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hard gchem rate problem =(
Started by Tina324
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?
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
OoO =) thanks man!
YQW, "Tina" is your s/n yet you are a guy?? thats funny.
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.
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