a)
ΔHo = 572.0 KJ
ΔSo = 179 J/K
= 0.179 KJ/K
use:
ΔGo = ΔHo - T*ΔSo
for reaction to be spontaneous, ΔGo should be negative
that is ΔGo<0
since ΔGo = ΔHo - T*ΔSo
so, ΔHo - T*ΔSo < 0
572.0- T *0.179 < 0
T *0.179 > 572.0
T > 3196 K
Answer: at high temperatures
b)
ΔHo = 286.0 KJ
ΔSo = -137 J/K
= -0.137 KJ/K
use:
ΔGo = ΔHo - T*ΔSo
for reaction to be spontaneous, ΔGo should be negative
that is ΔGo<0
since ΔGo = ΔHo - T*ΔSo
so, ΔHo - T*ΔSo < 0
286.0- T *-0.137 < 0
T *0.137 < -286.0
T < -2087.5912 K
Since kelvin temperature can't be negative
The reaction will never be spontaneous
Answer: at no temperatures
c)
ΔHo = -1274.0 KJ
ΔSo = 180 J/K
= 0.18 KJ/K
use:
ΔGo = ΔHo - T*ΔSo
for reaction to be spontaneous, ΔGo should be negative
that is ΔGo<0
since ΔGo = ΔHo - T*ΔSo
so, ΔHo - T*ΔSo < 0
-1274.0- T *0.18 < 0
T *0.18 > -1274.0
T > -7077.7778 K
Since kelvin temperature can't be negative
The reaction is spontaneous at all temperatures
The reaction will always be spontaneous
Answer: at all temperatures
d)
ΔHo = -844.0 KJ
ΔSo = -165 J/K
= -0.165 KJ/K
use:
ΔGo = ΔHo - T*ΔSo
for reaction to be spontaneous, ΔGo should be negative
that is ΔGo<0
since ΔGo = ΔHo - T*ΔSo
so, ΔHo - T*ΔSo < 0
-844.0- T *-0.165 < 0
T *0.165 < 844.0
T < 5115 K
Answer: at low temperatures
Only 1 question at a time please
Tedict and explain whether the follo ind explain whether the following reactions will be nonspontaneous at all temp...
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