Comment on the thermodynamic stability of NO(g), given,

($$\dfrac{1}{2}$$)N2(g) + ($$\dfrac{1}{2}$$)O2(g) → NO(g)$$\Delta _{r}H^{\Theta } = 90kJmol^{-1}$$

NO(g) + ($$\dfrac{1}{2}$$)O2(g) → NO2(g)$$\Delta _{r}H^{\Theta } = -74kJmol^{-1}$$

Asked by Abhisek | 1 year ago |  141

#### 1 Answer

##### Solution :-

The positive value of Δr​H indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants ( O2 and N2). Hence, NO(g) is unstable. The negative value of  Δr​H indicates that heat is evolved during the formation of $$NO_2$$ (g) from NO(g) and $$O_2$$ (g). The product, $$NO_2$$(g) is stabilized with minimum energy. Hence, unstable NO(g) changes to unstable $$NO_2$$(g) .

Answered by Pragya Singh | 1 year ago

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