The general mathematical expression for the half-life period () of an order kinetic reaction (where — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

For an $n^{\text{th}}$ order reaction ($n eq 1$), the integrated rate equation is $k(n-1)t = \frac{1}{(a-x)^{n-1}} - \frac{1}{a^{n-1}}$. At half-life, $x = a/2$. Substituting this yields $k(n-1)t_{1/2} = \frac{1}{(a/2)^{n-1}} - \frac{1}{a^{n-1}} = \frac{2^{n-1} - 1}{a^{n-1}}$. Rearranging gives $t_{1/2} = \frac{2^{n-1} - 1}{k(n-1)a^{n-1}}$.