One of the equations for 1st order rxn

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

JennyCKim1J
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

One of the equations for 1st order rxn

What is the equation d[A]/[A] for? Is this just change in [A] over time? What is the difference between d[A]=-kdt and d[A]/[A]?

jillian1k
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

Re: One of the equations for 1st order rxn

d[A]/[A] in itself is not an equation. Also, because it doesn't have the variable for time (t) in it, it couldn't be a change in time.
d[A]/[A] does come up when you're trying to find this change in concentration over time, though. Since a change in concentration over time is rate as a function of time, we have to use the integrated rate law (the differential rate law is just examining rate as a function of concentration). To find the integrated rate law, you have to set the differential rate law (rate=k[A]^n) equal to -1d[A]/adt. Then, you get [A] on one side and dt with k on the other. This gives d[A]/[A]=-kdt. From here you would integrate to get the integrated rate law and use that to find concentration after a certain change in time (or to find the specific time it takes to yield a specific concentration).

Salman Azfar 1K
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

Re: One of the equations for 1st order rxn

As the post above me pointed out, it isn't an equation. It is the derivative for concentration of A, which as you said is essentially change in concentration; however it is not over dt, so it is not with respect to time. The difference... well, that probably has to do more with derivations and integration than anything. When actually solving stuff, know that d[A]/[A] on its own isn't an equation.

Jennie Fox 1D
Posts: 66
Joined: Sat Jul 22, 2017 3:01 am

Re: One of the equations for 1st order rxn

d[A]/[A] stands for the rate of consumption of [A] during a first order reaction.