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  4. What will be the enthalpy of formation of NO2 from the given bond dissociation enthalpy values ? The bond dissociation enthalpy values for O2 , NO and NO2 are as follows, O2(g) : 0 kj/mol , NO(g) : 90.25 kj/mol and NO2(g) : 33.18 kj/mol respectively.

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Q. What will be the enthalpy of formation of $ NO_2 $ from the given bond dissociation enthalpy values ? The bond dissociation enthalpy values for $ O_2 $ , $ NO $ and $ NO_2 $ are as follows, $ O_{2(g)} $ : $ 0 \,kj/mol $ , $ NO_{(g)} $ : $ 90.25 \,kj/mol $ and $ NO_{2(g)} $ : $ 33.18 \,kj/mol $ respectively.

J & K CETJ & K CET 2018Thermodynamics

Solution:

$2 N O_{(g)}+O_{2(g)} \rightleftharpoons 2 N O_{2(g)}$
$\Delta H^{\circ}_{\text{reaction}} = \sum \Delta H_{f {\text(products)}}-\Delta H_{f {\text(reactants)}}$
$=\left(2\times33.18\right)-\left(2\times90.25+0\right)$
$\Delta H^{\circ}_{\text{reaction}}=-114.1\,KJ $
The values given are enthalpy of formation $\left(\Delta H_{f}^{\circ}\right)$