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Calculate ΔH∘ in kilojoules for the reaction of ammonia NH3 (ΔH∘f=−46.1kJ/mol) with O2 to yield nitric oxide (NO) (ΔH∘f=91.3 kJ/mol) and H2O(g) (ΔH∘f=−241.8kJ/mol), a step in the Ostwald process for the commercial production of nitric acid.


4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

Verified Answer :

Answer: \(\Delta H^o\) for the given reaction is -901.2 kJ.

Explanation: Enthalpy of the reaction is the amount of heat released or absorbed in a given chemical reaction.

Mathematically,

\(\Delta H_{rxn}=\Delta H_f_{(products)}-\Delta H_f_{(reactants)}\)

For  given reaction:

\(4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)\)

\(H_f_{(NH_3)}=-46.1kJ/mol\)

\(H_f_{(O_2)}=0kJ/mol\)

\(H_f_{(H_2O)}=-241.8kJ/mol\)

\(H_f_{(NO)}=-91.3kJ/mol\)

\(\Delta H_{rxn}=[6\Delta H_f_{(H_2O)}+4\Delta H_f_{(H_2O)}-[4\Delta H_f_{(NH_3)}+5\Delta H_f_{(O_2)}]\)

Putting values in above equation, we get:

\(\Delta H_{rxn}=[6mol(-241.8kJ/mol)+4mol(91.3kJ/mol)]-[4mol(-46.1kJ/mol)+5mol(0kJ/mol)]\)

\(\Delta H_{rxn}=-901.2kJ\)

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