Theloshua
  • Theloshua
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
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
aaronq
  • 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
  • Theloshua
oh hey best friend!! :D
Theloshua
  • Theloshua
listen, idk if youll do this but im in a hurry, i kinda need this one done for me....

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Theloshua
  • Theloshua
i needa turn it in.... can you maybe just do it for me?
aaronq
  • 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
  • Theloshua
okay *sigh* then walk me thru it please? i really really needa get this turned in...
aaronq
  • aaronq
yeah, sure. did you understand what I said in the beginning?
Theloshua
  • Theloshua
nope
aaronq
  • 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
  • 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
  • Theloshua
confuzzled as heck
aaronq
  • aaronq
Yeah, it's not a topic you can pick up in a few mins..
aaronq
  • 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
  • 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
  • Theloshua
this is hpw ive started out
Theloshua
  • Theloshua
the 0.10 can get crossed out right?
aaronq
  • aaronq
Yeah, but I what are you achieving with this?
Theloshua
  • Theloshua
so its 0.20/0.40
Theloshua
  • Theloshua
apparently this is how to solve the problem according to my printed out notes *shrug*
aaronq
  • 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
  • Theloshua
ok
aaronq
  • aaronq
Crap it was trial 2 and 3, not 1 and 3.
aaronq
  • 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
  • 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
  • Theloshua
2?
aaronq
  • aaronq
2 what? lol
Theloshua
  • Theloshua
i dont knoooowwwwwww :/
aaronq
  • aaronq
you're on the right track, the concentration doubles
Theloshua
  • Theloshua
ohhhh okay pfew, dont do that, i tought i was wrong :p
aaronq
  • 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
  • aaronq
How does the rate change?
Theloshua
  • Theloshua
it doubles.... -_-
aaronq
  • 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
  • aaronq
the second line is supposed to say \(\sf rate_3=[A]_3^m\) sorry, i'm just trying to answer fast
Theloshua
  • Theloshua
what the freak?? O.o idk...
aaronq
  • aaronq
you could use logarithms, but using common sense, what number multiplied by 2 gives you 2?
Theloshua
  • Theloshua
1
aaronq
  • aaronq
exponents work like this, btw \(3^1=3\) \(3^2=3*3=9\) \(3^3=3*3*3=27\)
aaronq
  • 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
  • aaronq
You need to do some algebra, it'll look like this: \(\sf k=\dfrac{rate}{[A]^1[B]^2}\)
aaronq
  • aaronq
I have to go, good luck!
Theloshua
  • Theloshua
wait
Theloshua
  • Theloshua
is the answer .26833?
aaronq
  • 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
  • Theloshua
oh pellet... guess I got the question wrong.... XD I was too tired. tanks anyway tho
aaronq
  • aaronq
lol damn. no problem still

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