A certain reaction is spontaneous at temperatures below 400. K but is not spontaneous at temperatures above 400. K. If ΔH° for t
he reaction is -20. kJ mol-1 and it is assumed that ΔH° and ΔS° do not change appreciably with temperature, then the value of ΔS° for the reaction is
1 answer:
Answer:
ΔS = 0.05 kJ /mol or 50 J /mol
Explanation:
As given that the reaction is spontaneous below 400 K and non spontaneous above 400K. Thus the reaction is at equilibrium at 400 K temperature.
At equilibrium the change in free energy ΔG is zero.
The relation between free energy change, enthalpy change, entropy change and temperature is
ΔG = ΔH - TΔS
Where
ΔH = enthalpy change
T = temperature
ΔS = entropy change
at equilibrium (400 K)
ΔG = 0 = (-20) + 400XΔS
ΔS = 0.05 kJ /mol or 50 J /mol
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