hard gchem rate problem =(

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Tina324

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trial # [A]o o rate
1 1 1 1
2 2 1 2
3 4 3 36

given these results how do you go about getting the order of the rxn with respect to B? Please EXPLAIN the method..i would be 4ever indebted to u =)
 
trial # [A]o o rate
1 1 1 1
2 2 1 2
3 4 3 36

given these results how do you go about getting the order of the rxn with respect to B? Please EXPLAIN the method..i would be 4ever indebted to u =)


Well by looking at trial number 1 and 2 we can calculate that the reaction is first order with respect to A. Now go to trial 2 and 3 and plug in your formula.
(4/2)^1*(3)^x=36/2
2*3^x=18
3^x=9
X=2
Answer is [A]^1^2
 
hey klutzy..thanks so much for responding.

i guess i don't have a firm understanding of the rate law..can u explain to me theoretically what u are doing with this formula...i only "know" how to do rate problems that have 2 trials that are constant for A and B respectively. Also, how do u know to choose trials 2 and 3, why not 1 and 3?
 
hey klutzy..thanks so much for responding.

i guess i don't have a firm understanding of the rate law..can u explain to me theoretically what u are doing with this formula...i only "know" how to do rate problems that have 2 trials that are constant for A and B respectively. Also, how do u know to choose trials 2 and 3, why not 1 and 3?

You could have chosen 1 and 3, watch.
(4/1)^1*(3/1)^x=36/1
4*3^x=36
3^x=36/4=9
3^x=9
X=2

You are just setting up your rate formula and substitute the information that you already have.
The rate formula is
[A]^x * ^y=rate
 
lolll! i am a girl! can't u tell by how many lol's and =)'s i use in my posts?
 
aww all this pre-dent love lol...well i'll be sending over a lot more gchem q's where that came from..so be ready guys =)
 
ok so with these rate problems..we are always given at least 2 trials in which one of the reactants remains constant, rite? because we have to make one of the reactants=1 in order to figure out the order of the other..correct?
 
ok so with these rate problems..we are always given at least 2 trials in which one of the reactants remains constant, rite? because we have to make one of the reactants=1 in order to figure out the order of the other..correct?

not always will one reactant rate remain constant. be careful.
 
uh oh crap...then how the heck do u figure out A with respect to B if there is no trial with the concentration of B remaining constant? can u give me an example?
 
i may be completely off, but doesnt rate = k(A^x)(B^y)

so wouldnt you need to know the rate constant, k, to work out this problem?
 
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