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Ammonia, NH3 (Delta.Hf = –46.2 kJ), reacts with oxygen to produce water (Delta.Hf = –241.8 kJ) and nitric oxide, NO (Delta.Hf = 91.3 kJ), in the following reaction: 4 upper N upper H subscript 3 (g) plus 5 upper O subscript 2 (g) right arrow 6 upper H subscript 2 upper O (g) plus 4 upper N upper O (g). What is the enthalpy change for this reaction? Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants.

Respuesta :

Answer: The enthalpy of reaction is -900.8 kJ

Explanation:

The chemical equation is as follows:

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

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[(6\times \Delta H^o_f_{(H_2O(g))})+(4\times \Delta H^o_f_{(NO(g))})]-[(4\times \Delta H^o_f_{(NH_3(g))})+(5\times \Delta H^o_f_{(O_2(g))})][/tex]

We are given:

[tex]\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(NH_3(g))}=-46.2kJ/mol\\\Delta H^o_{(NO(g))}=91.3kJ[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(6\times (-241.8))+(4\times (91.3))]-[(4\times (-46.2))})+(5\times (0))][/tex]

[tex]\Delta H^o_{rxn}=-1085.6kJ-(-184.8)kJ=-900.8kJ[/tex]

The enthalpy of reaction is -900.8 kJ

JayDifferent

Answer:

A on edge

Explanation:

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