CHM 3410 – Problem Set 3
Due date: Wednesday, September 16th
Do all of the following problems. Show your work.
"There are three things you do not discuss with your friends - politics, religion, and thermodynamics." - V. I. Arnold
1) The temperature of 1.000 mol of an ideal monatomic gas (CV,m = 3/2 R = 12.472 J/mol K) is changed from an initial value Ti = 300.0 K to a final volume Tf = 500.0 K. Find q, w, DU, and DH when the process is carried out by the following two methods:
a) Reversibly, under conditions of constant volume.
b) Reversibly, under conditions of constant pressure.
HINT: You can do this problem even though you do not know the value for volume (in a) or pressure (in b), a consequence of the gas being ideal.
2) Using the data in the appendix of Atkins, find DH°rxn for the following reaction at T = 298. K.
a) SF6(g) ® S(g) + 6 F(g) atomization of sulfur hexafluoride
b) Use your answer in a to find the average bond strength in a S - F bond in sulfur hexafluoride.
c) Estimate, based on your answer in a, the value for DU°rxn for the atomization of sulfur hexafluoride. What assumptions, if any, do you need to make to find the estimated value for DU°rxn?
3) Liquid natural gas, which is primarily methane (CH4, M = 16.04 g/mol) is a relatively clean energy source, as it generates less carbon dioxide per MJ (megajoule) energy produced than combustion of gasoline. The reaction may be written as follows (note this is not the combustion reaction, because the final state for water in rxn 3.1 is H2O(g), not H2O(l)).
CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) (3.1)
Using the data in the appendix of Atkins, find the following:
a) DH°rxn at T = 298. K.
b) DH°rxn at T = 1200. K, a typical temperature for methane combustion. You may assume the values for the constant pressure molar heat capacity are constant and equal to the values at T = 298. K
4) Consider the following two reactions (all data are at T = 298. K)
H2(g) + I2(s) ® 2 HI(g) DH°rxn = + 52.96 kJ/mol (4.1)
2 H2(g) + O2(g) ® 2 H2O(g) DH°rxn = - 483.64 kJ/mol (4.2)
Based on this information (and Hess' law) find DH°rxn for the following reaction:
4 HI(g) + O2(g) ® 2 I2(s) + 2 H2O(g) (4.3)
5) A 2.000 mol sample of air (CV,m = 5/2 R = 20.786 J/mol K is compressed reversibly and adiabatically from an initial pressure pi = 1.000 atm to a final pressure pf = 1.500 atm. The initial temperature of the air is Ti = 300.0 K.
a) What is Tf, the final temperature of the air?
b) What are q, w, DU, and DH for the process?
Solutions.
1) a) For any process involving an ideal gas, the changes in nternal energy and the change in enthalpy depend only on the initial and final temperatures of the gas
DU = òif n CV,m dT
DH = òif n Cp,m dT
Since the heat capacities are constant (CV,m = 3/2 R = 12.472 J/mol K, Cp,m = 5/2 R = 20.786 J/mol K) then
DU = òif n CV,m dT = n CV,m òif dT = n CV,m (Tf - Ti) = (1.000 mol) (12.472 J/mol K) (500.0 K - 300.0 K)
= 2494. J
DH = òif n Cp,m dT = n Cp,m òif dT = n Cp,m (Tf - Ti) = (1.000 mol) (20.786 J/mol K) (500.0 K - 300.0 K)
= 4157. J
The process is constant volume, and so w = 0. Therefore, from the first law, q = DU = 2494. J
b) The process involves the same ideal gas as in a, and has the same initial and final temperatures. Therefore the changes in internal energy and enthalpy are the same, and so DU = 2494. J, DH = 4157. J.
