See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

k1 k2 k3 3D 8C k1 : k2 : k3 = 4 : 2 : 1 3B A mole of A remain after 45 days      1/2 0 0 n T 45/15 3 t N N 1 1 1 8 2 2 2 2 moles of [A] convert into product 7 8 mol moles of       1 1 2 3 k 7 3 B 3 k k k 8 2 moles of       2 1 2 3 k 7 8 C 2 k k k 8 1 and moles of     1 7 3 D 3 7 8 8