Homework5 Answer (Due Friday, 02/23/2011)
Q4.1
P4.1, P4.2, P4.3, P4.5 P4.6 P4.8, P4.18, P4.19, P4.21, P4.29, P4.31, P4.32
Q4.1) Under what conditions are H and U for a reaction involving gases and/or liquids or solids
identical?
H = U + (p V). H ≈ U for reactions involving liquids and solids, since, in a good
approximation, the volume does not change in a chemical reaction. On the other hand, for chemical
reactions involving gases, (p V) = n R T, and H ≈ U if the number of moles of reactants and
products is identical.

P4.1) Calculate H reaction and U reaction
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)
Table 4.1 4.2 Appendix B
a) ΔH reaction  5 ΔH f N 2 , g   6 ΔH f H 2 O, g   4 ΔH f NH 3 , g   6 ΔH f NO, g 







ΔH reaction  5  0 kJ mol -1  6   241.8 kJ mol -1  4   45.9 kJ mol -1  6  91.3 kJ mol -1
 - 1815.0 kJ mol
-1





ΔU reaction  ΔH reaction  n R T   1815.0 kJ mol -1  1  8.314472 J K -1 mol -1  298.15 K 
  1817.5 kJ mol
-1
b) ΔH reaction  2 ΔH f NO 2 , g  - ΔH f O 2 , g   2 ΔH f NO, g 
ΔH reaction  2  33.2 kJ mol -1   0 kJ mol -1   2  91.3 kJ mol -1 
  116.2 kJ mol -1
ΔU reaction  ΔH reaction  n R T   116.2 kJ mol -1    1  8.314472 J K -1 mol -1  298.15 K 
  113.7 kJ mol -1
c) ΔH reaction  ΔH f TiO 2 , s   4 ΔH f HCl, g   ΔH f TiCl 4 ,    2 ΔH f H 2 O,  
ΔH reaction   944.0 kJ mol -1   4   92.3 kJ mol -1   804.2 kJ mol -1   2   285.8 kJ mol-1 
 62.6 kJ mol -1
ΔU reaction  ΔH reaction  n R T   60.8 kJ mol-1    4    8.314472 J K -1 mol -1    298.15 K 
 52.7 kJ mol -1
d) The overall reaction is:
2 OH  aq   2 H  aq  
 2 H 2 O  



ΔH reaction  2 ΔH f H 2 O,    2 ΔH f H  , aq  2 ΔH f OH  , aq

ΔH reaction  2   285.8 kJ mol -1   2  0 kJ mol -1   2   230.0 kJ mol -1 
  111.6 kJ mol -1
ΔU reaction  ΔH reaction  n R T   111.6 kJ mol -1   0
  111.6 kJ mol -1
e) CH4(g) + H2O(g)  CO(g) + 3H2(g)
ΔH reaction  ΔH f CO, g   3 ΔH f H 2 , g   ΔH f CH 4 , g   ΔH f H 2 O, g 
ΔH reaction   110.5 kJ mol -1   3  0 kJ mol -1    74.6 kJ mol -1    241.8 kJ mol -1 
 205.9 kJ mol -1
ΔU reaction  ΔH reaction  n R T  205.9 kJ mol -1   2   8.314472 J K -1 mol -1  298.15 K 
 200.9 kJ mol -1
f) CH3OH(g) + CO(g)  CH3COOH(l)
ΔH reaction  ΔH f CH 3 COOH,    ΔH f CH 3 OH, g   ΔH f CO, g 

 
 
ΔH reaction   484.3 kJ mol -1   201.0 kJ mol -1   110.5 kJ mol -1
  172.8 kJ mol

-1




ΔU reaction  ΔH reaction  n R T   172.8 kJ mol -1   2   8.314472 J K -1 mol -1  298.15 K 
  167.8 kJ mol -1

P4.2) Calculate  H reaction and U reaction
for the oxidation of benzene. Also calculate


H reaction  Ureaction

 Hreaction
The chemical equation for the oxidation of benzene is:
C 6 H 6 l   7 12 O 2 g   6 CO 2 g   3 H 2 Ol 
The standard enthalpy for this reaction is:



 

ΔH    ν i H f,i  6   393.5 kJ mol 1  3   285.8 kJ mol 1  49.1 kJ mol 1  3268 kJ mol 1
i
U is calculated as:
ΔU   ΔH   n R T 
 3268 10
3
 



