Question

Comment on thermodynamics of NO(g) and NO₂(g) on the basis of following reactions:

\((i)\frac{1}{2}N_{2(g) } + \frac{1}{2}O_{2(g)} \longrightarrow NO_{(g)};\ \Delta r\ H = 90\ kj\ mol^{-1}\)

\((ii)\ NO_{(g)} + \frac{1}{2}O_{2(g)}\longrightarrow NO_{2(g)};\ \Delta r\ H = 74.0\ kj\ mol^{-1}\)

Answer 1

\((i)\ \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}\longrightarrow NO_{(g)}\ ; \ \Delta r \ H = - 90\ kJ \ mol^{-1 }\)

Since, ∆r H is +ve. So, thermodynamically NO is not table. So it is an endothermic process.

\((ii)\ NO_{(g)} + \frac{1}{2}O_{2(g)}\longrightarrow NO_{2(g)}\ ;\ \Delta r\ H = 74\ kj\ mol^{-1}\)

Since ∆r H is negative. So it is thermodynamically table. The oxidation of NO to NO₂, is an exothermic process.