Question

Which of the following has the correct change in enthalpy and reaction description for the following reaction? NH3(g) + HCl(g) NH4Cl(s) Given: NH3: ∆H= -46.2 kJ HCl: ∆H= -92.3 kJ NH4Cl: ∆H= -314.43 kJ

Answer 1

 2HCl -----> H2 + Cl2 dH = +184.6kJ/mol (equation 1 reversed) 
2NH3 ------> 3H2 + N2 dH = +92.20kJ/mol ( equation 2 unchanged) 
N2 + 4H2 + Cl2 ------> 2NH4Cl dH = - 628.8kJ/mol (equation 3 doubled) 
--------------------------------------... 
Adding up: 2HCl + 2NH3 -----> 2NHCl4 dH = -352kJ/mol 
Hence HCl + NH3 --------> NH4Cl dH = -352/2 = - 176kJ/mol.

Answer 2

to find the change in enthalpy (ΔH), all you have to do is use the following formula:

ΔH reaction= [ΔH of products] - [ΔH of reactants]

remember that the products are the ones in the right side of the reaction, and reactants are in the left side. 

ΔH reaction= [ΔH NH4Cl] - [ΔH NH3 + ΔH HCl]

ΔH reaction= [-314.43] - [ -46.2 + -92.3]= -175.93 kJ