CHM 3400 – Fundamentals of Physical Chemistry
First Hour Exam
February 8, 2012
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
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1. (16 points) Scientists at Chibi University have recently been investigating the properties of a new substance, imaginarium. Over a particular range of pressure a sample of this substance is found to obey the following equation of state.
p = A + (BT/V2) (1.1)
where A and B are constants.
a) Find an expression for (¶p/¶V)T, the change in the pressure of the substance with respect to volume, under conditions of constant temperature.
b) Find an expression for w (work) when the volume of the above substance is changed iso-thermally and reversibly from an initial value Vi to a final value Vf.
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Thermochemical data for several substances at T = 298. K and p = 1.000 bar are given below. The data may be useful in answering problems 2 and 3 below.
substance M(g/mol) DH°f(kJ/mol) DG°f(kJ/mol) S°(J/mol.K) Cp,m(J/mol.K)
NO(g) 30.01 + 90.25 + 86.55 210.76 29.84
NO2(g) 46.01 + 33.18 + 51.31 240.06 37.20
O2(g) 32.00 0.00 0.00 205.14 29.36
O3(g) 48.00 + 142.7 + 163.2 238.93 39.20
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2. (30 points) Consider a 1.000 mol sample of molecular oxygen (O2) at a temperature T = 200.0 K.
a) What is c, the rms average speed of an oxygen molecule in the gas sample.
b) The temperature of the above oxygen sample is changed reversibly and at constant pressure from the initial value Ti = 200.0 K to a final value Tf = 350.0 K. Find q, w, DU, DH, DSsyst, DSsurr, and DSuniv for the process, or explain why there is insufficient information to find the values requested. You may assume that for these conditions O2(g) is an ideal gas, and that Cp,m, the constant pressure molar heat capacity, is constant over the temperature range of the process..
3. (18 points) At high temperatures nitrogen dioxide (NO2) will thermally decompose into nitrogen monoxide (NO) and molecular oxygen (O2). The stoichiometric reaction that takes place is
2 NO2(g) ® 2 NO(g) + O2(g) (3.1)
a) Find the value for DH°rxn for the above reaction at T = 298. K.
b) Find the value for DH°rxn for the above reaction at T = 500. K (a temperature at which the above reaction begins to be important).
4. (20 points) Maltose (C12H22O11(s)) is a disaccharide found in barley and other germinating seeds.
a) Write the correctly balanced formation reaction and combustion reaction for maltose.
b) The experimental value for the enthalpy of combustion of maltose is DH°comb = - 5649.2 kJ/mol. Based on this and the thermodynamic data in Atkins (pp 558-564) find the value DH°f, the enthalpy of formation of maltose.
5. (16 points) The constant pressure molar heat capacity of a substance is often written as
Cp,m = a + bT (5.1)
where a and b are constants found from experimental data.
a) Starting with eq 5.1 above, find an expression for DH and DS when the temperature of n moles of the above substance is changed reversibly and at constant pressure from an initial value Ti to a final value Tf.
b) Is there sufficient information above to find a general expression for DU? If your answer is yes, find the expression for DU. If your answer is no, justify your answer.
Solutions.
1) a) (¶p/¶V)T = ¶/¶p)T [p = A + (BT/V2)] = - 2BT/V3
b) w = - òif pex dV
The process is reversible, and so pex = p. Therefore
w = - òif [A + (BT/V2)] dV
The process is isothermal, and so T is constant. Therefore
w = - { AV – (BT/V) ]if = - { A(Vf – Vi) – BT [(1/Vf) – (1/Vi)] }
2) a) c = (3RT/M)1/2 = [3(8.314 J/mol.K)(200.0 K)/(32.00 x 10-3 kg/mol)]1/2 = 395. m/s
b) For any process involving an ideal gas
DU = òif nCV,m dT DH = òif nCp,m dT
For an ideal gas, Cp,m – CV,m = R
and so CV,m = Cp,m - R = 29.36 J/mol.K – 8.314 J/mol.K = 21.05 J/mol.K
Since both CV,m and Cp,m are constant
DU = n CV,m (Tf – Ti) = (1.000 mol) (21.05 J/mol.K) (350.0 K – 200.0 K) = 3158. J
DH = n Cp,m (Tf – Ti) = (1.000 mol) (29.36 J/mol.K) (350.0 K – 200.0 K) = 4404. J
The process is carried out at constant pressure, and so q = DH = 4404. J.
