TITLE.PM5

534 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th11-2.pm5

C 6 H 6 + 7.5O 2 → 6CO 2 + 3H 2 O (vapour)

nR = 1 + 7.5 = 8.5, nP = 6 + 3 = 9

Using the relation,

∆U 0 = ∆H 0 – (nP – nR)R 0 T 0

= – 3169100 – (9 – 8.5) × 8.314 × (25 + 273)

= – 3169100 – 1239 = – 3170339 kJ/mole

(It may be noted that ∆U 0 is negligibly different from ∆H 0 )

1 mole of C 6 H 6 = 6 × 12 + 1 × 6 = 78 kg

∴∆U 0 =
− 3170339
78
= – 40645 kJ/kg. (Ans.)
Example 11.33. ∆H 0 for CO at 60°C is given as – 285200 kJ/mole. Calculate ∆H 0 at 2500°C
given that the enthalpies of gases concerned in kJ/mole are as follows :
Gas 60°C 2500°C
CO 9705 94080
O 9696 99790
CO 2 10760 149100
Solution. The reaction equation is given by
CO +^12 O 2 → CO 2
Refer Fig. 11.7.
It can be seen from the property diagram of Fig. 11.7
that the enthalpy of combustion at temperature T, ∆HT can be
obtained from ∆H 0 and T 0 by the relationship

• ∆HT = – ∆H 0 + ()HHRRT− 0 – ()HHPPT− 0 ...(i)

where HHRRT0− = increase in enthalpy of the reactants
from T 0 to T
and HHPPT0− = increase in enthalpy of the products from
T 0 to T.
Now, from the given data, we have
HR 0 = 1 × 9705 +^12 × 9696 = 14553 kJ
HRT = 1 × 94080 +^12 × 99790 = 143975, kJ
HP 0 = 1 × 10760 = 10760 kJ
HPT = 1 × 149100 kJ = 149100 kJ
Using equation (i), we get

• ∆HT = + 285200 + (143975 – 14553) – (149100 – 10760)
  = 285200 + 129422 – 138340 = 276282
  ∴∆HT = – 276282 kJ/mole. (Ans.)

Fig. 11.7

T

H– HPPT 0

Products

Reactants

H

H– HRRT O

T 0 T

• ∆H 0