## Detail in the Rate Laws

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

Raul Garcia 3I
Posts: 17
Joined: Fri Jul 15, 2016 3:00 am

### Detail in the Rate Laws

Given the Zero order reactions to even the second, i noticed that the sign is negative (multiplied to the rate constant K); does our answer have to satisfy the negative, or does it not matter? In lecture and even in discussion the detail to whether to include the sign or not was particularly not mentioned (or i must have missed it) .
Thanks

Jerry Wang 1L
Posts: 10
Joined: Fri Jul 22, 2016 3:00 am

### Re: Detail in the Rate Laws

The negative signs in equations must be observed. I would recommend you to go through the derivation process again to see how each one came about (because some of the kinetics equations don't have negative signs because of how the mathematics works out) but from the foundational equations such as d[A]/dt = -k, or d[A]/dt = -k[A], or d[A]/dt = -k[A]^2, all of which have the situation you describe, the negative sign is to indicate that the concentration of reactant A is decreasing with time. If one is to neglect the negative sign, we will be assuming that the reactant will be increasing in concentration with time, which would be highly unconventional.

Wenqian_Deng_1L
Posts: 23
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Detail in the Rate Laws

What about when we're asked to find the temperature of a first order reaction? When we switch the first order integrated rate law to solve for temperature, the negative sign of the rate constant (K), disappears. Am I doing algebra wrong or does it just conveniently cancels itself somehow?