J mol 1  - 1.5  8.314472 J mol 1 K 1  298.15 K   3264 kJ mol 1
And:

 

ΔH   ΔU 
 3268 10 3 J mol 1   3264 10 3 J mol 1

 0.00122
ΔH 
 3268 10 3 J mol 1


P4.3) Use the tabulated values of the enthalpy of combustion of benzene and the enthalpies of
formation of CO2(g) and H2O(l) to determine H f for benzene.
3 H 2 O    6 CO 2 g  

15
2
O 2 g   C 6 H 6  
6 C s   6 O 2 g  
 6 CO 2 g 
- ΔH combustion C6 H 6 ,  
6 ΔH f CO 2 , g 
3 H 2 g   3 2 O 2 g  
 3 H 2 O  
3 ΔH f H 2 O,  
3 H 2 g   6 C s  
 C 6 H 6   - ΔH combustion C 6 H 6 ,    6 ΔH f CO 2 , g   3 ΔH f H 2 O,  






ΔH f C 6 H 6 ,      3268 kJ mol -1  6  - 393.5 kJ mol -1  3  - 285.8 kJ mol -1  49.6 kJ mol -1
P4.5) Several reactions and their standard reaction enthalpies at 25°C are given here:

 H reaction kJ mol1
CaC2(s) + 2H2O(l)  Ca(OH)2(s) + C2H2(g)
–127.9
Ca(s) + 1/2 O2(g)  CaO(s)
–635.1
CaO(s) + H2O(l)  Ca(OH)2(s)
–65.2
The standard enthalpies of combustion of graphite and C2H2(g) are –393.51 and –1299.58 kJ
mol–1, respectively. Calculate the standard enthalpy of formation of CaC2(s) at 25°C.
Ca OH 2 s   C 2 H 2 g  
 CaC 2 s   2 H 2 O  
CaO s   H 2 O   
 Ca OH 2 s 
127.9 kJ mol -1
- 65.2 kJ mol -1

3 CO 2 g   H 2 O   
 C 2 H 2 s   5 2 O 2 g 
1299.58 kJ mol -1
2 C s   2 O 2 g  
 2 CO 2 g 
2  - 393.51kJ mol -1 
C s   1 2 O 2 g  
 CaO s 
- 635.1 kJ mol -1
2 C s   Ca s  
 CaC 2 s 
ΔH f  59.8 kJ mol -1
P4.6) From the following data at 25°C, calculate the standard enthalpy of formation of FeO(s)
and of Fe2O3(s):

 H reaction kJ mol1

Fe2O3(s) + 3C(graphite)  2Fe(s) + 3CO(g)
492.6
FeO(s) + C(graphite)  Fe(s) + CO(g)
155.8
C(graphite) + O2(g)  CO2(g)
–393.51
CO(g) + 1/2 O2(g)  CO2(g)
–282.98
Fe s   CO g  
 FeO s   C graphite 
CO 2 g  
 CO g   1 2 O 2 g 
C graphite   O 2 g  
 CO 2 g 
Fe s   1 2 O 2 g  
 FeO s 
- 155.8 kJ mol -1
282.98 kJ mol -1
- 393.51kJ mol -1
ΔH f  266.3 kJ mol -1
2 Fe s   3 CO g  
 Fe 2 O 3 s   3 C graphite
- 492.6 kJ mol-1
3 CO 2 g  
 3 CO g   3 2 O 2 g 
3  282.98 kJ mol -1 
3 C graphite   3 O 2 g  
 3 CO 2 g 
3  - 393.51kJ mol -1
2Fe s   3 2 O 2 g  
 Fe 2 O 3 s 
ΔH f  824.2 kJ mol -1


P4.8) Calculate  H reaction at 650 K for the reaction 4NH3(g) +6NO(g)  5N2(g) + 6H2O(g)
using the temperature dependence of the heat capacities from the data tables.
H 650 K   H 298.15 K  

f

f
650
T T
C p   d
K K
298.15 K

ΔC p  5 C p,m N 2 , g   6 C p, m H 2 O, g   4 C p, m NH 3 , g   6 C p,m NO, g 
5  30.81  6  33.80  4  29.29  6  33.58

