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

Explanation:

The balanced chemical reaction is,

$NO(g)+\frac{1}{2}O_2(g)\rightarrow NO_2(g)$

The expression for enthalpy change is,

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H_{rxn}=[(n_{NO_2}\times \Delta H_f_{NO_2})]-[(n_{O_2}\times \Delta H_f_{O_2})+(n_{NO}\times \Delta H_f_{NO})]$

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

$-114.14=[(1\times 33.90)]-[(\frac{1}{2}\times 0)+(1\times \Delta H_f{NO})]$

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

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