Half Life of Second Order Reactions

[Student 1H]

I was wondering how we obtain the formula t1/2 = [A]0/(2k) from a second order reaction.
Given that 1/[A] = kt + 1/[A]0 and [A] = 2[A]0, then
1/(2[A]0) = kt + 1/[A]0
-kt = 1/[A]0 -1/(2[A]0) = 1/(2[A]0)
-kt = 1/(2[A]0)
However, this derivation looks really different. Could someone please help me understand why?

[Lena Jabourian 1C]

Hi! The half life formula you have written is actually for zero order. The 2nd order half life is t1/2=1/k[A]o. Also since we're talking about half life in which the concentration is half its original concentration, [A] at t=t1/2 would be equal to 1/2[A]o rather than 2[A]o. Hope that's helpful!

[806205379]

To derive the half-life expression for a second-order reaction without using numbers or symbols, we start with the fact that the half-life is the time it takes for the concentration of the reactant to decrease to half its initial value.In a second-order reaction, the relationship between the concentration of the reactant and time is given by an equation. By using this equation and considering that the concentration decreases to half its initial value after we can set up an equation to solve for half life Once we have the equation for half life we isolate it to find an expression that describes the half-life of the reaction without using specific numerical values or symbols. This expression represents the relationship between the half-life, the rate constant, and the initial concentration of the reactant in a general sense, applicable to any second-order reaction.