Gchem rate constant question

[joonkimdds]

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if there is an experiment 1,2,3, with [A],, and rate,
we know that we should find experiements that have the same
A or same B values so that we can cancel them out to find the power of A or B.

For example,
Experiment 1
A=0.181
B=0.148
rate=1.87

Experiment 2
A=0.181
B=0.300
rate=1.87

I set up the equation to cancel out A and K to find out m.
rate2/rate 1 = (K[A]^n^m of ex2 / K[A]^n^m of ex1) = 1.87/1.87

BUT! how do we know which rate should go to numerator and which one
goes to the denominator?
if i put rate1 for numerator, it becomes
rate1/rate2 = (0.148/0.3)^m = 1
if i put rate2 for numerator, it becomes
rate2/rate1 = (0.3/0.148)^m = 1
and so the m become a different value.

I want to know which one goes to the numerator.

 

[scottyhoop]

if there is an experiment 1,2,3, with [A],, and rate,
we know that we should find experiements that have the same
A or same B values so that we can cancel them out to find the power of A or B.

For example,
Experiment 1
A=0.181
B=0.148
rate=1.87

Experiment 2
A=0.181
B=0.300
rate=1.87

I set up the equation to cancel out A and K to find out m.
rate2/rate 1 = (K[A]^n^m of ex2 / K[A]^n^m of ex1) = 1.87/1.87

BUT! how do we know which rate should go to numerator and which one
goes to the denominator?
if i put rate1 for numerator, it becomes
rate1/rate2 = (0.148/0.3)^m = 1
if i put rate2 for numerator, it becomes
rate2/rate1 = (0.3/0.148)^m = 1
and so the m become a different value.

I want to know which one goes to the numerator.

Rate 2 would be on top and Rate 1 on bottom. The ratio of of concentrations is f and the ratio of rates is F. f^x=F. So for f=2 and F=1. 2^0=1, so your exponent for B is 0. I don't think you can solve for [A] with your data since [A] is not changing.
Scott

 

[UNMPDS]

Yeah, Rate 2 goes on top. However, do to time constraints on the actual DAT a quick hint for this kind of problem so you don't have to go through all the math is to just look at the rates compared to the concentration. As you can see when the concentration of B doubled, the rate stayed the same, and therefore is a zero order rxn, meaning it doesn't matter how much of B you put in, it doesn't effect the rate. Hope this helps.

 

[joonkimdds]

Rate 2 would be on top and Rate 1 on bottom. The ratio of of concentrations is f and the ratio of rates is F. f^x=F. So for f=2 and F=1. 2^0=1, so your exponent for B is 0.
Scott

wait...so what's the reason rate 1 can't be the numerator?

 

[joonkimdds]

Yeah, Rate 2 goes on top. However, do to time constraints on the actual DAT a quick hint for this kind of problem so you don't have to go through all the math is to just look at the rates compared to the concentration. As you can see when the concentration of B doubled, the rate stayed the same, and therefore is a zero order rxn, meaning it doesn't matter how much of B you put in, it doesn't effect the rate. Hope this helps.

wow...that really helps.
I didn't know that zero order means that.
could u also explain what the 1st and 2nd order can do?

 

[Chga]

the guy above being sarcastic?

 

[joonkimdds]

the guy above being sarcastic?

me? no man, i am serious...

 

[UNMPDS]

wow...that really helps.
I didn't know that zero order means that.
could u also explain what the 1st and 2nd order can do?

OK, first order means if the concentration doubles so will the rate, second order means if the concentration doubles, the rate will quadruple. Just simple exponential math.

 

[joonkimdds]

OK, first order means if the concentration doubles so will the rate, second order means if the concentration doubles, the rate will quadruple. Just simple exponential math.

thanks~