Question

Calculate the free energy change for the dissolution in water of four moles of NaBr and four moles of Nal at 298 K, given that the values of ΔΗ sin are-1 and-7 k/mol respectively, for NaBr and Nal. The corresponding values for AS soin are +57 and +74 /mol-K. ΔG° for NaBr -17.986 AGe for Nal 58.104

Answer 1

Question :

PART A

First we'll solve for NaBr -

We've,

∆H°(sol) = -1 kJ/mol

​​​​​​But, we've to calculate free energy change for dissolution of 4 moles of NaBr. Thus, ∆H°(sol) for 4 moles of NaBr = -4 kJ

Also,

∆S°(sol) = 57 J/mol.K

Thus, for 4 moles of NaBr,

∆S°(sol) = 4 moles × 57 J/mol.K

= 228 J/K or, 0.228 kJ/K

Also,

T = 298 K

Now,

∆G° = ∆H° - T∆S

= - 4 kJ - (298 K × 0.228 kJ/K)

= - 4 kJ - ( 67.944 kJ)

= -71.944 kJ (Answer) (part A)

(PART B) Now, we'll solve for NaI

We've,

∆H°(sol) = -7 kJ/mol

But, we've to calculate free energy change of 4 moles of NaI,

Thus,

∆H°(sol) = -7 kJ/mol × 4moles

= - 28 kJ

Also,

∆S°(sol) = 74 J/mol.K,

Thus for 4 moles,

∆S°(sol) = 4 moles × 74 J/mol.K

= 296 J/K

OR, = 0.296 kJ/K

Thus, applying formula:

∆G° = ∆H° - T∆S°

= - 28 kJ - (298 K × 0.296 kJ/K)

= - 28 kJ - (88.208 kJ)

= -116.208 kJ ( Answer ) (Part B)

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