For the following equilibrium, K_c = 6.3 xx 10^14 at1000 K. NO_((g)) + O_(3(g)) = NO_(2(g)) + O_(2(g)) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c for the reverse reaction ?

For the following equilibrium, K_c = 6.3 xx 10^14 at1000 K. NO_((g)) +

1.	For the following equilibrium, K_c = 6.3 xx 10^14 at1000 K. NO_((g)) + O_(3(g)) = NO_(2(g)) + O_(2(g)) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c for the reverse reaction ?
Answer» SOLUTION :(`K_c` of reverse REACTION of any reaction )= `1/(K_c "of forward reaction")` `THEREFORE K._c=1/(6.3xx10^14)=1.587xx10^(-15)`