 5  (0.01187)  6  (0.00795)  4  0.01103  6  (0.02593) T


 J K -1 mol -1


 5  2.3968  10 5  6  2.8228  10 5  4  4.2446  10 5  6  5.3326  10 5 T 2


8
8
8
8
3
 5  (1.0176  10 )  6  (1.3115  10 )  4  (2.7706  10 )  6  (2.7744  10 ) T 



 38.21  0.00441T  2.0053  10
650
650
298.15 K
298.15 K
 C p T dT 
4


T 2  1.4772  10 7 T 3 JK 1 mol 1 dT
650 2  298.15 2
2
3
3
650

298
.
15
650 4  298.15 4
 2.0053  10  4 
 1.4772  10 7 
Jmol 1
3
4
 13.444  0.736  16.585  6.30kJmol 1  3.895kJmol 1
 38.21  (650  298.15)  0.00441 


H f 298.15 K   5 H f N 2 , g   6 H f H 2 O, g   4 H f NH 3 , g   6 H f NO, g 






H f 298.15 K   6   241.8 kJ mol -1  4   45.9 kJ mol -1  6  91.3 kJ mol -1  1815 kJ mol -1
H f 650 K   (1815kJmol 1 )  (3.895kJmol 1 )   1811.1 kJ mol -1
P4.18) Given the data in Table 4.1 (Appendix B, Data Tables) and the following information,
calculate the single bond enthalpies and energies for Si–F, Si–Cl, C–F, N–F, O–F, H–F:
Substance

H f kJ mol1

SiF4(g)
SiCl4(g)
CF4(g)
NF3(g)
OF2(g)
HF(g)
–1614.9
–657.0
–925
–125
–22
–271
a) SiF4
The average Si-F single bond enthalpy is calculated as follows:
SiF4 g  
 Si s   2 F2 g 
ΔH reaction  1614.9 kJ mol -1

2 F2 g  
 4 F g 
ΔH reaction  4  79.4 kJ mol -1
Si s  
 Si g 
ΔH reaction  450.0 kJ mol -1
SiF4 g  
 Si g   4 F g 
ΔH reaction  2382 kJ mol -1

The average Si-F bond enthalpy is then:
bond
H avrg

2382 kJ mol   596 kJ mol
-1
-1
4
And the average Si-F bond free energy:
ΔU reaction  ΔH reaction  n R T   2382 kJ mol-1    4    8.314472 J K -1 mol-1    298.15 K 
 2372 kJ mol -1
bond
U avrg

2372 kJ mol   593 kJ mol
-1
-1
4
b) SiCl4
SiCl 4 g  
 Si s   2 Cl 2 g 
ΔH reaction  657.0 kJ mol -1
2 Cl 2 g  
 4 Cl g 
ΔH reaction  4  121.3 kJ mol -1 
Si s  
 Si g 
ΔH reaction  450.0 kJ mol -1
SiCl 4 g  
 Si g   4 Cl g 
ΔH reaction  1592 kJ mol -1
The average Si-Cl bond enthalpy is then:
bond
H avrg

1592 kJ mol   398.0 kJ mol
-1
-1
4
And the average Si-Cl bond free energy:
ΔU reaction  ΔH reaction  n R T  1592 kJ mol -1    4   8.314472 J K -1 mol -1    298.15 K 
 1582.0 kJ mol -1
bond
U avrg

1582.0 kJ mol   396.0 kJ mol
-1
-1
4
c) CF4
CF4 g  
 C s   2 F2 g 
ΔH reaction  925.0 kJ mol -1
2 F2 g  
 4 F g 
ΔH reaction  4  79.4 kJ mol -1

Cs  
 C g 
ΔH reaction  716.7 kJ mol -1
CF4 g  
 C g   4 F g 
ΔH reaction  1959 kJ mol -1
The average C-F bond enthalpy is then:
bond
H avrg


1959 kJ mol   489.7 kJ mol
-1
4
And the average C-F bond free energy:
-1




ΔU reaction  ΔH reaction  n R T  1959.0 kJ mol -1  4   8.314472 J K -1 mol -1  298.15 K 
 1949.0 kJ mol
bond
U avrg

-1
1949.0 kJ mol   487.2 kJ mol
-1
-1
4
d) NF3
NF3 g  
 1 2 N 2 g   3 2 F2 g 
3
1
ΔH reaction  125.0 kJ mol-1

