# Find units of rate constants and write reactions

1. Mar 13, 2013

### marimari

1. The problem statement, all variables and given/known data
Find the units of k1, k2, k3, k4, and k5. Write rate reactions for dA/dt

a. A$\leftrightarrow$B
k1 (forward)
k2 (backward)

b. A + B $\rightarrow$ C
k3 (forward)

c. A + R $\rightarrow$ 2R
k4 (forward)

d. A $\rightarrow$ B
k5 (forward)

2. Relevant equations

3. The attempt at a solution

a. $\frac{d[A]}{dt}$ = k1[A] - k2
units of k1 & k2 are in s$^{-1}$
b. $\frac{d[A]}{dt}$ = -k3[C]
units of k3 are in L/mol*s
c. $\frac{d[A]}{dt}$ = -k4[R]^2
units of k4 are in L/mol*s
d. $\frac{d[A]}{dt}$ = -k5
units of k5 are in s$^{-1}$

Are these correct? I'm a little confused since the problem specified finding dA/dt rather than each of the products. Thanks!

2. Mar 14, 2013

### epenguin

Looks all OK to me.

Except for first order I never remember the units of the top of my head. But you shouldn't be confused, it is quite easy:

[A] means moles.litre-1.

/dt is s-1

on the RHS, [A] is moles.litre-1.
[A]2 is moles2.litre-2
[A]ab[C]c... is molesa+b+c+....litre-(a+b+c+...)

So e.g. for a 2nd-order reaction like b

moles.litre-1.s-1 = k x moles2.litre-2

Isolate k by dividing both sides by moles2.litre-2 and after any cancellations you have it.

Do they insist nowadays, mnyah mnyah, you write out the cumbrous moles.litre-1 etc?

M used to be good enough. Or rather, better.

3. Mar 14, 2013

### Staff: Mentor

I would expect different signs for all "k", but that is probably just a convention issue.

At b, why does the change of A depend on C (and not on A and B), if C does not react at all? The same applies to c and d.
At b, how can you get units different from your answers at (a) and (d)?

4. Mar 14, 2013

### epenguin

It doesn't depend on C.

Sorry I read your answer for the units which were all correct, not your equation which is not right. The answer for the units is right for b, c and d somehow, but your equations for those are wrong.

I happen to have written a post recently about how simple it is to write the rate equations for a given mechanism.
https://www.physicsforums.com/showpost.php?p=4304238&postcount=16

Rate constants are always positive.

Last edited: Mar 14, 2013
5. Mar 14, 2013

### marimari

Thanks for all the help! So, my equations are incorrect for b, c, d? I'm not sure how to write the reaction in terms of dA/dt, then. I know how to write them for dC/dt or dProduct/dt, but not for a reactant.

6. Mar 14, 2013

### marimari

Wait a minute...
Am I just being an idiot?

Would
(b) simply be d[A]/dt = k3[A]

(c) d[A]/dt = k4[A][R]
actually I'm the most unsure about this one because of the 2R in the product. Would the rate be halved?

(d) d[A]/dt = k5[A]?

7. Mar 14, 2013

### Staff: Mentor

As A decreases in the reaction and the k-factors are always positive, those equations should have a minus sign. Apart from that, they are correct.
The reaction product (as long as it is not A) does not matter, even in case of (c). The amount of R will change in a more complicated way, of course.

8. Mar 14, 2013

### marimari

Awesome! Thank you so much

9. Mar 14, 2013

### marimari

Another quick question: Are my units correct as epenguin said?

10. Mar 14, 2013

### Staff: Mentor

With the fixed equation in (b), they are all correct now.

11. Mar 14, 2013

### epenguin

Given conservation of mass, or better, given that in b one molecule of A generates one of C what is the relation of dC/dt to dA/dt? Etc.

If you don't have a doh! moment read through the first page or so of whatever book chapter your course is using.

I initially thought your problems were with the dimensions or units, it did not occur to me that anyone would have a difficulty with formulating the equations so I did not even look at yours. (However their derivation is explained in the link I gave in last post.) In the same way I missed that your a is also wrong! - make it -d[A]/dt on the left. A is disappearing by reacting, the more A there is the faster it disappears! After you have corrected that, you have a correct first stage. But a second stage, for a more useful equation, is to express all in terms of A, without B appearing in the equation. Using the idea of my 1st line above.

12. Mar 14, 2013

### epenguin

b, c and d not quite, see my last post.