## The dependancy on initial concentration

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Peiliang Zhou 3K
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

### The dependancy on initial concentration

Why is the half-life equation for a first order reaction independent of the initial concentration, unlike the half-life for a 2nd or zero order reaction? Is it simply because the values cancel out mathematically? What is the chemical basis?

LinaLi2E
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

### Re: The dependancy on initial concentration

The equation for half-life of a 1st order reaction is t(1/2)=0.693/k, the equation for half-life of a 2nd order reaction is t(1/2)=1/(k*[A]0) and the equation for half-life of a zero order reaction is t(1/2)=[A]0/2*k. The key point here is that the equation for half-life of a 1st order reaction does NOT include the initial concentration whereas the equation for half-life of a 2nd and zero order reaction does include the initial concentration. Therefore the half-life of a 1st order reaction is independent of the initial concentration--initial concentration does NOT affect t(1/2) and k and the half-life of a 2nd and zero order reaction is dependent of the initial concentration.
The reason for this is probably because when the reaction is 2nd order, that means that if the initial concentration decrease by 1/2, then the rate decreases by 1/4 so it takes the reaction longer--due to the slower rate--to further decrease the concentration by 1/2 thus the half-life is dependent on the initial concentration. For a 1st order reaction, when the concentration is decreased by 1/2, the rate is decreased by 1/2. The proportion is the same so the half-life is independent.