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Q. $\Delta H_{f}^{o}$ of $CO_{2}\left(\right.g\left.\right),CO\left(\right.g\left.\right),N_{2}O\left(\right.g\left.\right)$ and $NO_{2}\left(\right.g\left.\right)$ in kJ/mol are respectively $-393,-110,81$ and $34$ . Calculate the $\Delta H$ in kJ of the following reaction
$2NO_{2}\left(\right.g\left.\right)+3CO\left(\right.g\left.\right) \rightarrow N_{2}O\left(\right.g\left.\right)+3CO_{2}\left(\right.g\left.\right)$

NTA AbhyasNTA Abhyas 2022

Solution:

$\Delta H_{r x n}=1.\Delta H_{f}^{o}\left(\right.N_{2}O,g\left.\right)+3.\Delta H_{f}^{o}\left(\right.CO_{2},g\left.\right)-\left[\right.2\Delta H_{f}^{o}\left(\right.NO_{2},g\left.\right)+3\Delta H_{f}^{o}\left(\right.CO,g\left.\right)\left]\right.$
$=81kJ+3\times \, \left(- 393\right) \, kJ-2\left[\times 34 kJ + 3 \left(- 110 kJ\right)\right]$
$=81-1179-68+330 \, kJ/mol$
$=-836 \, kJ/mol$