Look at the reaction: A (solid) <-> B (liquid) + C (liquid), in constant pressur

Question

Look at the reaction: A (solid) <-> B (liquid) + C (liquid), in constant pressure and temperature. Assume that the liquids cannot enter the solid (i.e., it's pure). Say that the molar free energy of A is 10 kJ/mole, while the chemical potentials of B and C have the following n(B), n(C) dependence:

?(B) = n(B) * 5 kJ/mole^2 + 1 kJ/mole
?(C) = n(C) * 5 kJ/mole^2

(i) Verify that the dimensions of ?(B), ?(C) are correct.
(ii) Obtain the total free-energy as a function of n(B),n(C),n(A).
(iii) Obtain the equilibrium conditions using the appropiate equations on the ?'s.
(iv) If we start the reaction with 1/2 mole of A, and no B or C, would the system be in material equilibrium? After letting it evolve, would any A remain? How much electrical work can we obtain from the reaction under such a condition?
(v) Like (iv), for starting with 20 moles of solid A.

Explanation / Answer

i) dimensions of mu = G/n=energy/mole

here mu(A)= Kj/mole=[energy]/mole

mu(B)= mole*Kj/mole^(2) +Kj/mole= Kj/mole=[energy]/[mole]

ii)total free energy =summation of n*mu = n(A)*10 + n(B)^(2)*5+n(B)+n(C)^(2)*5

iii)for equilibrium

mu(A)=mu(B)+mu(C)

n(A)*10 = n(B)^(2)*5+n(B)+n(C)^(2)*5

iv)no it will not be in equilibrium

x*10 =(0.5-x)^(2)*5 +(0.5-x)+(0.5-x)^(2)*5

gives x=0.15417 so this much of A will remain

electrical work =DelG=N(A)*mu(A)=5Kj

v)

solving

10*x=(20-x)^(2)*5 +(20-x) +(20-x)^(2)*5

gives x=16.044 so this much of A will remain

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