Engineering Chemistry S Datta

THERMODYNAMICS 95
Sol. A thermodynamic cycle is drawn:
H 1
3C( s) + 4H
2( g) + 5O
2( g)
CH(
3 8
g)+5O( 2
g)
H 2
H 3
3CO(
2
g)+4HO( 2
l)
According to Hess’s law – ∆H 1
= ∆H 2
– ∆H 3
∆H 1
= ∆H f
° (C 3
H 8
)
∆H 2
= 3∆H C
° (C) + 4 × ∆H C ° (H 2
).
= 3 × (– 393) + 4 × (– 286) kJ mol –1
= – 2323 kJ mol –1
∆H 3
= ∆H C
°
= – 2220 kJ mol –1
Hence, ∆H f ° (C 3
H 8
) = (– 2323) – (– 2220) kJ mol –1
= – 103 kJ mol –1 .
Example 9. Calculate the enthalpy change of transition from yellow P to Red P from the
given thermochemical cycle.
Sol. Thermochemical cycle:
Yellow P
2386 cals
H 2
H ( H )
T 1
HPO
3 4
H 3
Red P
2113 cals
According to Hess’s law – ∆H T
= ∆H 2
– ∆H 3
= 2386 – 2113 cals = 273 cals.
Example 10. The latent heat of evaporation of water at 100°C at constant pressure is
538 cal g –1 . If average C p
for water and steam are 1 cal g –1 and 8.1 cal mol –1 , respectively for the
interval, calculate the latent heat of water at 150°C.
Sol. Basis = 1 g.
∆H 150°
=?; ∆H 100°
= 538
C p
for water = 1; C p
for steam = 8.1
18 .
∆C p
= 0.45 – 1 = – 0.55.
T 2
– T 1
= 423 – 373 = 50 K.
We have, ∆H 2
– ∆H 1
= ∆C p
(T 2
– T 1
)
∆H 150°
– ∆H 100°
= – 0.55(50)