What is the value of ∆H˚(J) for ammonium chloride? NH3(g)+HCl(g)––>NH4Cl(s)?
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NH3(g)+HCl(g)––>NH4Cl(s) given the thermodynamically data of 298˚K ∆Hf˚(kJ/mol) S˚(J/K•mol) NH3(g): -46.19 192.5 HCl(g): -92.30 186.69 NH4Cl(s): -314.4 94.6
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Answer:
dH = (sum of reactants) - (sum of products) All ratios are 1/1 and moles cancel so your answer will be in kJ so multiply by 1000 J.
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