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NCERT Class XI Chemistry Thermodynamics Solutions

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Question : 21 of 22
Marks: +1, -0
Comment on the thermodynamic stability of NO(g)\mathrm{NO}_{(g)}, given
12N2(g)+12O2(g)\frac{1}{2}\mathrm{N}_{2(g)} + \frac{1}{2}\mathrm{O}_{2(g)} → NO(g)\mathrm{NO}_{(g)} ; ΔrH∘\Delta_{r} H^{\circ} = 90 kJ mol−1\mathrm{kJ\,mol}^{-1}
NO(g)+12O2(g)\mathrm{NO}_{(g)} + \frac{1}{2}\mathrm{O}_{2(g)} → NO2(g)\mathrm{NO}_{2(g)} ; ΔrH∘\Delta_{r} H^{\circ} = - 74 kJ mol−1\mathrm{kJ\,mol}^{-1}
Solution:  
Since ΔrH∘\Delta_{r}H^{\circ} is +ve, i.e., enthalpy of formation of NO is positive, therefore NO is unstable. But ΔrH∘\Delta_{r}H^{\circ} is negative for the formation of NO2\mathrm{NO}_2. So, NO2\mathrm{NO}_2 is stable.
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