2
F2 g  
 3 F g 
ΔH reaction  3  79.4 kJ mol -1
2
N 2 g  
 N g 
ΔH reaction  472.7 kJ mol -1
NF3 g  
 N g   3 F g 

ΔH reaction  835.9 kJ mol -1
The average N-F bond enthalpy is then:
bond
H avrg

835.9 kJ mol   278.6 kJ mol
-1
-1
3
And the average N-F bond free energy:




ΔU reaction  ΔH reaction  n R T  835.9 kJ mol -1  3  8.314472 J K -1 mol -1  298.15 K 
 828.5 kJ mol
bond
U avrg

-1
828.5 kJ mol   276.2 kJ mol
-1
-1
3
e) OF2
OF2 g  
 1 2 O 2 g   F2 g 
F2 g  
 2 F g 
1
2
O 2 g  
 O g 
ΔH reaction  22.0 kJ mol -1

ΔH reaction  2  79.4 kJ mol -1

ΔH reaction  249.2 kJ mol -1
OF2 g  
 O g   2 F g 
ΔH reaction  430.0 kJ mol -1
The average O-F bond enthalpy is then:
bond
H avrg

430.0 kJ mol   215.0 kJ mol
-1
-1
2
And the average O-F bond free energy:




ΔU reaction  ΔH reaction  n R T  430.0 kJ mol -1  2   8.314472 J K -1 mol -1  298.15 K 
 425.0 kJ mol
bond
U avrg

-1
425.0 kJ mol   213.0 kJ mol
-1
-1
2
f) HF
HF g  
 1 2 H 2 g   1 F2 g 
2
1
1
2
H 2 g  
 H g 
2
F2 g   F g 
ΔH reaction  271.0 kJ mol -1
ΔH reaction  218.0 kJ mol -1
ΔH reaction  79.4 kJ mol -1
HF g  
 H g   F g 
ΔH reaction  568.4 kJ mol -1
The average H-F bond enthalpy is then:
bond
H avrg
 568.4 kJ mol -1
And the average H-F bond free energy:




ΔU reaction  ΔH reaction  n R T  568.4 kJ mol -1  1  8.314472 J K -1 mol -1  298.15 K 
 565.9 kJ mol
-1
P4.19) Given the data in Table 4.3 and the data tables, calculate the bond enthalpy and energy of
the following:
a. The CH bond in CH4
b. The CC single bond in C2H6
c. The CC double bond in C2H4
Use your result from part (a) to solve parts (b) and (c).
a) CH 4 g  
 C g   4 H g 
H reaction  4 H f H, g   H f C, g   H f CH 4 , g 

 
 

 4   218.0 kJ mol -1  716.7 kJ mol -1   74.6 kJ mol -1  1663.3 kJ mol -1
The average C-H bond enthalpy is then:
bond
H avrg

1663.30 kJ mol   415.8 kJ mol
-1
-1
4
And the average C-H bond free energy:
ΔU reaction  ΔH reaction  n R T  1663.3 kJ mol -1   4   8.314472 J K -1 mol -1  298.15 K 
 1653.4 kJ mol -1
ΔU bond
ave 
1653.4kJmol -1
 413.4kJ mol -1
4
b) C 2 H 6 g  
 2 C g   6 H g 
H reaction  6 H f H, g   2 H f C, g   H f C 2 H 6 , g 



 

