Question

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

Answer 1

Answer: -134 kJ

Step-by-step explanation:

The balanced chemical reaction is,

`3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq)+NO(g)`

The expression for enthalpy change is,

`\Delta H=\sum [n* \Delta H_f(product)]-\sum [n* \Delta H_f(reactant)]`

`\Delta H=[(n_(HNO_3)* \Delta H_(HNO_3))+(n_(NO)* \Delta H_(NO))]-[(n_(H_2O)* \Delta H_(H_2O))+(n_(NO_2)* \Delta H_(NO_2))]`

where,

n = number of moles

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

`\Delta H=[(2* -207)+(1* 90)]-[(1* -286)+(3* 32)]`

`\Delta H=-134kJ`

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