Question

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

Answer 1

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.

Answer 2

oh hey best friend!! :D

Answer 3

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

Answer 4

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

Answer 5

Can't man, sorry :S

it's against the "rules", but more importantly, it's against my belief of doing your own work lol

Answer 6

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

Answer 7

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

Answer 8

nope

Answer 9

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 \(k\).

Answer 10

Theres an error in "\(\sf rate_2=1=k[B]^n=4=k[2]^n\)"

Ignore the 1, i mustve hit the key accidentally

Answer 11

confuzzled as heck

Answer 12

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

Answer 13

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)?

Answer 14

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

Answer 15

this is hpw ive started out

Answer 16

the 0.10 can get crossed out right?

Answer 17

Yeah, but I what are you achieving with this?

Answer 18

so its 0.20/0.40

Answer 19

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

Answer 20

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

Answer 21

ok

Answer 22

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

Answer 23

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

Answer 24

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)?

Answer 25

2?

Answer 26

2 what? lol

Answer 27

i dont knoooowwwwwww :/

Answer 28

you're on the right track, the concentration doubles

Answer 29

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

Answer 30

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

Answer 31

How does the rate change?

Answer 32

it doubles.... -_-

Answer 33

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 \(m\)?

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

1

Answer 38

exponents work like this, btw

\(3^1=3\)
\(3^2=3*3=9\)
\(3^3=3*3*3=27\)

Answer 39

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 \(k\)

Answer 40

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

Answer 41

I have to go, good luck!

Answer 42

wait

Answer 43

is the answer .26833?

Answer 44

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

Answer 45

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

Answer 46

lol damn. no problem still