Ba(OH)2 (aq) + 2 NH4NO3 (s) → Ba(NO3)2 (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.

Ba(OH)2        Mass: g
NH4NO3         Mass: g
Ba(NO3)2       Mass: g
H2O            Mass: g
NH3            Mass: g  Volume: L

Enthalpy of Reaction

[1ΔHf(Ba(NO3)2 (aq)) + 2ΔHf(H2O (l)) + 2ΔHf(NH3 (g))] - [1ΔHf(Ba(OH)2 (aq)) + 2ΔHf(NH4NO3 (s))]
[1(-952.32) + 2(-285.83) + 2(-46.11)] - [1(-997.58) + 2(-365.56)] = 112.5 kJ
112.50 kJ     (endothermic)

Entropy Change

[1ΔSf(Ba(NO3)2 (aq)) + 2ΔSf(H2O (l)) + 2ΔSf(NH3 (g))] - [1ΔSf(Ba(OH)2 (aq)) + 2ΔSf(NH4NO3 (s))]
[1(302.48) + 2(69.91) + 2(192.34)] - [1(-11.9) + 2(151.08)] = 536.72 J/K
536.72 J/K     (increase in entropy)

Free Energy of Reaction (at 298.15 K)

From ΔGf° values:
[1ΔGf(Ba(NO3)2 (aq)) + 2ΔGf(H2O (l)) + 2ΔGf(NH3 (g))] - [1ΔGf(Ba(OH)2 (aq)) + 2ΔGf(NH4NO3 (s))]
[1(-783.48) + 2(-237.18) + 2(-16.48)] - [1(-875.36) + 2(-184.01)] = -47.4200000000003 kJ
-47.42 kJ     (spontaneous)

From ΔG = ΔH - TΔS:
-47.52 kJ     (spontaneous)

Equilibrium Constant, K (at 298.15 K)

203276341.264123
When K = 1, the reaction is at equilibrium. The larger K is, the more product favored the reaction is at this temperature. The smaller K is, the more reactant favored the reaction is at this temperature. K is calculated from Grxn = -RTlnK, where R = 8.314 J K-1 mol-1.