Physical Chemistry II
Chem 402
Spring 2012
Chapter 4 (4, 6, 8, 9, 10, 16, 17, 18, 25)
P4.4) Calculate H for the process NO(g), 1 bar, 298.15 K → NO(g), 1 bar, 550. K assuming
that the heat capacities of reactants and products are constant over the temperature interval at
their values at 298.15 K
550.
H  H

f
 NO, g , 298.15K   
CP T '  dT '
298.15
1
1
CP  CP ,m  NO, g   CP ,m  N 2 , g   CP ,m  O 2 , g 
2
2
  29.86  0.5  29.13  0.5  29.38  J K -1mol-1
 0.605 J K -1mol-1
 550.

H  H f  NO, g , 298.15 K     0.605 dT ' J mol-1
 298.15


-1
 H f  NO, g , 298.15 K   0.327 kJ mol  91.3 kJ mol-1  0.328 kJ mol-1  91.6 kJ mol-1
P4.6) Derive a formula for H R T  for the reaction CO(g) + 1/2O2(g)  CO2(g) assuming that
the heat capacities of reactants and products do not change with temperature.
H R T   H R  298.15 K   [C rxn P ]T
1
 H f  CO 2  g    H f  CO  g    [CP , m  CO 2  g    CP ,m  CO  g    CP ,m  O 2  g  ]T
2
P4.8) Use the following data at 298.15 K to complete this problem:
H R (kJ mol–1)
½H2(g) + ½O2(g)  OH(g)
38.95
H2(g) + ½O2(g)  H2O(g)
–241.814
H2(g)  2H(g)
435.994
O2(g)  2O(g)
498.34
Calculate H R for
a. OH(g)  H(g) + O(g)
b. H2O(g)  2H(g) + O(g)
c. H2O(g)  H(g) + OH(g)
Assuming ideal gas behavior, calculate H R and U R for all three reactions.

H reaction
 kJ mol-1 
a)
OH(g) → ½H2(g) + ½O2(g)
–38.95
½H2(g) → H(g)
½  435.994
½O2(g) → O (g)
½  498.34
____________________________________________________________
OH(g)  H(g) + O(g)

= 428.22 kJ mol-1
H reaction


= H reaction
U reaction
 nRT
= 428.22 kJ mol-1 – 8.314 J mol-1 K-1  298.15 K = 425.74 kJ mol-1

H reaction
 kJ mol-1 
b)
H2O(g)  H2(g) + ½O2(g)
H2(g)  2H(g)
½ O2(g) → O (g)
241.814
435.994
½  498.34
_______________________________________________________
H2O(g)  2H(g) + O(g)

= 926.98 kJ mol-1
H reaction


= H reaction
U reaction
 nRT
= 926.98 kJ mol-1 – 2  8.314 J mol-1 K-1  298.15 K = 922.02 kJ mol-1

H reaction
 kJ mol-1 
c)
H2O(g)  H2(g) + ½O2(g)
241.814
½H2(g) + ½O2(g)  OH(g)
38.95
½H2(g)  H(g)
½  435.994
_________________________________________________________
H2O(g)  H(g) + OH(g)

= 498.76 kJ mol-1
H reaction


= H reaction
U reaction
 nRT
= 498.76 kJ mol-1 – 8.314 J mol-1 K-1  298.15 K = 496.28 kJ mol-1
P4.9) Calculate the standard enthalpy of formation of FeS2(s) at 300.°C from the following data
at 298.15 K. Assume that the heat capacities are independent of temperature.
Substance
Fe(s)
FeS2(s)
S(rhombic)
–824.2
H f (kJ mol–1)
3.02
CP , m /R
Fe2O3(s)
7.48
SO2(g)
–296.81
2.72
You are also given:

