A community for students.
Here's the question you clicked on:
 0 viewing
Theloshua
 one year ago
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 × 102 M/min
2 0.10 M 0.40 M 4.8 × 102 M/min
3 0.20 M 0.40 M 9.6 × 102 M/min
Theloshua
 one year ago
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 × 102 M/min 2 0.10 M 0.40 M 4.8 × 102 M/min 3 0.20 M 0.40 M 9.6 × 102 M/min

This Question is Closed

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4For 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
 one year ago
Best ResponseYou've already chosen the best response.0oh hey best friend!! :D

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0listen, idk if youll do this but im in a hurry, i kinda need this one done for me....

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0i needa turn it in.... can you maybe just do it for me?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4Can't man, sorry :S it's against the "rules", but more importantly, it's against my belief of doing your own work lol

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0okay *sigh* then walk me thru it please? i really really needa get this turned in...

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4yeah, sure. did you understand what I said in the beginning?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4Okay i'll do one, you do the other. Trial [A] [B] Rate 1 0.10 M 0.20 M 1.2 × 102 M/min 2 0.10 M 0.40 M 4.8 × 102 M/min 3 0.20 M 0.40 M 9.6 × 102 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 4fold, \(\sf \dfrac{4.8*10^{2}}{1.2*10^{2}}=4\) 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
 one year ago
Best ResponseYou've already chosen the best response.4Theres an error in "\(\sf rate_2=1=k[B]^n=4=k[2]^n\)" Ignore the 1, i mustve hit the key accidentally

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4Yeah, it's not a topic you can pick up in a few mins..

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4I'll give you a hand. Trial [A] [B] Rate 1 0.10 M 0.20 M 1.2 × 102 M/min 3 0.20 M 0.40 M 9.6 × 102 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
 one year ago
Best ResponseYou've already chosen the best response.0rate = k[h2]x[I2]y 1.2 × 102 M/min/4.8 × 102 M/min = [0.10]x[0.20]y/[0.10]x[0.40]y

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0this is hpw ive started out

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0the 0.10 can get crossed out right?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4Yeah, but I what are you achieving with this?

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0apparently this is how to solve the problem according to my printed out notes *shrug*

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4Hm, 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

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4Crap it was trial 2 and 3, not 1 and 3.

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4Trial [A] [B] Rate 2 0.10 M 0.40 M 4.8 × 102 M/min 3 0.20 M 0.40 M 9.6 × 102 M/min Do it with this

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4In 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
 one year ago
Best ResponseYou've already chosen the best response.0i dont knoooowwwwwww :/

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4you're on the right track, the concentration doubles

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0ohhhh okay pfew, dont do that, i tought i was wrong :p

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4haha 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
 one year ago
Best ResponseYou've already chosen the best response.4How does the rate change?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4Okay, 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
 one year ago
Best ResponseYou've already chosen the best response.4the second line is supposed to say \(\sf rate_3=[A]_3^m\) sorry, i'm just trying to answer fast

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0what the freak?? O.o idk...

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4you could use logarithms, but using common sense, what number multiplied by 2 gives you 2?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4exponents work like this, btw \(3^1=3\) \(3^2=3*3=9\) \(3^3=3*3*3=27\)

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4yes! 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
 one year ago
Best ResponseYou've already chosen the best response.4You need to do some algebra, it'll look like this: \(\sf k=\dfrac{rate}{[A]^1[B]^2}\)

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4I have to go, good luck!

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0is the answer .26833?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4the 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
 one year ago
Best ResponseYou've already chosen the best response.0oh pellet... guess I got the question wrong.... XD I was too tired. tanks anyway tho

aaronq
 one year ago
Best ResponseYou've already chosen the best response.4lol damn. no problem still
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.