Question

the value of DrH0 for NH3 is -.8kj/mol. What will be the enthalpy change for the following reaction?

2NH3(g)- N2(g) + 3H2(G)

Answer 1

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N2(g) + 3H2(g) ------> 2NH3(g) ; ∆rH° = -92.4 KJ/mol

enthalpy of formation of NH3 means heat released in the formation of 1 mole of NH3 .

so, divide both sides by 2 in above reaction

1/2N2(g) + 3/2H2(g) ---> NH3(g) ;

Therefore , ∆fH° = ∆rH°/2 = -92.4/2 = 46.2 KJ/mol

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