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 equilbrium are elementary 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(g)+

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 equilbrium are elementary bimolecular reactions what is K_C for the reverse reaction?
Answer» Solution :For the REVERSE reaction `K_(C)= (1)/(K_C)= (1)/(6.3xx 10^(14)) = 1.59 xx 10^(-15)`