## Second Order/Zero Order Half-Life

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Kyle Sheu 1C
Posts: 87
Joined: Fri Jun 23, 2017 11:39 am

### Second Order/Zero Order Half-Life

Why is it that the half-lives for zero order and second order reactants depend on initial concentration (but not of first order)?

I understand the mathematical derivation, so a theoretical/conceptual answer would be appreciated !

Clara Rehmann 1K
Posts: 53
Joined: Fri Sep 29, 2017 7:03 am

### Re: Second Order/Zero Order Half-Life

It possibly may have to do with the way molecules of these reactants interact with each other? I think we'd have to learn more about the mechanics of 0, 1st, and 2nd order reactions in order to understand the exact ways in which they differ, but I would guess that more of these molecules being in a smaller space (ie. at a higher concentration) could possibly make them decay more quickly as it is more likely that they would interact with each other.

Scott Chin_1E
Posts: 55
Joined: Sat Jul 22, 2017 3:00 am

### Re: Second Order/Zero Order Half-Life

I believe this is because for a zero order reaction, the rate at which the reactant is consumed is constant, thus the time it takes for the concentration of the substance to reduce by half would be dependent on the initial concentration.

I could imagine that this explanation could almost be applied to 2nd Order reactions as well, but I'm not too sure.