Ca(OH)2 (aq) + 2 NH4Cl (s) → CaCl2 (aq) + 2 H2O (ℓ) + 2 NH3 (g)

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Reaction Type:

Double Displacement/Decomposition of Ammonium Hydroxide

Stoichiometry

Enter a mass or volume in one of the boxes below. Upon hitting submit, the stoichiometric equivalents will be calculated for the remaining reactants and products. All gases are assumed to be at STP.

Ca(OH)2        Mass: g  or Solution Volume: mL of Concentration: mol/L
NH4Cl          Mass: g
CaCl2          Mass: g  or Solution Volume: mL of Concentration: mol/L
H2O            Mass: g
NH3            Mass: g  or Gas Volume: L

Enthalpy of Reaction

[1ΔHf(CaCl2 (aq)) + 2ΔHf(H2O (ℓ)) + 2ΔHf(NH3 (g))] - [1ΔHf(Ca(OH)2 (aq)) + 2ΔHf(NH4Cl (s))]
[1(-877.13) + 2(-285.83) + 2(-46.11)] - [1(-1002.81) + 2(-314.43)] = 90.6600000000001 kJ
90.66 kJ     (endothermic)

Entropy Change

[1ΔSf(CaCl2 (aq)) + 2ΔSf(H2O (ℓ)) + 2ΔSf(NH3 (g))] - [1ΔSf(Ca(OH)2 (aq)) + 2ΔSf(NH4Cl (s))]
[1(59.82) + 2(69.91) + 2(192.34)] - [1(-74.64) + 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(CaCl2 (aq)) + 2ΔGf(H2O (ℓ)) + 2ΔGf(NH3 (g))] - [1ΔGf(Ca(OH)2 (aq)) + 2ΔGf(NH4Cl (s))]
[1(-816.04) + 2(-237.18) + 2(-16.48)] - [1(-868.1) + 2(-202.97)] = -49.3200000000002 kJ
-49.32 kJ     (spontaneous)

From ΔG = ΔH - TΔS:
-49.42 kJ     (spontaneous)

Equilibrium Constant, K (at 298.15 K)

437492845.956168
This process is favorable at 25°C.