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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)

Respuesta :

Answer: [tex]\Delta H^o[/tex] 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,

[tex]\Delta H_{rxn}=\Delta H_f_{(products)}-\Delta H_f_{(reactants)}[/tex]

For  given reaction:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

[tex]H_f_{(NH_3)}=-46.1kJ/mol[/tex]

[tex]H_f_{(O_2)}=0kJ/mol[/tex]

[tex]H_f_{(H_2O)}=-241.8kJ/mol[/tex]

[tex]H_f_{(NO)}=-91.3kJ/mol[/tex]

[tex]\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)}][/tex]

Putting values in above equation, we get:

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

[tex]\Delta H_{rxn}=-901.2kJ[/tex]

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