<meta http-equiv="refresh" content="1; url=/nojavascript/"> Determining the Rate Law from Experimental Data ( Study Aids ) | Chemistry | CK-12 Foundation
Dismiss
Skip Navigation

Determining the Rate Law from Experimental Data

%
Best Score
Practice Determining the Rate Law from Experimental Data
Practice
Best Score
%
Practice Now
Determining Rate Law From Experimental Data
 0  0  0

Feel free to modify and personalize this guide by clicking "Customize."

When determining rate law from experimental data, you are essentially trying solve for the exponents of reactant concentrations by isolating two experiments with equal concentrations of one reactant.

Sounds a bit confusing, but see if the following solution approach helps clear things up. Remember this is just one way to solve this type of problem.

Let's say you're given the following table of experimental data.

Experiment [NO] [H ] Initial Rate (M/s)
1 0.0050 0.0020 1.25 \times 10^{-5}
2 0.010 0.0020 5.00 \times 10^{-5}
3 0.010 0.0040 1.00 \times 10^{-4}

We begin by creating a generic rate law with variables x and y as exponents of reactant concentrations. Our goal in this procedure is to solve for x,y, and k (the specific rate constant).

rate= k[NO]x[H2]y

Next, we choose a variable to isolate and divide the experiments with the same concentration of the non-isolated reactant by each other and solve. What this means is, if we choose to isolate and solve for y (the reaction order for [H2], we look for two experiments where [NO] is held constant so that when we divide the two experiments, [NO] is canceled out.

It looks like [NO] is held constant in experiment 2 and 3, so divide experiment 2 by 3

 5.00 \times 10^{-5}= k (0.010)(0.0020)

1.00 \times 10^{-4} = k (0.010)(0.0040)y

= .500 = .500y

= y= 1

So now we know y=1. We then repeat the same method listed above to solve for x. 

You should get that x= 2.

Now we can piece together the rate law is rate= k[NO]2[H2].

To solve for k, simply choose any of the experiments and plug the data in for the rate law we derived above, rate= k[NO]2[H2], and solve.

Beware! When solving for k, it is crucial that you pay attention to units since there is no "specific rate constant unit". Rather, units are unique to each reaction.

How do we write the rate law of a reaction determined to be of 0 order?

For guidance in answering the above question or for more information on determining rate law from experimental data, look here.

Image Attributions

Reviews

Email Verified
Well done! You've successfully verified the email address .
OK
Please wait...
Please wait...
ShareThis Copy and Paste

Original text