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Consider the following data for compound A: (delta)Hf(kj/mol) S(J/K*mol) A(s) -224.57 32.36 A(l) -172.45 109.55 Use...

Consider the following data for compound A:

(delta)Hf(kj/mol) S(J/K*mol)

A(s) -224.57 32.36

A(l) -172.45 109.55

Use these data to determine the normal freezing point of compound A. Report your answer to 3 significant figures and degrees Celsius)

Freezing point =

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Answer #1

we have the Balanced chemical equation as:

A(s) ---> A(l)

we have:

Hof(A(s)) = -224.57 KJ/mol

Hof(A(l)) = -172.45 KJ/mol

deltaHo rxn = 1*Hof(A(l)) - 1*Hof( A(s))

deltaHo rxn = 1*(-172.45) - 1*(-224.57)

deltaHo rxn = 52.12 KJ/mol

we have:

Sof(A(s)) = 32.36 J/mol.K

Sof(A(l)) = 109.55 J/mol.K

we have the Balanced chemical equation as:

A(s) ---> A(l)

deltaSo rxn = 1*Sof(A(l)) - 1*Sof( A(s))

deltaSo rxn = 1*(109.55) - 1*(32.36)

deltaSo rxn = 77.19 J/mol.K

deltaSo rxn = 0.07719 KJ/mol.K

FOR EQUILIBRIUM:

deltaGo rxn = 0

we have below equation to be used:

deltaGO = deltaHO - T*deltaSO

0.0 = 52.12 - T *0.07719

T = 675 K

= (675 - 273) oC

= 402 oC

Answer: 402 oC

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