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```211
1
O (g) â¯â¯
â O(g) ;
2 2
aH O 2 = 249.0kJ molâ1
Solution :
180g
heat
heat
ice â¯â¯â¯â¯
â liquid â¯â¯â¯
â vapours
(m.c.Ît)
H 2 O(g) â¯â¯
â 2H(g) + O(g) ;
â´
ÎH = (â44.0 + 285.8 + 436.0 +
249.0)KJ molâ1
ÎH = 926.8k]
HâOâH molecule contains two (OâH)
bonds
926.8
Bond enthalpy of OâH bonds =
=
2
463.4 kJ molâ1 â Bond enthalpy of
O â H bond = 4634 kJmolâ1 in H2O.
0ÂºC
100ÂºC
100ÂºC
180 Ã 4.18 Ã 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. Ît =
â´
=
=
*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 0ÂºC, heat it to 100ÂºC and
of vaporization of ethanol?
then vaporize it at that temperature. Given :
Î fusH(ice) = 6.01 kJ mol â1 at 0ÂºC,
ÎHÂº = 5.89kJ (Heat of vaporisation),
â1
ÎvapH(H2O) = 40.7 kJ mol at 100ÂºC.
mass of ethanol = 6.24g
â1 â1
Specific heat of water = 4.18 J g K . To find :
Given :
ÎvapH = ?
(i) number of moles of (ice) H2O(s)
Solution :
gram molecular weight of C2H5OH
Wt
180
=
=
= 10 moles.
= 2 Ã 12 + 1 Ã 5 + 16 + 1 = 46g
Mol.Wt.
18
â´ 6.24 g ethanol are vaporised
(ii) ÎfusH (ice) = 6.01 kJâ1 at 0ÂºC
= 5.89 kJ of heat energy
â´ For 10 moles, ÎfusH ice = 60.1 J at 0ÂºC
â´ 46g of ethanol are vaporised
...(i)
46 Ã 5.89
(iii) ÎvapH (H2O) = 40.7 kJ molâ1 at 100Âº C
=
6.24
â´ For 10 moles, ÎvapH(H2O) = 407 kJ
...(ii)
= 43.5 kJ molâ1
(iv) Heat absorbed to raise the temperature
46g of ethanol are vaporised
from 0ÂºC to 100ÂºC
= 43.5 kJ mol â1.
â1
â1
sp. heat of water = 4.18 Jg K
To find :
Total heat required = ?
Chapter - 3 Chemical Thermodynamics And Energetics
```
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