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

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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
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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.
oh hey best friend!! :D
listen, idk if youll do this but im in a hurry, i kinda need this one done for me....

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i needa turn it in.... can you maybe just do it for me?
Can't man, sorry :S it's against the "rules", but more importantly, it's against my belief of doing your own work lol
okay *sigh* then walk me thru it please? i really really needa get this turned in...
yeah, sure. did you understand what I said in the beginning?
nope
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 \(k\).
Theres an error in "\(\sf rate_2=1=k[B]^n=4=k[2]^n\)" Ignore the 1, i mustve hit the key accidentally
confuzzled as heck
Yeah, it's not a topic you can pick up in a few mins..
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)?
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
this is hpw ive started out
the 0.10 can get crossed out right?
Yeah, but I what are you achieving with this?
so its 0.20/0.40
apparently this is how to solve the problem according to my printed out notes *shrug*
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
ok
Crap it was trial 2 and 3, not 1 and 3.
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
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)?
2?
2 what? lol
i dont knoooowwwwwww :/
you're on the right track, the concentration doubles
ohhhh okay pfew, dont do that, i tought i was wrong :p
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
How does the rate change?
it doubles.... -_-
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 \(m\)?
the second line is supposed to say \(\sf rate_3=[A]_3^m\) sorry, i'm just trying to answer fast
what the freak?? O.o idk...
you could use logarithms, but using common sense, what number multiplied by 2 gives you 2?
1
exponents work like this, btw \(3^1=3\) \(3^2=3*3=9\) \(3^3=3*3*3=27\)
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 \(k\)
You need to do some algebra, it'll look like this: \(\sf k=\dfrac{rate}{[A]^1[B]^2}\)
I have to go, good luck!
wait
is the answer .26833?
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
oh pellet... guess I got the question wrong.... XD I was too tired. tanks anyway tho
lol damn. no problem still

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