## Second Order

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

Alexandra Carpenter 1G
Posts: 32
Joined: Sat Jul 22, 2017 3:00 am

### Second Order

There is a problem in a workbook that asks, "Consider the reaction, A + B --> C + D, with rate = k[A]^2. True or false, the time it takes [A] to decrease from 1.0 to 0.50 M is the same as the time it takes for [A] to decrease from 0.50 to 0.25 M.

Why is this false?

vicenteruelos3
Posts: 42
Joined: Fri Sep 29, 2017 7:07 am

### Re: Second Order

false because the rate is dependent on the concentration of the reaction
r=k[A]^2
so if you change [A] then the rate also changes

Nicole 1F
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Re: Second Order

Since 0.25 is half of 0.50, you can use the second order half life equation to help you with this problem. For half life, it only depends on your initial concentration, k, and t. You can see how there are different initial concentrations and because they are different you know that they will have different half lives and times.

Connor Kelligrew 2D
Posts: 39
Joined: Fri Sep 29, 2017 7:07 am

### Re: Second Order

Both changes in concentration can be described by the half-life of a second order reaction:

$t_{1/2} = \frac{1}{k[A]_{0}}$

Since the initial concentration of A is in the denominator, the time it takes different initial concentrations to decay to half of their original concentration is dependent upon the initial concentration of A. Thus, the answer would be false.

By contrast, for first order reactions the half life can be modeled by:

$t_{1/2} = \frac{ln 2}{k}$

Since $[A]_{0}$ does not show up in that equation, the half-life is not dependent upon the initial concentration of A, and the answer to this question applied to first order reactions would be true.