= –1655 kJ mol-1
2FeS2(s) + 11/2O2(g)  Fe2O3(s) + 4SO2(g) ; H reaction
1655 kJ mol 1  H f  Fe 2 O3 ,s   4H f  SO 2 , g   2 H f  Fe 2S2 , s 
H

f
 Fe2S2 , s,298 K  
1655 kJ mol-1  H f  Fe 2 O3 , s   4H f SO 2 , g 
2
1655  824.2  4  296.81 kJ mol -1
=
2
-1
 178.2 kJ mol
The enthalpy of formation at 300.  C is given by
573 K
H

f
 FeS  s  ,573 K   H  FeS  s  , 298 K   
2

f
2
C p T ' dT '
298 K
Because the heat capacities are assumed to be independent of T,
H f  FeS2  s  ,573 K   H f  FeS2  s  , 298 K 
 CP ,m  FeS2 , s   CP ,m  Fe, s   2CP ,m  S, s   573 K  298 K 
 178.2 kJ mol-1  8.314 J K -1 mol-1   7.48  3.02  2  2.70    573 K  298 K 
 180.0 kJ mol-1
P4.10) The data below are a DSC scan of a solution of a T4 lysozyme mutant. From the data
determine Tm. Determine also the excess heat capacity CP at T = 308 K. Determine also the
intrinsic  C Pint and transition  C Ptrs excess heat capacities at T = 308 K. In your calculations use
the extrapolated curves, shown as dotted lines in the DSC scan, where the y axis shows CP.
Cptrs = 0.83 J K-1 g-1
Cp = 1.25 J K-1 g-1
Cpint = 0.42 J K-1 g-1
Tm = 304 K
P4.16) The total surface area of the earth consisting of forest, cultivated land, grass land, and
desert is 1.49x108 km2. Every year, the mass of carbon fixed by photosynthesis by vegetation
covering
this
land
surface
according
to
the
reaction
6CO2  g   6H 2O  l   C6 H12O6  s   6O2  g  is about 450. metric
Tons km-2. Calculate the annual enthalpy change resulting from photosynthetic carbon fixation
over the land surface per mole of carbon given the data above. Assume P = 1 bar and T = 298 K.
HR = H f (C6H12O6 (s)) ─6 H f (H2O(l)) ─ 6 H f (CO2(g))
 1273.1 kJ mol 1  6  393.5 kJ mol1  6  285.5 kJ mol1
 2802.7 kJ mol1
The number of moles of fixed carbon is:
nC , fixed 
mC , fixed
MC
 4.50 10 kg km year   3.747 10

12.0110 kg mol 
5
-2
-3
-1
7
-1
mol km -2 year -1
We calculate the enthalpy change per mole of C using the above result:
7
-2
-1
-1
8
2
H R  3.747  10 mol km year    2802 kJ mol   1.49  10 km 
H 

6
6
18
-1
 2.60  10 kJ year
P4.17) Calculate H R and U R at 298.15 K for the following reactions:
a. 4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(g)
b. 2NO(g) + O2(g) → 2NO2(g)
c. TiCl4(l) + 2H2O(l) → TiO2(s) + 4HCl(g)
d. 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) Assume complete dissociation of NaOH,
H2SO4, and Na2SO4
e. CH4(g) + H2O(g) → CO(g) + 3H2(g)
f.
CH3OH(g) + CO(g) → CH3COOH(l)
a) 4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(g)

H reaction
 5H f  N 2 , g   6H f  H 2 O, g   4H f  NH 3 , g   6H f  NO, g 
 0  6  241.8 kJ mol-1  4  45.9 kJ mol-1  6  91.3 kJ mol-1
 1815.0 kJ mol-1


U reaction
 H reaction
 nRT
 1815.0 kJ mol-1  8.314 J K -1mol-1  298.15 K  1817.5 kJ mol-1
b) 2NO(g) + O2(g) → 2NO2(g)

H reaction
 2H f  NO 2 , g   H f  O 2 , g   2H f  NO, g 
 2  33.2 kJ mol-1  0  2  91.3 kJ mol-1
 116.2 kJ mol-1


U reaction
 H reaction
 nRT
 116.2 kJ mol-1  8.314 J K -1mol-1  298.15 K  113.7 kJ mol-1
c) TiCl4(l) + 2H2O(l) → TiO2(s) + 4HCl(g)

