1. Tardigrade
  2. Question
  3. Chemistry
  4. Given,NO(g) +O3 (g) → NO2(g) +O2 (g) Δ H1 =-198.9 KJ/mol O3( g)arrow 3 2 O2(g); Δ H =-142.3 KJ mol O2(g)arrow O2 (g); Δ H3 =+495.0 KJ mol The enthalpy change (Δ H) for the following reaction is NO(g)+O(g) arrow NO2(g)

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Q. Given,$NO\left(g\right) +O_{3} \left(g\right) \to NO_{2}\left(g\right) +O_{2} \left(g\right)$
$\Delta H_{1} =-198.9 KJ/$mol
$ O_{3}\left( g\right)\rightarrow 3 2 O_{2}\left(g\right); \Delta H =-142.3 KJ$ mol
$O_{2}\left(g\right)\rightarrow O2 \left(g\right); \Delta H_{3} =+495.0 KJ $mol
The enthalpy change $ \left(\Delta H\right)$ for the following reaction is
$NO\left(g\right)+O\left(g\right) \rightarrow NO_{2}\left(g\right)$

KVPYKVPY 2016Thermodynamics

Solution:

Given
$NO\left(g\right) +O_{3} \left(g\right) \to NO_{2}\left(g\right) +O_{2} \left(g\right)$
$\Delta H_{1} =-198.9\, KJ/$mol
$O_{3} \rightarrow \frac{3}{2} O_{2}\left(g\right) \Delta H_{2} $
$=-142.3\, KJ/$ mol
$O_{2}\rightarrow \frac{3}{2} O_{2} \left(g\right) \Delta H_{3} +=495.0\, KJ /$mol
For the reaction
$NO\left(g\right)+O\left(g\right) \rightarrow NO_{2}\left(g\right) $
$\Delta H=\Delta H_{1} -\Delta H_{2} -\frac{1}{2} \Delta H_{3}$
$=-198.9-\left(-142.3\right)- \frac{1}{2}\times495$
$=-304.1\, kJ/$ mol