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Given :
a)
2H 3BO 3 (aq) â¯â¯
â B 2 O 3 (s) + 3H 2O(l )
;
ÎHÂº = +14.4 kJ
b)
H 3BO 3 (aq) â¯â¯
â HBO 2 (aq) + H 2 O (l )
;
ÎHÂº = â0.02 kJ
;
ÎHÂº = 17.3 kJ
c) H 2 B 4 O 7 (s) â¯â¯
â 2B 2O 3 (s) + H 2O(l )
To find :
ÎHÂº(reaction) = ?
Solution :
Required equation : H 2 B 4 O 7 (s) + H 2O(l ) â¯â¯
â 4HBO 2 (aq)
Treatment : eq (iii) + 2 Ã reverse eq (i) + 4 Ãeq (ii)
H 2 B 4O 7 (S) â¯â¯
â 2B 2O 3 (s) + H 2 O(l )
; ÎHÂº = 17.3 kJ
2B 2O 3 (s) + 6H 2 O(l ) â¯â¯
â 4H 3BO 3 (aq)
; ÎHÂº = â28.8 kJ
4 H 3BO 3 (aq) â¯â¯
â 4HBO 2 (aq) + 4H 2O(l ) ; ÎHÂº = â0.08 kJ
H 2 B 4 O 7 (s) + H 2O(l ) â¯â¯
â 4HBO 2 (aq)
Î HÂº = â11.58 kJ
*24.
; ÎHÂº = +17.3 kJ â 28.8 kJ â 0.08 kJ
ENTROPY
Calculate ÎStotal and hence, show whether the following reaction is spontaneous at
â Hg(l ) + SO 2 (g) ;ÎHÂº = â238.6 kJ and ÎSÂº = +36.7 JKâ1
25ÂºC. HgS (s) + O 2 (g) â¯â¯
Given :
ÎHÂº = â238.6 kJ ÎSÂº = +36.7 JKâ1
To find :
ÎSâ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.
â(238.6) (kJ)
ÎHÂº
ÎSsurr = â
=
Log calculations
298 (K)
T
2.3777
ÎSsurr = +0.8007 kJ Kâ1
â1
â2.4742
= + 800.7 JK
ÎStotal =
ÎSsys =
ÎStotal =
ÎStotal =
Î Stotal is
1.9035
ÎSsys + ÎSsurr
AL
â1
ÎSÂº = +36.7 JK
0.8007
36.7 (JKâ1) + 800.7 (JKâ1)
837.4 JKâ1
positive, the reaction is spontaneous at 298 K.
Unique Solutions Â®
S.Y.J.C. Science - Chemistry - Part I
```
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