Engineering Chemistry S Datta

94 ENGINEERING CHEMISTRY
Sol. Given,
(i) C 2
H 4
+ 3O 2
⎯⎯→ 2CO 2
+ 2H 2
O,
(ii) H 2
+ 1 2 O 2 ⎯⎯→ H 2 O,
∆H = – 337.2 Kcal.
∆H = 68.3 Kcal.
(iii) C 2
H 6
+ 7 2 O 2 ⎯⎯→ 2CO 2 + 3H 2 O,
∆H = – 372.8 Kcal
or, (iv) 2CO 2
+ 3H 2
O ⎯⎯→ C 2
H 6
+ 7 2 O 2
, ∆H = + 372.8 Kcal.
Adding the equations (i), (ii) and (iv) we have,
C 2
H 4
+ 3O 2
+ H 2
+ 1 2 O 2 + 2CO 2 + 3H 2 O ⎯⎯→ 2CO 2 + 2H 2 O + H 2 O + C 2 H 6 + 7 2 O 2
∴ C 2
H 4
+ H 2
→ C 2
H 6
, ∆H = – 32.7 Kcal.
(∆H = – 32.7 Kcal)
Example 7. Calculate O—H bond energy from the following heats of reactions.
H 2
(g) → 2H(g), ∆H = 104 Kcal ...(i)
O 2
(g) → 2O(g), ∆H = 120 Kcal ...(ii)
H 2
(g) + 1 2 O 2 (g) → H 2O(g), ∆H = – 58 Kcal ...(iii)
Sol. From equation (ii)
1
2 O 2
(g) → O(g), ∆H = 60 Kcal.
From equation (iii), H 2
O(g) → H 2
(g) + 1 2 O 2 (g)
∆H = 58 Kcal.
(i) + (ii) – (iii) gives = H 2
(g) + 1 2 O 2 (g) + H 2 O(g) → 2H(g) + O(g) + H 2 (g) + 1 2 O 2 (g)
i.e., H 2
O(g) → 2H(g) + O(g), ∆H = 222 Kcal.
We know that, there are two O—H bonds in a molecule of water, and for breaking of
these two bonds energy requirement is 222 Kcal.
So, for dissociation of one O—H bond energy requirement is 222 ÷ 2 = 111 Kcal.
∴ O—H bond energy is 111 Kcal.
Example 8. Calculate the enthalpy of formation of propane (C 3
H 8
) at 298 K. Given the
standard enthalpies of combustion:
∆H C
°
(C 3
H 8
) = – 2220 kJ mol –1
∆H C
° (C) = – 393 kJ mol –1
∆H C
°
(H 2
) = – 286 kJ mol –1 .