Question

Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen monoxide, given that ΔH°f of NO2(g) is 33.90 kJ/mol. Answers: 181.9 kJ/mol -35.64 kJ/mol 91.04 kJ/mol 148.0 kJ/mol -114.1 kJ/mol

Answer 1

Answer: ΔH°f of gaseous nitrogen monoxide is 148.0 kJ/mol

Step-by-step explanation:

The balanced chemical reaction is,

`NO(g)+(1)/(2)O_2(g)\rightarrow NO_2(g)`

The expression for enthalpy change is,

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

`\Delta H_(rxn)=[(n_(NO_2)* \Delta H_f_(NO_2))]-[(n_(O_2)* \Delta H_f_(O_2))+(n_(NO)* \Delta H_f_(NO))]`

where,

n = number of moles

`\Delta H_(O_2)=0` (as heat of formation of substances in their standard state is zero

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

`-114.14=[(1* 33.90)]-[((1)/(2)* 0)+(1* \Delta H_f{NO})]`

`\Delta H_f{NO}=148.0kJ/mol`

Therefore, ΔH°f of gaseous nitrogen monoxide is 148.0 kJ/mol