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Pure acetic acid is often called glacial acetic acid because it is purified by fractional freezing at its melting point of 16.6 °C. At 1 atm, a flask containing several moles of acetic acid at 16.6 °C is briefly lowered into an infinitely large ice-water bath. When it is removed it is found that exactly 1 mol of acetic acid has frozen. Given: ΔH_fusion,acetic acid = 11.45kJ/mol and ΔH_fusion, H2O= 6.01 kJ/mol.

Required:
a. What is the change in entropy of the acetic acid?
b. What is the change in entropy of the water bath?
c. Now consider that the water bath and acetic acid are the same system. What is the entropy change for the combined system? Is the process reversible or irreversible? Why?

Sagot :

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 .

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