MHT CET Chemistry Question (2026) — Solution

Step-by-step solution

For a first-order reaction, the concentration after n half-lives is given by [A]_t = [A]_0 2^n . Given [A]_0 = 1.0\ M and [A]_t = 0.25\ M , we have: 0.25 = 1.0 2^n 2^n = 4 n = 2 Thus, two half-lives have passed in 276\ s . 2 t_ 1/2 = 276\ s t_ 1/2 = 138\ s The rate constant k is given by: k = 0.693 t_ 1/2 = 0.693 138 k 0.0050\ s ^ -1