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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: kelaklub on May 11, 2006, 11:23:36 AM

Title: Rate Law
Post by: kelaklub on May 11, 2006, 11:23:36 AM
I have uploaded the sample test problem (really bad scan)...

They answer given is A, I think it's typo and should be E. What do you think? Thanks.
Title: Re: Rate Law
Post by: Borek on May 11, 2006, 11:38:55 AM
Looks like typo.

Make it B ;)
Title: Re: Rate Law
Post by: kelaklub on May 11, 2006, 11:51:23 AM
Hmm, I thougt it was E because I determined y from y from trails 2 and 3, and setting up my equation as...

0.5y = 1

=> y = 0

Is this not right?
Title: Re: Rate Law
Post by: kelaklub on May 11, 2006, 11:54:23 AM
Doooh, I misread the rates from trials 2 and 3, you are right, it is B.

Thanks.
Title: Re: Rate Law
Post by: kelaklub on May 11, 2006, 12:06:42 PM
Oh now that my question is answered, can I remove that scan from the server or leave it on for a day in case a few people want to check out this problem? What would be considered good etiquette?



Title: Re: Rate Law
Post by: Borek on May 11, 2006, 12:53:53 PM
I have edited your post and added image here, so you can delete it from the outside server. Next time click on Additional options... below the edit field and select image to upload.
Title: Re: Rate Law
Post by: Will on May 11, 2006, 01:16:14 PM
I've got a similar problem that I wasn't bothered to start a new thread about, but I might as well post it now that this thread has started!

The question is: A reaction is first order with respect to species A and zero order with respect to B. For each mechanism [see pic], explain whether or not it is consistent with the rate equation (rate=k[A]).

My teacher says Mechanism 2 is consistent with the rate equation, although I would've thought that because two steps containing B happen before the step with A in it (which is the rate determining step), that Mechanism 2 isn't consistent with the rate equation (as it would be second order wrt to B?).
Mechanism 3 might be OK because although the first equation involves two species of A (2nd order), A is regenerated in the next equation, so it may only be first order wrt to A, but I am completely unsure about this and I would appreciate a second (or 3rd, 4th etc!) opinion. :)
Title: Re: Rate Law
Post by: Donaldson Tan on May 11, 2006, 07:53:52 PM
Assuming all elementary reactions with A as a reactant as the rate determining step

Consider mechanism 1, the first equation is the rate determining step. Its rate equation is thus rate = k [ A ] [ B ]

Consider mechanism 2, the last equation is the rate determing step. its rate equation is thus rate = k[ A ] [ S ]

Consider mechanism 3, the first equation is the rate determining step. Its rate equation is rate = k[ A ]2

We require the rate equation to be first order with respect to A, so mechanism 3 is rejected.

We require the rate equation to be zero order with respect to B, so mechanism 1 is rejected.

In order to make sure that mechanism 2 fits the stipulated requirement

1. B + Q -> R     r1 = k1[ B ][ Q ]
2. R + B -> S      r2 = k2[ R][ B ]
3. S + A -> Z      r3 = k3[ S ][ A ]

Steps 1 and 2 are very fast reactions. S is quickly formed from B and Q, so the rate of formation of S has no effect on the rate determining step. This means the rate equation is zero order with respect to B. Mechanism 2 is therefore the only mechanism that meets the requirement.
Title: Re: Rate Law
Post by: Will on May 11, 2006, 08:03:08 PM
Thanks geodome ;).

The one bit I am still slightly confused about is if S is part of the rate determining step, surely it has to be in the rate equation? If you double the conc. of B then I think you'd quadruple the rate of formation of S, which would increase the overall rate, but the rate is zero order wrt to B ???.
Title: Re: Rate Law
Post by: tennis freak on May 11, 2006, 10:14:33 PM
hey guys i to have a question that i can't seem to comprehend, help would be greatly appreciated,.
It asks us for the half life and the order of the reaction

here is the data

Time (days)   0     1   2   3   4  5   6   7  10  20
%Reactant    100 79 63 50 40 31 25 20 10   1
Remaining

we know the half life is 3 days but why is it first order?
Title: Re: Rate Law
Post by: JZ_1 on May 11, 2006, 10:37:47 PM
Plot it

Zero order:
    |
[]  |           Slope = - K
    |_______
       Time

First order
      |
ln[] |              Slope = - K
      |_______
           Time

Second order
       |
1/[] |                 Slope = K
       |________
             Time

Whichever graph gives you a straight line is the order of the rxn.

Also you could use the half-life formula
First order
T1/2= .693/k

K is negative for zero and 1st order and positive for second order
Title: Re: Rate Law
Post by: tennis freak on May 11, 2006, 11:39:42 PM
awsome thanks a bunch, finally somebody who can explain it! :P
Title: Re: Rate Law
Post by: mrdeadman on May 12, 2006, 08:52:49 AM
Quote from: tennis freak on May 11, 2006, 11:39:42 PM
awsome thanks a bunch, finally somebody who can explain it! :P
that's how i explained it to you, when you asked that question in class!  :P
Title: Re: Rate Law
Post by: tennis freak on May 13, 2006, 07:42:33 PM
you did not nobody knew what they should do, and if you did i don't remember, sorry
Title: Re: Rate Law
Post by: rctrackstar2007 on May 14, 2006, 01:13:11 PM
Quote from: tennis freak on May 13, 2006, 07:42:33 PM
you did not nobody knew what they should do, and if you did i don't remember, sorry

ya that's def how we explained it in class lol
Title: Re: Rate Law
Post by: tennis freak on May 14, 2006, 02:24:23 PM
whatever maybe i was just not paying attention, my bad dudes