Kinetics question

[ASDIC]

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Given the following data for the reaction A + B ==> C, what is the rate-law expression?

Trial Initial [A] Initial Initial rate of formation of [C]
1

0.10 M​

0.10 M​

2.0 x 10^-4 M/s​

2

0.20 M​

0.10 M​

8.0 x 10^-4 M/s​

3

0.40 M​

0.20 M​

2.56 x 10^-2M/s​

a) rate = k[A]25
b) rate = k[A]32
c) rate = k[A]23
d) rate = k[A]22

 

[dnelsen]

I'm pretty sure it is c

If you look at trial one and two you can easily derive [A]:
you double A and the rate goes up four-fold so it is second order

If you look at trial two and three you can get :
you double A again so you expect a four-fold rate increase due to A, so you factor that out and then you see what B is responsible for and it turns out to be an 8 fold increase so it is third order in respect to B


Anyone?

 

[imsotired]

C

 

[bgreet]

C

For the rate law of B, 5 just seems like way too much and its definantly more than 2

 

[princessd3]

Isn't the rate law 2nd order with respect to A. Look at trial 1 and 2. When A was doubled the rate went up by 2^2. From 2 to 8.

 

[princessd3]

Disregard my post.

 

[MoosePilot]

Don't think there's the right info. They never hold A constant and vary B, so how can you derive the rate law for B?

 

[captain hazel]

look at dnelson's explanation. even though they don't ever hold [A] constant, you know it's second order for A. so you can compare any two trials where varies (1&3 or 2&3); you just have to divide out however much of the change in rate the change in [A] is responsible for.

so like comparing 1 & 3:
trial 3 was 128 times faster
[A] was up by 4, so caused rate to increase by 4^2=16
so increase due to B doubling is 128/16=8
increasing by 2 causes rate to increase by factor of 8=2^3

So it's third order with respect to B

👍

 

[MoosePilot]

look at dnelson's explanation. even though they don't ever hold [A] constant, you know it's second order for A. so you can compare any two trials where varies (1&3 or 2&3); you just have to divide out however much of the change in rate the change in [A] is responsible for.

so like comparing 1 & 3:
trial 3 was 128 times faster
[A] was up by 4, so caused rate to increase by 4^2=16
so increase due to B doubling is 128/16=8
increasing by 2 causes rate to increase by factor of 8=2^3

So it's third order with respect to B

👍

Ah, I see. I never saw one where you needed to apply the rate you determined for A to find the rate for B. I think that's more work than will be on the test.

 

[ASDIC]

This question is tricky because they dont hold A constant. The correct answer is C.

Kudos to the ones who got it rite!

captain hazel has the correct explanatio

 

[willthatsall]

I got C. You know the ones I reall suck at though? The ones that are show like four steps to a reaction and then ask what the overall rate is. And you have to plug in different equilibrium reactions for the intermediate products. Anyone know what I'm talking about? I've seen it a few times in the Kaplan stuff.

 

[DieselPetrolGrl]

I got C. You know the ones I reall suck at though? The ones that are show like four steps to a reaction and then ask what the overall rate is. And you have to plug in different equilibrium reactions for the intermediate products. Anyone know what I'm talking about? I've seen it a few times in the Kaplan stuff.

dude..can u post a sample?

feeling scared/ lost


thanx <3

 

[DieselPetrolGrl]

Ps: i am in kaplan but have been cutting class/ find the lesson book a wasteful simplification
plus personal problems as of late so if its lesson 3 anything im sorry - i will look it up in kaplan book if u dont want to post it

 

[willthatsall]

It's on full length 6 problem #51. I can't figure out how to copy and paste off the pdf (if it's possible) so I'm not going to copy all the reactions and everything. I've seen it in another place too, but don't remember where.

 

[MoosePilot]

It's on full length 6 problem #51. I can't figure out how to copy and paste off the pdf (if it's possible) so I'm not going to copy all the reactions and everything. I've seen it in another place too, but don't remember where.

If it's what I'm thinking (although I've shot my credibility in this subject, so watch out), then you just determine the rate for each step and set up the generalized equation. Then substitute numbers in from any experiment on the chart, any one of them will work.

Make sense?

 

[imsotired]

This one?

Consider a reaction that proceeds by four steps:
(fast) A2 + BC &#8594; A2C+ B
(slow) B+ BC &#8594; B2C
(fast) A2C + B2C &#8594; A2C2+ B2
(fast) A2C2 + B2 &#8594; AB2 + AC2
What is the expression for the rate of the overall reaction?
A. k[A2][BC]2
B. k[BC]
C. k[A2][BC]
D. k[A2C][A2][BC]

if you want to copy with a pdf file in adobe, just click on the select text button on the tool bar. 😉

edit: hm.....thought i had to right answer until you said something about intermediates....I think it's A...not sure (I don't have the answers)

 

[willthatsall]

If it's what I'm thinking (although I've shot my credibility in this subject, so watch out), then you just determine the rate for each step and set up the generalized equation. Then substitute numbers in from any experiment on the chart, any one of them will work.

Make sense?

Nah, that's not what it is, there are no numbers. It's just a bunch of letters, say it's like 4 different steps to a reaction. And it will tell you which one is the slow step (rate determining) and I guess that is the overall reaction rate. But since the slow step has intermediate products involved as reactants, you have to replace them with original reactants and you have to do that by substituting equilibrium equations. That's not a great description, but like I said, I'm not good at those. Anyone know the ones I'm talking about?

 

[dnelsen]

This one?

Consider a reaction that proceeds by four steps:
(fast) A2 + BC &#8594; A2C+ B
(slow) B+ BC &#8594; B2C
(fast) A2C + B2C &#8594; A2C2+ B2
(fast) A2C2 + B2 &#8594; AB2 + AC2
What is the expression for the rate of the overall reaction?
A. k[A2][BC]2
B. k[BC]
C. k[A2][BC]
D. k[A2C][A2][BC]

if you want to copy with a pdf file in adobe, just click on the select text button on the tool bar. 😉

edit: hm.....thought i had to right answer until you said something about intermediates....I think it's A...not sure (I don't have the answers)

I'm pretty sure it is A as well. You want to look at the reactants in the slow step; however, in one of the reactants (B) is a product of the first step, so you have to take that into account. B if formed by A2 + BC, so the overall rate reaction is [A2][BC][BC] (you get this by substituting in A2 and BC for B) which is the same as option A.