Question

Enthalpy of solution (ΔH) for BaCl2 . 2H2O & BaCl2 are 8.8 & -20.6kJ mol-1 reapectively. Calculate the heat of hydration of BaCl2 to BaCl2 . 2H2O................

Answer 1

ENTHALPY OF REACTION

[1ΔHf(BaCl2 (aq)) + 2ΔHf(H2O (ℓ)) + 2ΔHf(NH3 (g))] - [1ΔHf(Ba(OH)2 (aq)) + 2ΔHf(NH4Cl (s))]
[1(-871.9) + 2(-285.83) + 2(-46.11)] - [1(-997.58) + 2(-314.43)] = 90.6600000000001 kJ
90.66 kJ     (endothermic)

ENTROPY CHANGE

[1ΔSf(BaCl2 (aq)) + 2ΔSf(H2O (ℓ)) + 2ΔSf(NH3 (g))] - [1ΔSf(Ba(OH)2 (aq)) + 2ΔSf(NH4Cl (s))]
[1(122.56) + 2(69.91) + 2(192.34)] - [1(-11.9) + 2(94.56)] = 469.84 J/K
469.84 J/K     (increase in entropy)

FREE ENERGY OF REACTION (AT 298.15 K)

From ΔGf° values:
[1ΔGf(BaCl2 (aq)) + 2ΔGf(H2O (ℓ)) + 2ΔGf(NH3 (g))] - [1ΔGf(Ba(OH)2 (aq)) + 2ΔGf(NH4Cl (s))]
[1(-823.3) + 2(-237.18) + 2(-16.48)] - [1(-875.36) + 2(-202.97)] = -49.3199999999999 kJ
-49.32 kJ     (spontaneous)
From ΔG = ΔH - TΔS:
-49.42 kJ     (spontaneous)

Answer 2

Answer:

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Explanation:

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