From the first law, DU = q + w, and so w = DU – q = (3158. J) – (4404. J) = - 1246. J
The entropy change for the system is
DSsyst = òif đqrev/T
Since the process in the problem is reversible we can use it to find DSsyst. The process is contant pressure and reversible, and so đqrev = n Cp,m dT. Therefore
DSsyst = òif (nCp,m dT)/T
But n and Cp,m are constants and so can be taken outside the integral, to give
DSsyst = n Cp,m òif (dT)/T = n Cp,m ln(Tf/Ti)
= (1.000 mol) (29.36 J/mol.K) ln(350.0 K/200.0 K) = 16.43 J/K.
The process is reversible, and so DSuniv = 0. But DSuniv = DSsyst + DSsurr
So DSsurr = - DSsyst = - 16.43 J/K.
3) a) At T = 298. K, the enthalpy of reaction is
DH°rxn = [2 DH°f(NO(g)) + DH°f(O2(g))] – [2 DH°f(NO2(g))]
= [2 (90.25 kJ/mol) + (0.0 kJ/mol)] – [2 (33.18 kJ/mol)]
= 114.14 kJ/mol
b) To find the reaction enthalpy at T = 500.0 K we may use Kirchoff’s law
DH°(T2) = DH°(T1) + òT1T2 DCp dT
where DCp = [2 Cp,m(NO(g)) + Cp,m(O2(g))] – [2 Cp,m(NO2(g))]
= [2 (29.84 J/mol.K) + (29.36 J/mol.K)] – [2 (37.20 J/mol.K)]
= 14.64 J/mol.K = 14.64 x 10-3 kJ/mol.K
where we have used the fact that all of the constant pressure molar heat capacities can be assumed constant. Therefore
DH°(T2) = DH°(T1) + DCp òT1T2 dT = DH°(T1) + (DCp) (Tf – Ti)
= 114.14 kJ/mol + (14.64 x 10-3 kJ/mol.K) (500. K – 298. K) = 117.10 kJ/mol
4) a) formation 12 C(s) + 11 H2(g) + 11/2 O2(g) ® C12H22O11(s)
combustion C12H22O11(s) + 12 O2(g) ® 12 CO2(g) + 11 H2O()
b) DH°comb = [12 DH°f(CO2(g)) + 11 DH°f(H2O())] – [DH°f(C11H22O11(s)) + 12 DH°f(O2(g))]
Rearranging this gives
DH°f(C11H22O11(s)) = [12 DH°f(CO2(g)) + 11 DH°f(H2O())] –
[DH°comb(C11H22O11(s)) + 12 DH°f(O2(g))]
= [12 (- 393.51 kJ/mol) + 11 (- 285.83 kJ/mol)]
- [(- 5649.2 kJ/mol) + 12 (0.00 kJ/mol)]
= - 2217.0 kJ/mol
5) a) The process is reversible and constant pressure. Therefore
DH = òTiTf n Cp,m dT DS = òTiTf (n Cp,m dT)/T
So DH = n òTiTf (a + bT) dT = n { a (Tf – Ti) + (b/2) (Tf2 – Ti2) }
DS = n òTiTf [(a/T) + b) dT = n { a ln(Tf/Ti) + b (Tf – Ti) }
b) Since we don’t know whether the substance is an ideal gas, and we don't know the pathway, there is insufficient information to find DU for the process.