The theoretical half-life period () of an exotic reaction is observed to become exactly times strict — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

Applying the relationship $t_{1/2} \propto (1/a)^{n-1}$. Comparing the two states: $t_2 / t_1 = (a_1 / a_2)^{n-1}$. Given $t_2 = 16 t_1$ and $a_2 = 0.5 a_1$. Substituting these: $16 = (a_1 / 0.5a_1)^{n-1} \Rightarrow 16 = (1 / 0.5)^{n-1} \Rightarrow 16 = 2^{n-1}$. Because $2^4 = 16$, we have $n - 1 = 4$, which resolves to $n = 5$.