 6   218.0 kJ mol -1  2   716.7 kJ mol -1   84.0 kJ mol -1  2825.4 kJ mol -1
H reaction  6  C - H bond enthalpy  C - C bond enthalpy
C - C bond enthalpy  H reaction  6  C - H bond enthalpy  (2825.4kJmol -1 ) - 6  (415.8kJmol -1 )
 330.6kJmol 1
ΔU reaction  ΔH reaction  n R T  2825.4 kJ mol -1   7   8.314472 J K -1 mol -1  298.15 K 
 2808.0 kJ mol -1
U reaction  6  C - H bond energy  C - C bond energy
C - C bond energy  U reaction  6  C - H bond energy  (2808.0kJmol -1 ) - 6  (413.4kJmol -1 )
 327.6kJmol 1
 2 C g   4 H g 
c) C 2 H 4 g  
H reaction  4 H f  H,g   2 H f  C,g   H f  C2 H 4 ,g 
  4    218.0 kJ mol-1    2    716.7 kJ mol-1    52.4 kJ mol-1   2253 kJ mol-1
H reaction  4  C - H bond enthalpy  C  C bond enthalpy
C  C bond enthalpy  H reaction  4  C - H bond enthalpy  (2253kJmol -1 ) - 4  (415.8kJmol -1 )
 589.8kJmol 1
ΔU reaction  ΔH reaction  n R T   2253 kJ mol -1    5  8.314472 J K -1 mol -1    298.15 K 
 2240 kJ mol -1
U bond  4  C - H bond energy  C  C bond energy
C  C bond energy  (2240kJmol -1 ) - 4  (413.4kJmol -1 ) 586.4kJmol 1
P4.21) Benzoic acid, 1.35 g, is reacted with oxygen in a constant volume calorimeter to form
H2O(l) and CO2(g). The mass of the water in the inner bath is 1.240  103 g. The temperature of
the calorimeter and its contents rise 3.45 K as a result of this reaction. Calculate the calorimeter
constant.
The reaction is, with n = ½:
C 6 H 5 COOH s  
15
O 2 g  
 7 CO 2 g   3 H 2 O  
2
ΔU combustion  ΔH combustion  n R T




  3227 kJ mol -1   1 2   8.314472 J K -1 mol -1  298.15 K   3225.76 kJ mol -1

 ms
m H 2O

ΔU R 
C H 2O ΔT 
 Ms

M H 2O

C calorimeter  
ΔT

1.350 g    3225.76 kJ mol -1  1.240  10 3 g  75.291 J K -1 mol -1  3.45 C
 
18.02 g mol -1
122.13 g mol -1

3.45  C


 5.15  10 J C
3

 




 

 


-1
P4.29) A good yield of photosynthesis for agricultural crops in bright sunlight is 2000. kg of
carbohydrate (e.g., sucrose) per km2 per hour. The net reaction for sucrose formation in
photosynthesis is:


light


12CO2 g  11H 2 O l 
 C12 H22O11 s  12O2 g
a. Use standard enthalpies of formation to calculate H reaction for the production of 1 mol of
sucrose at 298 K by the preceding reaction.
b. Calculate the rate at which energy is stored in carbohydrates (e.g., sucrose) per km2 as a
result of photosynthesis. (Note: 1 watt = 1 joule/s.)
c. Bright sunlight corresponds to radiation flux at the surface of the earth of about 1 kW/m2.
What percentage of this energy can be stored in the form of carbohydrates as a result of
photosynthesis?
a) The enthalpy of reaction at 298 K is:
ΔH reaction 298 K   ΔH f C12 H 22 O11 , s    11 ΔH f H 2 O,     12 ΔH f CO 2 , g 





  2226.1 kJ mol -1   11   285.8 kJ mol -1   12  - 393.5 kJ mol -1
 5639.7 kJ mol

-1
b) The stored energy is obtained as:
E stored 
2000 kg km h  5639.7  10 J mol   9.155  10
342.23  10 kg mol  3600 s h 
-2
-1
3
3
1
-1
1
6
J s -1 km -2  9.155  10 6 W km -2
c) The percentage of energy that can be stored is:
% stored,Energy 
9.155  10 W km   9.155 W m   0.9137
1 10 W m  1 10 W m 
6
3
-2
-2
-2
3
-2
%
P4.31) The figure below shows a DSC scan of a solution of a T4 lysozyme mutant. From the
DSC data, determine Tm, the excess heat capacity CP and the intrinsic and transition excess heat
capacities at T = 308 K. In your calculations, use the extrapolated curves, shown as dashed lines
in the DSC scan.
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
P4.32) Using the protein DSC data from Problem P 4.31, calculate the enthalpy change between
the T=288 K and T=318 K. Give your answer in units of kilojoules per mole. Assume the
molecular weight of the protein is 14000. grams. 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 P 4.31. 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.
Dividing the area up into small rectangles and summing the areas of the rectangles as
 ΔTK  ΔC J K
p
-1

