who knows GENCHM

[xDent09]

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this rate law is hard for me please help:
concentrationof A conc of B rate
1 1*10^10-2 3*10^-3 2*10^-3
2 1*10^-2 6*10^-3 8*10^-3
3 2*10^-2 1.2*10^-2 3.2*10^-2

with kaplan I know u switch to this
concentrationof A conc of B rate
1 1 1 1
2 1 2 4
3 2 4 16

now what is the easiest way to find line 4???????

 

[SayAahh]

You are trying to find the reactant order with respect to which reactant?

 

[SayAahh]

this rate law is hard for me please help:
concentrationof A conc of B rate
1 1*10^10-2 3*10^-3 2*10^-3
2 1*10^-2 6*10^-3 8*10^-3
3 2*10^-2 1.2*10^-2 3.2*10^-2

with kaplan I know u switch to this
concentrationof A conc of B rate
1 1 1 1
2 1 2 4
3 2 4 16

now what is the easiest way to find line 4???????

The way I solved it was like this; So it's much easier to eyeball the order with respect to reactant B... (Lines 1 and 2) Hold A constant, and while B doubles, rate quadruples. This makes B 2nd order.
Now for A, (lines 1 and 3) you see that as concentration of A doubles, reaction rate increases by a factor of 16. But you don't know, yet, what the contribution of reactant B is because none of your data holds B constant. So since we now know reactant order of B (which is 2), you determine that B, which quadruples from lines 1 to 3 would increase by factor of 16 (4^2 = 16). Therefore, ALL the effect on rate was contributed to by B and you can conclude that A is 0 order (i.e. changes in concentration of A do not influence the reaction rate). Makes sense?

 

[klutzy1987]

this rate law is hard for me please help:
concentrationof A conc of B rate
1 1*10^10-2 3*10^-3 2*10^-3
2 1*10^-2 6*10^-3 8*10^-3
3 2*10^-2 1.2*10^-2 3.2*10^-2

with kaplan I know u switch to this
concentrationof A conc of B rate
1 1 1 1
2 1 2 4
3 2 4 16

now what is the easiest way to find line 4???????

I do not understand the question you are asking at all can yah try and ask it a little but clearer??

 

[xDent09]

sorry yeah the question was not clear, here it is please help! Also here is a full practice test for Genchem enjoy it, and the problem im looking for is number 2?http://www.kaptest.com/pdf_files/webcaf/tests/514t.pdf

 

[Willem]

Is this part of a study guide you paid for? If so I would remove it. I know its just a link (and thats weird since it seems to be freely available) but it is copyrighted.

 

[Streetwolf]

#2

When [A] doubles you get the rate doubling as well which means [A] is to the first power.

Now look at the bottom two rows. You see that [A] doubles yet again and the rate is multiplied by 18. You know that when [A] doubles the rate doubles so really the effect of [B.] must be to multiply the rate by 9 ( [A] takes care of the other 2 and 9x2 = 18).

So when [B.] is tripled the rate goes up by 9 which is a square relationship. So [B.] gets squared.

You have [A]^1 [B.]^2 which is third order.


PS I write [B.] because the forum would mistaken it for the BOLD font without the period.

 

[SayAahh]

#2

When [A] doubles you get the rate doubling as well which means [A] is to the first power.

Now look at the bottom two rows. You see that [A] doubles yet again and the rate is multiplied by 18. You know that when [A] doubles the rate doubles so really the effect of [B.] must be to multiply the rate by 9 ( [A] takes care of the other 2 and 9x2 = 18).

So when [B.] is tripled the rate goes up by 9 which is a square relationship. So [B.] gets squared.

You have [A]^1 [B.]^2 which is third order.


PS I write [B.] because the forum would mistaken it for the BOLD font without the period.

I would have to disagree unless you're addressing another question. When conc. of A does NOT change (lines 1 to 2), the rate still quadruples; A cannot be 1st order. Additionally, when A doubles (lines 1 to 3), rate increases by factor of 16; if it were indeed 1st order, you'd expect it to increase by a factor of 4.

