## Concentration and Rate Relationship

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

Amanda Mac 1C
Posts: 58
Joined: Fri Sep 29, 2017 7:05 am

### Concentration and Rate Relationship

In a second order reaction, if one reactant alone is doubled does that mean the rate is quadrupled? What happens if two reactants in a second order reaction are both doubled?

Sean Monji 2B
Posts: 66
Joined: Fri Sep 29, 2017 7:06 am

### Re: Concentration and Rate Relationship

I think so. This can be shown mathematically.
Lets say rate = k[A]^2
If A is doubled,
rate = k[2A]^2 = 4k[A]^2, meaning rate was quadrupled
in rate = k[A][B]
k[2A][2B] = 4k[A][B]
Though in the second reaction, if you only double A or B, the rate only doubles
k[2A][B] = 2k[A][B]
This has to do with how likely the molecules will collide to form the product.
Hope that makes sense

Posts: 88
Joined: Fri Sep 29, 2017 7:03 am

### Re: Concentration and Rate Relationship

For the most part, it would be easier to see the changes if you had the rate of reaction and the corresponding orders. From there, you could just plug in new concentrations to see what kind of effect it would have on the overall rate.

Harjas Sabharwal 1G
Posts: 42
Joined: Sat Jul 22, 2017 3:01 am

### Re: Concentration and Rate Relationship

The answer to your question depends on the specific rate law for the reaction you are discussing. Generally if you plug in the change and then compare the resulting rate laws, you should be able to figure out what the change in rate is.

Christina Bedrosian 1B
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

### Re: Concentration and Rate Relationship

for multiple reactions, see the effect of each reactant on the rate. so if one reactant is 1st order and another is 2nd order, the 1st would double the rate if the concentration is doubled and the 2nd one would quadruple the rate if the concentration was doubled. you usually view them separately like that