Consider reversibility in free radical substitution reaction of alkane then Kc value is smallest for:

Explanation

In a free radical substitution reaction of an alkane, the reaction can proceed through several steps involving free radicals (species with unpaired electrons). For this type of reaction, reversibility is more likely to occur in the termination step.The termination step involves the combination of two free radicals, leading to the formation of a stable molecule. This step can be reversible in certain cases, resulting in the reformation of free radicals. For example:Initiation step: Formation of free radicals (R·) by the homolytic cleavage of a bond in the alkane due to the presence of an initiator (e.g., light or heat). Propagation steps: Free radicals react with the alkane to form new free radicals and alkyl radicals, continuing the chain reaction.Termination step: Two free radicals combine to form a stable molecule.Since the termination step involves the recombination of free radicals and is less likely to be fully irreversible, the equilibrium constant (Kc) for this step would likely be smaller compared to the propagation steps, which involve the generation of new free radicals.It's important to note that the overall reaction is usually considered irreversible, but individual steps within the reaction mechanism may exhibit different degrees of reversibility. The termination step is generally considered to have a smaller Kc value due to its potential for reversibility.