## Rate Law

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

melodyzaki2E
Posts: 61
Joined: Fri Sep 28, 2018 12:18 am

### Rate Law

How do we determine the Rate Law for a reaction when the experimental data isn't clear on what is doubled?

2c_britneyly
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am

### Re: Rate Law

When it's not clear if the concentration is doubled, then you need to divide one rate law equation by the other, with the respective experimental numbers for each variable. This only works if everything else stays constant besides the one reactant you are working with. You'll end up with a number on the left hand side of the equation and another number raised to some power. You can then take the log to find that missing exponent, which will then be the order of your reactant.
You repeat this for each reactant until you've found all the orders of the reactants. Then you can plug in the exponents you just found with the experimental data, and you should be able to solve for kr, the rate constant.

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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### Re: Rate Law

Yes, or in equations:

Supposing $r=[A]^x[B]^y[C]^z$

$\frac{r_1}{r_2}=\frac{[A]_1^x[B]_1^y[C]_1^z}{[A]_2^x[B]_2^y[C]_2^z}$

$\frac{r_1}{r_2}=\Big(\frac{[A]_1}{[A]_2}\Big)^x\Big(\frac{[B]_1}{[B]_2}\Big)^y\Big(\frac{[C]_1}{[C]_2}\Big)^z$

Supposing [B] and [C] don't change:

$\frac{r_1}{r_2}=\Big(\frac{[A]_1}{[A]_2}\Big)^x(1)^y(1)^z$

$log\Big(\frac{r_1}{r_2}\Big)=x * log\Big(\frac{[A]_1}{[A]_2}\Big)$

$x = \frac{log(\frac{r_1}{r_2})}{log\Big(\frac{[A]_1}{[A]_2}\Big)}$