Comment on the thermodynamic stability of NO_((g)), given: (1)/(2) N_(2(g)) , + (1)/(2) O_(2(g)) to NO_((g)), Delta_(r) H^( Theta ) = 90 "kJ mol"^(-1) NO_((g)) + (1)/(2) O_(2(g)) to NO_(2(g)), Delta_(r) H^( Theta ) = -74"kJ mol"^(-1)

Comment on the thermodynamic stability of NO_((g)), given: (1)/(2) N_

1.	Comment on the thermodynamic stability of NO_((g)), given: (1)/(2) N_(2(g)) , + (1)/(2) O_(2(g)) to NO_((g)), Delta_(r) H^( Theta ) = 90 "kJ mol"^(-1) NO_((g)) + (1)/(2) O_(2(g)) to NO_(2(g)), Delta_(r) H^( Theta ) = -74"kJ mol"^(-1)
Answer» SOLUTION :Heat is absorbed in FIRST reaction. So, `NO_((g))` unstable. In second reaction, heat is released, So, `NO_(2(g))` is stable. So, unstable `NO_((g))` is CONVERTED into stable `NO_(2(g))`.