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Use the following reactions to find H when 1 mole of HCl gas forms from its elements: N2(g) + 3 H2(g) 2 NH3(g)  ...

Use the following reactions to find \DeltaH when 1 mole of HCl gas forms from its elements:

N2(g) + 3 H2(g) \longrightarrow 2 NH3(g)   LaTeX: \DeltaH = LaTeX: -91.8 kJ

N2(g) + 4 H2(g) + Cl2(g) \longrightarrow 2 NH4Cl(s) \DeltaH = LaTeX: -628.8 kJ

NH3(g) + HCl(g) \longrightarrow NH4Cl(s) \DeltaH = LaTeX: -176.2 kJ

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Answer #1

we need:

1/2H2(g) + 1/2Cl2(g) = HCl(g)

therefore...

invert rxn 3

NH4Cl = NH3 + HCl H = 176.2

divide rxn2 so we can cancel NH4Cl

1/2N2 + 2H2 +1/2Cl2 = NH4Cl H = -628.8/2 = - 314.4

add previous 2 reactions

1/2N2 + 2H2 +1/2Cl2 + NH4Cl = NH3 + HCl + NH4Cl H = 176.2-314.4 = - 138.2

cancel common terms

1/2N2 + 2H2 +1/2Cl2 = NH3 + HCl H = - 138.2

now.. invert and divide by 2 rxn 1

NH3 = 1/2N2 + 3/2H2 H= -1/2*(-91.8) = 45.9

now add previous 2

NH3 + 1/2N2 + 2H2 +1/2Cl2 = NH3 + HCl +  1/2N2 + 3/2H2 H = - 138.2+45.9

cancel commont erms

1/2H2 +1/2Cl2 = HCl + H = -92.3 kJ/mol

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