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211
1
O (g) ⎯⎯
→ O(g) ;
2 2
aH O 2 = 249.0kJ mol–1
Solution :
180g
heat
heat
ice ⎯⎯⎯⎯
→ liquid ⎯⎯⎯
→ vapours
(m.c.Δt)
H 2 O(g) ⎯⎯
→ 2H(g) + O(g) ;
∴
ΔH = (–44.0 + 285.8 + 436.0 +
249.0)KJ mol–1
ΔH = 926.8k]
H–O–H molecule contains two (O–H)
bonds
926.8
Bond enthalpy of O–H bonds =
=
2
463.4 kJ mol–1 ≈ Bond enthalpy of
O – H bond = 4634 kJmol–1 in H2O.
0ºC
100ºC
100ºC
180 × 4.18 × 100
1000
= 75.24 kJ ...(iii)
Total heat required
60.1 kJ + 407 kJ + 75.24 kJ
542.34 kJ.
Total heat required = 542.34 kJ.
q = m.c. Δt =
∴
=
=
*16.
6.24 g of ethanol are vaporized by supplying
*15. Calculate the total heat required to melt
5.89 kJ of heat energy. What is enthalpy
180 g of ice at 0ºC, heat it to 100ºC and
of vaporization of ethanol?
then vaporize it at that temperature. Given :
Δ fusH(ice) = 6.01 kJ mol –1 at 0ºC,
ΔHº = 5.89kJ (Heat of vaporisation),
–1
ΔvapH(H2O) = 40.7 kJ mol at 100ºC.
mass of ethanol = 6.24g
–1 –1
Specific heat of water = 4.18 J g K . To find :
Given :
ΔvapH = ?
(i) number of moles of (ice) H2O(s)
Solution :
gram molecular weight of C2H5OH
Wt
180
=
=
= 10 moles.
= 2 × 12 + 1 × 5 + 16 + 1 = 46g
Mol.Wt.
18
∴ 6.24 g ethanol are vaporised
(ii) ΔfusH (ice) = 6.01 kJ–1 at 0ºC
= 5.89 kJ of heat energy
∴ For 10 moles, ΔfusH ice = 60.1 J at 0ºC
∴ 46g of ethanol are vaporised
...(i)
46 × 5.89
(iii) ΔvapH (H2O) = 40.7 kJ mol–1 at 100º C
=
6.24
∴ For 10 moles, ΔvapH(H2O) = 407 kJ
...(ii)
= 43.5 kJ mol–1
(iv) Heat absorbed to raise the temperature
46g of ethanol are vaporised
from 0ºC to 100ºC
= 43.5 kJ mol –1.
–1
–1
sp. heat of water = 4.18 Jg K
To find :
Total heat required = ?
Chapter - 3 Chemical Thermodynamics And Energetics