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Using the standard enthalpies of formation for the chemicals involved, calculate the enthalpy change for the following reaction.

(note: show the math clearly and provide units in your set up) ( Hf values in kJ/mol are as follows: NO2 32, H2O 286, HNO3 207, NO 90.)

3NO2(g) H2O(l) 2HNO3(aq) NO(g) g

Respuesta :

Answer: -134 kJ

Explanation:

The balanced chemical reaction is,

[tex]3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq)+NO(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{HNO_3}\times \Delta H_{HNO_3})+(n_{NO}\times \Delta H_{NO})]-[(n_{H_2O}\times \Delta H_{H_2O})+(n_{NO_2}\times \Delta H_{NO_2})][/tex]

where,

n = number of moles

Now put all the given values in this expression, we get

[tex]\Delta H=[(2\times -207)+(1\times 90)]-[(1\times -286)+(3\times 32)][/tex]

[tex]\Delta H=-134kJ[/tex]

Therefore, the enthalpy change for this reaction is, -134 kJ

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