Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided.
A + B yields products
Trial [A] [B] Rate
1 0.10 M 0.20 M 1.2 × 10-2 M/min
2 0.10 M 0.40 M 4.8 × 10-2 M/min
3 0.20 M 0.40 M 9.6 × 10-2 M/min

- Theloshua

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- aaronq

For each of the reactants, isolate two trials where one of the concentrations is kept constant. You can now ignore the contribution of that reactant to the rate - that is ignore it from the rate law.
Now examine how the rate changes and figure out the order for the reactant that IS changing in concentration (between the two trials). You can assign it's order (exponent) by using some simple math.

- Theloshua

oh hey best friend!! :D

- Theloshua

listen, idk if youll do this but im in a hurry, i kinda need this one done for me....

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Theloshua

i needa turn it in.... can you maybe just do it for me?

- aaronq

Can't man, sorry :S
it's against the "rules", but more importantly, it's against my belief of doing your own work lol

- Theloshua

okay *sigh*
then walk me thru it please? i really really needa get this turned in...

- aaronq

yeah, sure. did you understand what I said in the beginning?

- Theloshua

nope

- aaronq

Okay i'll do one, you do the other.
Trial [A] [B] Rate
1 0.10 M 0.20 M 1.2 × 10-2 M/min
2 0.10 M 0.40 M 4.8 × 10-2 M/min
3 0.20 M 0.40 M 9.6 × 10-2 M/min
The rate law is: \(\sf \large rate=k[A]^m[B]^n\)
In trial 1 and 2, [A] is constant but [B] changes.
It fact doubles 0.2 M to 0.4 M
The rate changes in 4-fold, \(\sf \dfrac{4.8*10^{-2}}{1.2*10^{-2}}=4\)
rate law: \(\sf rate=k[B]^n\)
I'll use simple numbers since we know that the concentration doubles, and the rate quadruples:
\([B]_1=1 ~and~ [B]_1=2\), \(rate_1=1 ~and ~rate_2=4\)
\(\sf rate_1=k[B]^n=1=k[1]^n\)
\(\sf rate_2=1=k[B]^n=4=k[2]^n\)
Ignore the constant, k. We have
\(\sf 1=[1]^n\) and \(\sf 4=[2]^n\)
The only way for the above to be true is for n=2
so far, we have the rate law: \(\sf rate=k[A]^m[B]^2\)
now figure out the exponent "m" in the same fashion and FINALLY, plug in any one set of data to figure out \(k\).

- aaronq

Theres an error in "\(\sf rate_2=1=k[B]^n=4=k[2]^n\)"
Ignore the 1, i mustve hit the key accidentally

- Theloshua

confuzzled as heck

- aaronq

Yeah, it's not a topic you can pick up in a few mins..

- aaronq

I'll give you a hand.
Trial [A] [B] Rate
1 0.10 M 0.20 M 1.2 × 10-2 M/min
3 0.20 M 0.40 M 9.6 × 10-2 M/min
In trial 1 and 3, [B] is constant but [A] changes.
How does the concentration of A change, and how does that affect the reaction (by what factor does the rate change)?

- Theloshua

rate = k[h2]x[I2]y
1.2 × 10-2 M/min/4.8 × 10-2 M/min = [0.10]x[0.20]y/[0.10]x[0.40]y

- Theloshua

this is hpw ive started out

- Theloshua

the 0.10 can get crossed out right?

- aaronq

Yeah, but I what are you achieving with this?

- Theloshua

so its 0.20/0.40

- Theloshua

apparently this is how to solve the problem according to my printed out notes *shrug*

- aaronq

Hm, there are other ways to solve these types of problems, but the way i'm showing you simplifies everything a lot. Read what I wrote last about [B] being constant in trial 1 and 3

- Theloshua

ok

- aaronq

Crap it was trial 2 and 3, not 1 and 3.

- aaronq

Trial [A] [B] Rate
2 0.10 M 0.40 M 4.8 × 10-2 M/min
3 0.20 M 0.40 M 9.6 × 10-2 M/min
Do it with this

- aaronq

In trial 2 and 3, [B] is constant but [A] changes.
How does the concentration of A change, and how does that affect the reaction (by what factor does the rate change)?

- Theloshua

2?

- aaronq

2 what? lol

- Theloshua

i dont knoooowwwwwww :/

- aaronq

you're on the right track, the concentration doubles

- Theloshua

ohhhh okay pfew, dont do that, i tought i was wrong :p

- aaronq

haha you should elaborate on your answers, i know what you mean but if I didn't know the answer already I wouldnt know what youre talking about

- aaronq

How does the rate change?

- Theloshua

it doubles.... -_-

- aaronq

Okay, so we can say \([A]_2=1\), \([A]_3=2\) and then \(rate_2=1\) \(rate_3=2\)
Now ignoring k and ignoring [B] (because the concentration is constant)
\(\sf rate_2=[A]^m\rightarrow 1=[1]^m \)
\(\sf rate_2=[A]^m \rightarrow 2=[2]^m \)
what is \(m\)?

- aaronq

the second line is supposed to say \(\sf rate_3=[A]_3^m\) sorry, i'm just trying to answer fast

- Theloshua

what the freak?? O.o idk...

- aaronq

you could use logarithms, but using common sense, what number multiplied by 2 gives you 2?

- Theloshua

1

- aaronq

exponents work like this, btw
\(3^1=3\)
\(3^2=3*3=9\)
\(3^3=3*3*3=27\)

- aaronq

yes!
so the rate law so far is: \(\sf rate=k[A]^1[B]^2\)
now just plug in the rate and the concentrations from one of the trials to get \(k\)

- aaronq

You need to do some algebra, it'll look like this: \(\sf k=\dfrac{rate}{[A]^1[B]^2}\)

- aaronq

I have to go, good luck!

- Theloshua

wait

- Theloshua

is the answer .26833?

- aaronq

the answer was supposed to be the rate law, not just a number. The last thing (solving for k) was to complete it.
It should've looked like: \(\sf rate=k[A][B]^2\)
with k as whatever the value was

- Theloshua

oh pellet... guess I got the question wrong.... XD I was too tired. tanks anyway tho

- aaronq

lol damn. no problem still

Looking for something else?

Not the answer you are looking for? Search for more explanations.