H reaction
 H f  TiO 2 , s   4H f  HCl, g   H f  TiCl4 , l   2H f  H 2 O, l 
 944  4  92.3 kJ mol-1  804.2 kJ mol-1  2  285.8 kJ mol-1
 62.6 kJ mol-1


U reaction
 H reaction
 nRT
 62.6 kJ mol-1  4×8.314 J K -1mol-1  298.15 K  52.7 kJ mol-1
d) 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) +2H2O(l)
Assume that NaOH(aq) and H2SO4(aq) are completely dissociated.
The overall reaction is 2OH -  aq  + 2H +  aq   2H 2 O(l )

H reaction
 2H f  H 2 O, l   2H f  H + , aq   2H f  OH - , aq 
 2  285.8 kJ mol-1  0  2  230.0 kJ mol-1
 111.6 kJ mol-1


U reaction
 H reaction
 nRT
 111.6 kJ mol-1  0  111.6 kJ mol-1
e) CH4(g) + H2O(g) → CO(g) + 3H2(g)

H reaction
 H f  CO, g   3H f  H 2 , g   H f  CH 4 , g   H f  H 2 O, g 
 110.5 kJ mol-1  0  74.6 kJ mol-1  241.8 kJ mol-1
 205.9 kJ mol-1


U reaction
 H reaction
 nRT
 205.9 kJ mol-1  2  8.3145 J K -1mol-1  298.15 K  200.9 kJ mol-1
f) CH3OH(g) + CO(g) → CH3COOH(l)

H reaction
 H f  CH 3 COOH, l   H f  CH 3 OH, g   H f  CO, g 
  484.3 kJ mol-1  201.0 kJ mol-1 110.5 kJ mol-1
 172.8 kJ mol-1


U reaction
 H reaction
 nRT
 172.8 kJ mol-1  2  8.3145 J K -1mol-1  298.15 K  167.8 kJ mol-1
P4.18) A sample of Na2SO4(s) is dissolved in 225 g of water at 298 K such that the solution is
0.250 molar in Na2SO4. A temperature rise of 0.112°C is observed. The calorimeter constant is
330. J K–1. Calculate the enthalpy of solution of Na2SO4 in water at this concentration. Assume
that the solution volume is the same as the solvent volume. Compare your result with that
calculated using the data in Table 4.1 (Appendix B, Data Tables).
0
mH 2O
ms
H solution ,m 
CH O ,m T  Ccalorimeter T
Ms
M H 2O 2
H solution ,m


225 g
 75.3 J K -1mol-1  0.112o C  330 J K -1  0.112o K 

-1
18.02 g mol

 
0.225 kg
0.250 mol L1 
0.997 kg L1
 2.52 103 J mol-1
H solution ,m  2H solution ,m  Na  , aq   H solution ,m  SO 24 , aq   H f  Na 2SO 4 , s 
 2  240.1103 J mol-1  909.3103 J mol-1  1387.1103 J mol-1
 2.4 103 J mol-1
2.52 kJ mol-1  2.4 kJ mol-1
The relative error is
 4.8%
2.52 kJ mol-1
P4.25) Using the protein DSC data in Problem 4.10, calculate the enthalpy change between the
T = 288 K and T = 318 K. Give your answer in units of kJ per mole. Assume the molar mass of
the protein is 14000. grams per mole. Hint: You can perform the integration of the heat capacity
by estimating the area under the DSC curve and above the dotted baseline in problem 4.28. This
can be done by dividing the area up into small rectangles and summing the areas of the
rectangles. Comment on the accuracy of this method.
There are approximately 624 squares in the yellow area. On the horizontal axis, 1 square = 0.50
K. On the vertical axis, 1 square = 0.0418 J K-1 g-1. we obtain:
H m  624  0.50 K  0.0418 J K 1 g 1  14000. g mol 1  1.8  10 2 kJ mol 1
The method is reasonably accurate because the number of squares can be counted quite
accurately.