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NH3 (g) ––� NH3 (aq) ΔH = �4640 J/mol
HCl (g) ––� HCl (aq) ΔH = �5140 J/mol
NH4Cl (s) ––� NH4Cl (aq) ΔH = -15428.82J/mol
NH3 (aq) + HCl (aq) � NH4Cl (aq) ΔH= - 44718.6J/mol
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Cancelling NH3(aq),HCl(aq),NH3(aq) and HCl(Aq) gives me:
NH3(g)+HCl(g)+NH4Cl(s)------->2[NH4Cl(aq)
Combination of Delta H=�4640 J/mol +(�5140 J/mol) + 2(-15428.8J/mol)+ 2(- 44718.6J/mol) =-199217.2J/mol
The answer is :
NH3(g)+HCl(g)+NH4Cl(s)------->2[NH4Cl(aq) DeltaH=-199217.2J/mo
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