g -1 we obtain:
ΔH318 K - 288 K   15.3 J g -1  14000 g mol -1   214.2 kJ mol -1
The method is not very accurate since it depends on the size of the rectangles chosen by eye,
which is not small enough to give an accurate answer.
Answers (Q1-Q3 first, then textbook problems):
Q1.
Enthalpy is a state function and therefore an infinitesimal change in H can be written:
δH = (δH/δT)p,n δT + (δH/δP)T,n δP + (δH/δn)T,p δn
(δH/δP)T,n = -CpμJ-T and (δH/δT)p,n = Cp
The term (δH/δP)T,n is the pressure dependence of Enthalpy at constant temperature and constant
number of particles. It can be found with the heat capacity at constant pressure, Cp, and the
Joule-Thompson coefficient. Likewise, (δH/δT)p,n is the definition of heat capacity at constant
pressure.
Q2.
We calculate the reaction enthalpy at temperature T and P = 1 bar for the following reaction:
2B + C  X+3Y
The path for the reaction is given in the following 4 steps:
(1) 2B + C (at T)  (2) 2B+C (at 298.15K)  (3) X+3Y ( at 298.15K)
 (4) X+3Y (at T)
For the “reaction” (or the change of the state) from (1) to (2), 2 moles of B and 1 mole of C is
subject to the temperature change from T to 298.15K at 1 bar. Thus, the enthalpy change in this
process is given by
T
298.15K
ΔH(1  2) 

(2C P , B ,m  C P ,C ,m )dT  

(2C P , B ,m  C P ,C ,m )dT
298.15K
T
For the change of the state from (2) to (3), the reaction takes place at the standard condition (T
=298.15K and 1 bar) as 2B + C  X + 3Y. The reaction enthalpy at the standard condition is
given by
ΔH(2  3)
 ΔH reaction, 298.15K
 ΔH X  ΔH Y  ΔH B  ΔH C
0
0
,
0
0
 H f (X)n X  H f (Y)n Y  H f (B)n B  H f (C)n C
0
0
0
0
 H f (X)  3H f (Y)  2 H f (B)  H f (C)
Now, our system is made of 1 mole of X and 3 mole of Y (X + 3Y) at T = 298.15K.
Then, for a change of the state from (3) to (4),
T
ΔH(3  4) 

(C P , X ,m  3C P ,Y ,m ) dT
298.15K
ΔH reaction, T  ΔH(1  2)  ΔH(2  3)  ΔH(3  4)
0
0
0
0
 H f (X)  3H f (Y)  2 H f (B)  H f (C) 
T
T
T
 T


   C PX ,m dT  3  C PY ,m dT  2  C PB ,m dT   C PC ,m dT 
298.15
298.15
298.15
 298.15

T
 ΔC
 ΔH reaction, 298.15K 
P
(T' )dT '
298.15
where ΔC P (T' )   n j C Pj ,m (T ' )
k
Q3.
The path for the reaction is given in the following 4 steps:
(1) 2B + C (at P)  (2) 2B+C (at 1 bar)  (3) X+3Y ( at 1 bar)
 (4) X+3Y (at P)
For the the change of the state from (1) to (2), 2 moles of B and 1 mole of C is subject to pressure
change from P to 1 bar at 298.15 K. Thus, the enthalpy change in this process is given by
ΔH(1  2) 
1 bar
P
P
1 bar
 -(2C P,B,m  C P,C ,m )  J T dP'    -(2C P,B,m  C P,C ,m )  J T dP'  (2C P ,B,m  C P,C ,m ) J T P
For the change of the state from (2) to (3), the reaction takes place at the standard condition (T
=298.15K and 1 bar) as 2B + C  X + 3Y. The reaction enthalpy at the standard condition is
given by
ΔH(2  3)
 ΔH 0 reaction, 298.15K
0
,
0
0
0
 H f (X)  3H f (Y)  2 H f (B)  H f (C)
Now, our system is made of 1 mole of X and 3 mole of Y (X + 3Y) at T = 298.15K.
Then, for a change of the state from (3) to (4),
P
ΔH(3  4) 
 -(C
P , X ,m
 3C P ,Y ,m )  J T dP '  (C P , X ,m  3C P ,Y ,m )  J T P
1bar
ΔH reaction, T  ΔH(1  2)  ΔH(2  3)  ΔH(3  4)
0
0
0
0
 H f (X)  3H f (Y)  2 H f (B)  H f (C) 


 C PX ,m  3C PY ,m  2C PB ,m  C PC ,m  J T P
 ΔH reaction, 298.15K  ΔC P (298.15K)  J T P
where ΔC P (T' )   n j C Pj ,m (T ' )
k