The process is constant pressure, and so q = DH = 4157. J
Finally, from the first law, DU = q + w, and so w = DU - q = (2494. J) - (4157. J) = - 1663. J
2) a) DH°rxn = [ DH°f(S(g)) + 6 DH°f(F(g)) ] - [ DH°f(SF6(g)) ]
= [ (278.81 kJ/mol) + 6 (78.99 kJ/mol) ] - [ - 1209. kJ/mol ] = 1962. kJ/mol
b) There are six equivalent S - F bonds in SF6, and so the average bond enthalpy for an S - F bond is
S - F bond enthalpy = (1962. kJ/mol)/6 = 327. kJmbond
c) As discussed in class, a good approximation for the relationship between DH and DU is
DH @ DU + DngRT where Dng is the change in the number of moles of gas per mole of reaction
So DU @ DH - DngRT = 1962. kJ/mol - (6) (8.3145 x 10-3 kJ/mol K) (298. K) = 1947. kJmol
Note that the difference between DU and DH is small. In fact, saying DHrxn @ DUrxn for a chemical reaction is also usually not a bad approximation.
3) a) At T = 298. K
DH°rxn = [ DH°f(CO2(g)) + 2 DH°f(H2O(g)) ] - [ DH°f(CH4(g)) + 2 DH°f(O2(g)) ]
= [ ( - 393.51 kJ/mol) + 2 ( - 241.82 kJ/mol) ] - [ ( - 74.81 kJ/mol) + 2 (0.00 kJ/mol) ]
= - 802.34 kJ/mol
Note that this is not the value listed for DH°c(CH4(g)) because we have written the final state for water as H2O(g), instead of H2O(l)).
b) In general DH°rxn(T2) = DH°rxn(T1) + òT1T2 DCp,m dT
In this case we can choose T1 = 298. K and T2 = 1200. K. We are told to assume all values of Cp,m are constant over this temperature range and equal to the values at T = 298. K, and so
DH°rxn(T2) = DH°rxn(T1) + DCp,m òT1T2 dT = DH°rxn(T1) + DCp,m (T2 - T1)
DCp,m = [ Cp,m(CO2(g) + 2 Cp,m(H2O(g)) ] - [ Cp,m(CH4(g)) + 2 Cp,m(O2(g)) ]
= [ (37.11 J/mol K) + 2 (33.58 J/mol K) ] - [ (35.31 J/mol K) + 2 (29.355 J/mol K) ]
= 10.25 J/mol K
and so DH°rxn(1200. K) = (- 802.34 kJ/mol) + (10.25 x 10-3 J/mol K)(1200. K - 298. K) = - 793.09 kJ/mol
4) Hess' law tells us that any process or combination of processes with the same initial and final state will have the same value for the change in enthalpy (or change in the value for any other state function). So
4 HI(g) ® 2 H2(g) + 2 I2(s) DH = (- 2) (52.96 kJ/mol)
2 H2(g) + O2(g) ® 2 H2O(g) DH = (1) (- 483.64 kJ/mol)
______
4 HI(g) + O2(g) ® 2 I2(s) + 2 H2O(g) DH = - 589.56 kJ/mol
5) a) To find Tf we may use the general relationship (discussed in class) that applies to adiabatic reversible processes on ideal gases with constant heat capacity
(Tf/Ti) = (pf/pi)(g-1)/g
Tf = Ti (pf/pi)(g-1)/g
For air, g = Cp,m/CV,m = (7/2 R)/(5/2 R) = 7/5 = 1.400
And so (g-1)/g = (1.400 - 1)/(1.400) = 0.2857
Tf = (300.0 K) (1.500 atm/1.000 atm)0.2857 = 336.8 K
b) The process is adiabatic, and so q = 0. As in problem 1, we have an ideal gas, and so
DU = òif n CV,m dT = n CV,m òif dT = n CV,m (Tf - Ti) = (2.000 mol) (20.786 J/mol K) (336.8 K - 300.0 K)
= 1530. J
DH = òif n Cp,m dT = n Cp,m òif dT = n Cp,m (Tf - Ti) = (2.000 mol) (29.101 J/mol K) (336.8 K - 300.0 K)
= 2142. J
Since q = 0, it follows from the first law that w = DU = 1530. J