All the effect on rate is contributed to by reactant B.

 

[iverson3504]

wouldn't the answer to this question be C, second order? the order with respect to B would be 2nd

 

[bigstix808]

Nm

 

[Streetwolf]

I would have to disagree unless you're addressing another question. When conc. of A does NOT change (lines 1 to 2), the rate still quadruples; A cannot be 1st order. Additionally, when A doubles (lines 1 to 3), rate increases by factor of 16; if it were indeed 1st order, you'd expect it to increase by a factor of 4.

All the effect on rate is contributed to by reactant B.

A moderator merged two threads by the OP. The second thread had a link to a PDF file with apparently a different question. Here's the post and the link to the question (#2):

sorry yeah the question was not clear, here it is please help! Also here is a full practice test for Genchem enjoy it, and the problem im looking for is number 2?http://www.kaptest.com/pdf_files/webcaf/tests/514t.pdf

 

[2009er]

this rate law is hard for me please help:
concentrationof A conc of B rate
1 1*10^10-2 3*10^-3 2*10^-3
2 1*10^-2 6*10^-3 8*10^-3
3 2*10^-2 1.2*10^-2 3.2*10^-2

with kaplan I know u switch to this
concentrationof A conc of B rate
1 1 1 1
2 1 2 4
3 2 4 16

now what is the easiest way to find line 4???????

once u have switched the numbers u can see that in trial 2 when only the conc of B is doubled (2) the rate quadruples(4), there for B is second order, and from trial three u know ur doubling the con of B vs. trial 2 therefore the rate should quadruple (from 4 to 16), since the rate is 16 that means that conc of A plays no role in the rate law and there for the rate law is:
= k [A]^0 ^2 = k ^2

 

[xDent09]

however do u come with line 4

 

[hoss19]

however do u come with line 4

xdent - what do you mean when you say "line 4"?

do you mean to ask how kaplan the 4th vertical line (representing rxn rates) that you entered on the first post.....

1
4
16

Not sure what you mean here.....

 

[hoss19]

this rate law is hard for me please help:
concentrationof A conc of B rate
1 1*10^10-2 3*10^-3 2*10^-3
2 1*10^-2 6*10^-3 8*10^-3
3 2*10^-2 1.2*10^-2 3.2*10^-2

with kaplan I know u switch to this
concentrationof A conc of B rate
1 1 1 1
2 1 2 4
3 2 4 16

now what is the easiest way to find line 4???????

i am not a great with math, but sometimes these problems require a few simple calculations. it didnt take too long, less than 1 min...here is how i did it (once i found that b is 2nd order)

# A B RxnRate
1 10 3 2
2 10 6 8
3 20 12 32

Line 3 / Line 1:

32 = k*20^y*12^2
___________________

8 = k*10^y*6^2

factoring out, cancellng k, and cross multiplying gives you:

32 / 32 = 20^y / 10^y

y must therefore be 0

 

[Streetwolf]

The simplest way is to find 2 lines where A stays the same and see what happens to B and what happens to the rate. If B changes and the rate stays the same then B is 0 order. If B and the rate change equally then it's 1st order. If the rate changes as a function of B^2 then B is second order (e.g. if B triples and the rate is multiplied by 9).

Once you figure out B, try to find 2 lines where B stays the same and see what happens to A and the rate. Use the same method as above.

The only tricky part is when you can't find a line like that. For example say B is 2nd order which means when B increases by some amount, the rate increases by the square of that amount. Then say each line has a different value of B so that you can't hold it constant. You just have to factor in the effect of B on the rate. If B doubles between two lines (for example) then the rate must quadruple as a result. So if the rate between two lines is multiplied by 16 (for example) and B is multiplied by 2, that means B's effect on the rate is multiplying it by 4. Since the rate was actually multiplied by 16, that means A must have the effect of multiplying the rate by 4 as well (4x4 = 16). So then look to see what A does but keep in mind that A only quadruples the rate.

I know it's confusing but go through that slowly.