For the following equilibrium,K_c=6.3xx10^14 at 1000 K NO(g)+O_3(g)iffNO_2(g)+O_2(g) What is K_c for the reverse reactions?

For the following equilibrium,K_c=6.3xx10^14 at 1000 K NO(g)+O_3(g)if

1.	For the following equilibrium,K_c=6.3xx10^14 at 1000 K NO(g)+O_3(g)iffNO_2(g)+O_2(g) What is K_c for the reverse reactions?
Answer» SOLUTION :`K_c` for the REVERSE REACTION =`1/K_c=1/(6.3xx10^4)=0.159xx10^-14=1.59xx10^-15`