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Transcript
216
Given :
a)
2H 3BO 3 (aq) ⎯⎯
→ B 2 O 3 (s) + 3H 2O(l )
;
ΔHº = +14.4 kJ
b)
H 3BO 3 (aq) ⎯⎯
→ HBO 2 (aq) + H 2 O (l )
;
ΔHº = –0.02 kJ
;
ΔHº = 17.3 kJ
c) H 2 B 4 O 7 (s) ⎯⎯
→ 2B 2O 3 (s) + H 2O(l )
To find :
ΔHº(reaction) = ?
Solution :
Required equation : H 2 B 4 O 7 (s) + H 2O(l ) ⎯⎯
→ 4HBO 2 (aq)
Treatment : eq (iii) + 2 × reverse eq (i) + 4 ×eq (ii)
H 2 B 4O 7 (S) ⎯⎯
→ 2B 2O 3 (s) + H 2 O(l )
; ΔHº = 17.3 kJ
2B 2O 3 (s) + 6H 2 O(l ) ⎯⎯
→ 4H 3BO 3 (aq)
; ΔHº = –28.8 kJ
4 H 3BO 3 (aq) ⎯⎯
→ 4HBO 2 (aq) + 4H 2O(l ) ; ΔHº = –0.08 kJ
H 2 B 4 O 7 (s) + H 2O(l ) ⎯⎯
→ 4HBO 2 (aq)
Δ Hº = –11.58 kJ
*24.
; ΔHº = +17.3 kJ – 28.8 kJ – 0.08 kJ
ENTROPY
Calculate ΔStotal and hence, show whether the following reaction is spontaneous at
→ Hg(l ) + SO 2 (g) ;ΔHº = –238.6 kJ and ΔSº = +36.7 JK–1
25ºC. HgS (s) + O 2 (g) ⎯⎯
Given :
ΔHº = –238.6 kJ ΔSº = +36.7 JK–1
To find :
ΔS–total = ?
Solution :
The heat evolved in the reaction is 238.6 kJ. The same quantity of heat is absorbed by the
surroundings. Hence, entropy change of the surroundings is given by.
–(238.6) (kJ)
ΔHº
ΔSsurr = –
=
Log calculations
298 (K)
T
2.3777
ΔSsurr = +0.8007 kJ K–1
–1
–2.4742
= + 800.7 JK
ΔStotal =
ΔSsys =
ΔStotal =
ΔStotal =
Δ Stotal is
1.9035
ΔSsys + ΔSsurr
AL
–1
ΔSº = +36.7 JK
0.8007
36.7 (JK–1) + 800.7 (JK–1)
837.4 JK–1
positive, the reaction is spontaneous at 298 K.
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I