Question

A scientist measures the standard enthalpy change for this reaction to be - 172.0 kJ/mol: HCl(g) + NH3(g) →NH CH(s) Based on
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Answer #1

Given:

ΔHo rxn = -172.0 KJ/mol

Hof(NH3(g)) = -46.11 KJ/mol

Hof(NH4Cl(s)) = -314.43 KJ/mol

Balanced chemical equation is:

HCl(g) + NH3(g) ---> NH4Cl(s)

ΔHo rxn = 1*Hof(NH4Cl(s)) - 1*Hof( HCl(g)) - 1*Hof(NH3(g))

-172.0 = 1*(-314.43) - 1*Hof(HCl(g)) - 1*(-46.11)

Hof(HCl(g)) = -96.32 KJ/mol

Answer: -96.32 KJ/mol

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