Comment on the thermodynamic stability of NO ( g ) , given 1 2 N 2 ( g ) + 1 2 O 2 ( g ) → NO ( g ) ; ∆ r H ° = 90 kJ mol - 1 NO ( g ) + 1 2 O 2 ( g ) → NO 2 ( g ) ; ∆ r H ° = - 74 kJ mol - 1

1 2 N 2 ( g ) + 1 2 O 2 ( g ) → NO ( g ) ; ∆ r H ° = 90 kJ mol - 1 The positive value of ∆ r H in the fist equation indicates that heat is absorbed during the formation of NO ( g ) . This means that NO ( g ) has higher energy than the reactants ( N 2 and O 2 ) . Hence, NO ( g ) is unstable. NO ( g ) + 1 2 O 2 ( g ) → NO 2 ( g ) ; ∆ r H ° = - 74 kJ mol - 1 The negative value of ∆ r H in the second equation 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 stabilised with minimum energy. Hence, unstable NO ( g ) changes to stable NO 2 ( g ) .

E M B I B E

Intermediate First Year Chemistry>Chapter 6 - Thermodynamics>QUESTIONS>Q 34

HARD

11th Telangana Board

IMPORTANT

Earn 100

Equilibrium constant ' $K$ ' is measured accurately in the laboratory at given temperature. Is it possible to calculate $∆ G °$ at any other temperature? How?

Important Questions on Thermodynamics

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Chemistry>Chapter 6 - Thermodynamics>QUESTIONS>Q 35

Comment on the thermodynamic stability of

NO (g)

, given

\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) \to NO (g); ∆_{r} H ° = 90 kJ {mol}^{- 1}

NO (g) + \frac{1}{2} O_{2} (g) \to {NO}_{2} (g); ∆_{r} H ° = - 74 kJ {mol}^{- 1}

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Chemistry>Chapter 6 - Thermodynamics>QUESTIONS>Q 36

Calculate the entropy change in surroundings when $1.00 mol$ of $H_{2} O (l)$ is formed under standard conditions. $∆_{f} H ° = - 286 kJ {mol}^{- 1}$ .