Question
Comment on the thermodynamic stability of NO(g), given
`1/2` N2(g) + `1/2` O2(g) → NO(g) ; ΔrHθ = 90 kJ mol–1
NO(g) +`1/2` O2(g) → NO2(g) : ΔrHθ= –74 kJ mol–1
Solution 1
The positive value of ΔrH indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (N2 and O2). Hence, NO(g) is unstable.
The negative value of ΔrH indicates that heat is evolved during the formation of NO2(g) from NO(g)and O2(g). The product, NO2(g) is stabilized with minimum energy.
Hence, unstable NO(g) changes to stable NO2(g).
Solution 2
For `NO(g); triangle_rH^(Theta)` = +ve : Unstable in nature
For `NO_2 (g); triangle_rH^(Theta)` = - ve : Stable in nature
Is there an error in this question or solution?
Solution Comment on the Thermodynamic Stability of NO(g), Given Concept: Measurement of ∆U and ∆H Calorimetry - ∆H Measurements.
