Answer :
Answer:
Explanation:
a ) ΔH_fusion of acetic acid is 11.45 kJ / mol . It means 11.45 kJ of heat will be released when 1 mole of acetic acid will freeze . During this process temperature of acetic acid will be constant .
Temperature of acetic acid T = 16.6⁰C = 273 + 16.6 K = 289.6 K .
By definition change in entropy
ΔS = dQ / T
= dH / T
As temperature remains constant during withdrawal of heat
ΔS = ΔH / T
T = 289.6 K .
ΔH = - 11.45kJ
ΔS = - 11.45kJ / 289.6
= - 11450 / 289.6
= - 39.53 J
It will be negative as heat is released by acetic acid .
b )
Heat amounting ΔH will be absorbed by water bath so its entropy will be increased . As water bath is very large , temperature of bath also will remain constant at 289.6 K .
Increase in entropy of water bath =
ΔS = ΔH / T
T = 289.6 K .
ΔH = 11.45kJ
ΔS = 11.45kJ / 289.6
= 11450 / 289.6
= 39.53 J
c )
consider that the water bath and acetic acid are the same system : --
Total change in the heat content of the system
= 289.6 - 289.6 = 0
So for the whole system
ΔQ =ΔH = 0
Change in entropy is zero .
Δ G = ΔH - TΔS
= ΔH - ΔH = 0
ΔG = 0
Change in entropy is zero .