)(400.0 K - 320.0 K) = 6659.52 J
H = n Cp,m (Tf - Ti) = (4.000 mol)(29.125 J/molK)(400.0 K - 320.0 K) = 9320.00 J
To find the entropy change for the system we need a reversible pathway connecting the initial and final states of the
system. One such pathway is the following:
Step 1
Isothermal reversible expansion of the gas, from pi = 3.000 atm to pf = 1.000 atm, at T = 320. K
Step 2
Constant pressure heating of the gas from Ti = 320.0 K to Tf = 400.0 K, at p = 1.000 atm
Step 1
S1 = if (dqrev/T)
The process is isothermal and so T can be taken out of the integral. This gives
S1 = if (dqrev/T) = (1/T) if (dqrev) = qrev/T
The gas is ideal, and so U = 0. From the first law, U = q + w, and so qrev = - wrev.
The work done by the hypothetical reversible expanion is
w = - if pex dV
The gas is ideal and the process is reversible and isothermal, and so p ex = p = nRT/V. Substituting, we get
wrev = - if (nRT/V) dV = - nRT ln(Vf/Vi)
From Boyle's law, for an isothermal process involving an ideal gas p iVi = pfVf, or (Vf/Vi) = (pi/pf)
wrev = - nRT ln(pi/pf)
qrev = - wrev = nRT ln(pi/pf)
S1 = qrev/T = (1/T) nRT ln(pi/pf) = nR ln(pi/pf)
= (4.000 mol)(8.314 J/molK) ln(3.000/1.000) = 36.54 J/K
Step 2
S2 = if (dqrev/T)
The process is constant pressure, and so dqrev = n Cp,m dT
S2 = if (dqrev/T) = if (nCp,m/T)dT
Since Cp,m is constant over the temperature range of the process it can be taken outside of the integral, and so
S2 = if (nCp,m/T) dT = nCp,m if (dT/T) = nCp,m ln(Tf/Ti)
And so S2 = (4.000 mol)(29.125 J/molK) ln(400.0/320.0) = 26.00 J/K
Since our hypothetical two step process has the same initial and final state as the actual process that takes
place, and since entropy is a state function, it follows that
S = S1 + S2 = 36.54 J/K = 26.00 J/K = 62.54 J/K
3)
a) Since the process occurs in an insulated container, the heat lost by the hot water must equal the heat
gained by the cold water. This means
qh + qc = 0
where qh is q for the hot (initial 80.0 C) water and qc is q for the cold (initial 20.0 C). water. It is also convenient,
though not required, to work with the constant pressure specific heat (sp), or heat capacity per gram of water. Note
sp = Cp,m/M = (75.29 J/molK)/(18.01 g/mol) = 4.180 J/gK = 4.180 J/gC
where the last way of expressing specific heat makes use of the fact that the size of one degree Kelvin and one degree
Centigrade is the same.
We first find Tf, the final temperature of the mixture.
q = 0 = qh + qc = mh sp(Tf - T0,h) = mc sp(Tf - T0,c)
where mh is the mass of hot water, T0,h is the initial temperature of the hot water, mc is the mass of the cold water, T0,c
is the initial temperature of the cold water, sp is the specific heat of the water, and Tf is the final temperature of the
mixture.
We can divide both sides of the above equation by sp and eliminate it from the equation. We then have
mh(Tf - T0,h) + mc(Tf - T0,c) = 0
mhTf - mhT0,h + mcTf - mcT0,c = 0
mhTf + mcTf = mhT0,h + mcT0,c
(mh + mc) Tf = mhT0,h + mcT0,c
Tf = mhT0,h + mcT0,c = (20.00)(80.0) + (80.00)(20.0) = 32.0 C
(mh + mc)
(20.00 + 80.00)
b) To find the entropy change of the process it is convenient to break the process into two parts.
Step 1 - Reversibly cool 20.00 g of water from 80.0 C to 32.0 C
Step 2 - Reversibly heat 80.00 g of water from 20.0 C to 32.0 C
Note the combination of the above two processes have the same initial and final state as the process actually taking
place, and so S = S1 + S2
For changing the temperature of a substance under conditions of constant pressure
S = if (dqrev/T) = if (nCp,m/T)dT = if (msp/T)dT
where sp is the specific heat of the substance. If sp is independent of temperature, as is the case in this problem, then
S = if (msp/T)dT = msp if (dT/T) = msp ln(Tf/Ti)
Therefore
S1 = (20.00 g)(4.180 J/gK) ln(305.2 K/353.2 K) = - 12.21 J/K
S2 = (80.00 g)(4.180 J/gK) ln(305.2 K/293.2 K) = 13.41 J/K
S = S1 + S2 = (- 12.21 J/K) + (13.41 J/K) = 1.20 J/K
Note that S > 0 as expected, since Ssurr = 0, and so Suniv = Ssyst = S.
4)
For heating a pure substance under conditions of constant pressure, we showed in class that
S = if (dqrev/T) = if (nCp,m/T)dT +  n Hpt/Tpt
where the first term on the right accounts for heating the substance in the absence of phase transitions, and the
second term (the summation) accounts for the entropy change for the phase transitions that take place.
We can divide the process of adding heat to Br2 into three steps, with S = S1 + S2 + S3
Step 1 - Heat Br2(l) from 300.0 K to 332.4 K under conditions of constant pressure
Step 2 - Convert Br2(l) into Br2(g) at 332.4 K and constant pressure
Step 3 - Heat Br2(g) from 332.4 K to 400.0 K under conditions of constant pressure
Since Cp,m is constant for both Br2(l) and Br2(g), we can use the result from step 2 of problem 2 for step 1 and step 3
above
S1 = n[Cp,m(Br2(l))]ln(Tf/Ti) = (1.000 mol)(75.69 J/molK) ln(332.4/300.0) = 7.763 J/K
S2 = n Hvap/Tvap = (1.000 mol) [(29450. J/mol)/(332.4 K)] =88.60 J/K
S3 = n[Cp,m(Br2(g))]ln(Tf/Ti) = (1.000 mol)(36.02 J/molK) ln(400.0/332.4) = 6.668 J/K
And so S = S1 + S2 + S3 = 7.763 J/K + 88.60 J/K + 6.668 J/K = 103.03 J/K