ThermodynamicsMediumBloom L3
Question
In conversion of lime-stone to lime,
CaCO3(s) → CaO(s) + CO2 (g)
the vales of ᐃHo and ᐃSo are + 179.1 kJ mol-1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ᐃHo do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is
CaCO3(s) → CaO(s) + CO2 (g)
the vales of ᐃHo and ᐃSo are + 179.1 kJ mol-1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ᐃHo do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is
Options
A.1008 K
B.1200
C.845 K
D.1118 K
Solution
We know, ᐃG = ᐃH - TᐃS
So, lets find the equilibrium temperature, i.e. at which
ᐃG = 0
ᐃH = TᐃS
T =
= 1118 K
So, at temperature above this, the reaction will become spontaneous.
Hence, (4) is correct answer.
So, lets find the equilibrium temperature, i.e. at which
ᐃG = 0
ᐃH = TᐃS
T =
= 1118 KSo, at temperature above this, the reaction will become spontaneous.
Hence, (4) is correct answer.
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