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```218
= 10.428 Ã 103 K
= 10428 K
As both Î HÂº and Î SÂº are negative
therefore, the reaction will be
spontaneous below 10428K.
Determine whether the following reaction
is spontaneous under standard conditions.
2H 2 O(l ) + O 2 (g) â¯â¯
â 2H 2 O 2 (l ) ÎHÂº
= +196 kJ, ÎSÂº = â126 JKâ1
Does it have a cross-over temperature?
Given :
ÎHÂº = +196 kJ, ÎSÂº = â126o JKâ1
To find :
State the reaction condition = ?
Solution :
As ÎHÂº is positive and ÎSÂº is negative
therefore, the reaction will be
nonspontaneous at all temperatures.
No, it has no cross over temperature.
=
â´
*28.
*29.
What is the value of ÎSsurr for the following
reaction at 298 K?
â C6H12O6(s) +
6CO2(g) + 6H2O(l) â¯â¯
6O2(g), ÎGÂº = 2879 kJ molâ1, ÎSÂº = â210
JKâ1 molâ1.
Given :
ÎGÂº = 2879 kJmolâ1,
ÎSÂº = â210 JKâ1 molâ1
= â0.210 kJ Kâ1molâ1
T
= 298 K
To find :
ÎSsurr = ?
Solution :
ÎGÂº = ÎHÂº â T â ÎSÂº
2879 kJmolâ1
= ÎHÂº â 298 K Ã (â0.210kJ Kâ1 molâ1)
Unique Solutions Â®
ÎHÂº + 62.58 kJ molâ1
ÎHÂº = (2879 â 62.58) kJ = 2816.42 kJ.
ÎSsurr = â
ÎHÂº
T
ÎSsurr = â
2816.42
= â9.45 kJ Kâ1
298
Î Ssurr = â9.45 kJ K â1
Calculate ÎSsurr when one mole of methanol
(CH3OH) is formed from its elements
under standard conditions if Î f HÂº
(CH3OH) = â238.9 kJmolâ1
Given :
The heat evolved in the reaction is 23.89
kJ. The same quantity of heat is absorbed
by the surroundings. Hence, entropy
change for the surroundings is given by
T = 25ÂºC = 289K
To find :
ÎSsurr = ?
Solution :
*30.
ÎSsurr = â
â238.9 (kJ)
ÎHÂº
=â
298 (K)
T
= 0.8017 kJ Kâ1
= 801.7 JKâ1
Î Ssurr = 801.7 JKâ1
*31.
GIBBS FREE ENERGY
Calculate K p for the reaction,
C 2 H 4 (g) + H 2 (g) â¯â¯
â C 2 H 6 (g) , Î GÂº
= â100 kJ molâ1 at 25ÂºC.
Given :
ÎGÂº = â100kJmolâ1 = â100000 Jmolâ1
T = 25ÂºC + 273K = 298K
S.Y.J.C. Science - Chemistry - Part I
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