See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

A 1 A dN N dt  , B 1 A 2 B dN 2λ N N dt   , B N = maximum  B dN 0 dt   max 1 A 2 B 2 N N    max 1 B A 2 2 N N     1 max t 1 B 0 2 2 N N e    = 2. SECTION – D