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A reaction between the substances A and B has been found to give the following data: 3A + 2B -----> 2C + D?

Asked by sillymichelleyoung (217points) February 19th, 2010

I am having trouble figuring out rate laws.

A reaction between the substances A and B has been found to give the following data:

3A + 2B——-> 2C + D

1.0 X 10⁻²
1.0 X 10⁻²
2.0 X 10⁻²
2.0 X 10⁻²
3.0 X 10⁻²


Rate of Appearance of C
0.3 X 10^-6
8.1 X 10^-6
3.24 X 10^-5
1.20 X 10^-6
7.30 X 10^5

Using the data above, determine the order of the reaction with respect to A and B and the rate law and calculate the specific rate constant.

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9 Answers

Cruiser's avatar

My reaction is “HUH??”

Tropical_Willie's avatar

Should this be under ======== HOMEWORK ========

Zen_Again's avatar

There is a new website for teens and homework at

Next (real) question?

lucillelucillelucille's avatar

This makes my head hurt, Ow.

Harp's avatar

[Mod says] A reminder to all that just because it’s a homework question doesn’t mean it doesn’t belong on Fluther. While it would be inappropriate to supply homework answers, giving someone guidance in how to approach a problem like this is perfectly appropriate.

Response moderated
BhacSsylan's avatar

So, this is a very general rate law problem using what’s known as proportionation. So, first thing you need to do is set up a general rate law. This is done by using Rate=k, then adding the concentration of each reactant. So, if I had x->y, it’d be Rate=k[x]. Or, if x+y->z, then Rate=k[x][y].

Next, you need to find the ‘order’ of each reactant. This is a power that each concentration term is raised to. If y in above was second order, it would look like Rate=k[x][y]^2. But how do we find it? That’s where proportionation comes in. Since k generally stays constant, if you take a ratio of two rate laws, were only rates and concentrations change, you get something like this:

if Rate1=k[x1][y1] and Rate2=k[x2][y2], then Rate2/Rate1=(k/k)([x2]/[x1])([y2]/[y1]). The k terms cancel, and you’re left with Rate2/Rate1=([x2]/[x1])([y2]/[y1]).

Now, you’d look at your data, and chose a pair of experiments where one reactant is constant, and the other is not. If one is constant, that term drops out. Say y is constant, we get Rate2/Rate1=([y2]/[y1]). Now, just feed in the numbers. Lets take a case where Rate2=16, Rate1=4, [y2]=8, and [y1]=4. Putting that in gets this:

16/4=(8/4), so 4=(2)

Now, we know 4 doesn’t equal 2, so we know that the part on the right has to be squared, since 4=(2)^2. So, we just determined that the order of y must be 2.

If you repeat this process for every reactant, you can find the order. If we let y stay the same, and take Rate2=8, Rate1=4, x2=4, x1=2, then we get

8/4=4/2 so 2=2

Since that equation is fine, we know x is first order, and exponent of 1. So, we solved the order of y and x, we can put that in and get our total rate law:


Now comes the last and easiest part. You have the whole equation, so you can simply plug previous values in for rate, [x], and [y] to find k. if Rate=4, [x]=2, and [y]=4, then

4=k*2*4, so k=4/(2*4)=½

And, now you’ve found k.

So, this is the very long but hopefully general way to find k, or a rate law. Hope I helped.

sillymichelleyoung's avatar

thank you bhacssylan

i was having difficulty figuring out what order everything was thus preventing me from finishing the problem

BhacSsylan's avatar

No problem. Glad I could help. I’m actually a TA for Gen Chem so I just taught this a few weeks ago ^_^.

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