For the following equilibrium , K_(c) = 6*3 xx 10^(14)" at "1000 K NO(g) +O_(3) (g) hArr NO_(2) (g) + O_(2) (g) Both the forward and reverse reactions in the equilibrium in the equilibrium are elementry bimolecular reactions. What is K_(c) for the reverse reaction ?

For the following equilibrium , K_(c) = 6*3 xx 10^(14)" at "1000 K NO(

1.	For the following equilibrium , K_(c) = 6*3 xx 10^(14)" at "1000 K NO(g) +O_(3) (g) hArr NO_(2) (g) + O_(2) (g) Both the forward and reverse reactions in the equilibrium in the equilibrium are elementry bimolecular reactions. What is K_(c) for the reverse reaction ?
Answer» SOLUTION :For the REVERSE REACTION , ` K' _(c) = 1/K_(c) = 1/(63 XX 10^(14) )= 159 xx 10^(-15) `