Pre-Equilibrium Help  [ENDORSED]

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Ryan McDonough 4E
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Joined: Fri Sep 25, 2015 3:00 am

Pre-Equilibrium Help

Postby Ryan McDonough 4E » Sat Mar 12, 2016 4:54 pm

The pre-equilibrium approach still confuses me. I am trying to work on a past quiz, and I need to use the pre-equilibrium approach to justify that the rate law for CHF3(g)+F2(g) (yields) CF4+HF equals: Rate=k[F2][CHF3]. Can somebody please help me figure this out? Thank you.

Jessica Manzano 1B
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Joined: Fri Sep 25, 2015 3:00 am

Re: Pre-Equilibrium Help  [ENDORSED]

Postby Jessica Manzano 1B » Sat Mar 12, 2016 6:13 pm

My quiz has a slightly different version but it's the same idea:
CHBr3(g) + Br2(g) -> CBr4(g) + HBr(g)

Use the pre-equilibrium approach to justify if the following mechanism is consistent with the observed rate law:
Rate= k[Br2][CHBr3]

Step 1 (fast): Br2(g) -> 2Br(g)

Step 2 (slow): CHBr3(g) + 2Br(g) -> CBr4(g) + HBr(g)

Steps for the pre-equilibrium approach:

1) Write the overall rate law for the reaction. The slow step is the rate determining step, so you would write the rate law for the slow step:
rate = k1[CHBr3][Br]2

However, this rate law has the intermediate [Br]2 in it, and the rate law cannot have the intermediate in it. So...

2) Set the rates of the forward and reverse reactions of the fast reaction equal to one another:

Notice that this equality has the intermediate [Br]2 in it, so you can solve for it:

3) Plug the substitute for the intermediate you just found back into the slow step's rate and you have now justified the observed rate law:

Zane Mills 1E
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Re: Pre-Equilibrium Help

Postby Zane Mills 1E » Sat Mar 12, 2016 6:46 pm

yewww so pre eq approach is only when the fast step is first and slow (rate determining) is second [which we have here]. so you have the observed rate of rate = k [F2] [CHF3] (and by looking at the RXN's, you can determine that your intermediate is 2F. Basically what you're trying to do is find the rate for step 1 and step 2, solve for your indeterminate, plug it back in, and make sure it matches the observed rate law, like this:

Step 1: F2 (g) --> 2F (g) (fast) Rate=k1[F2]
Step 2: CHF3 (g) + 2F (g) --> CF4 (g) + HF (g) (slow) Rate=k2[CHF3][F]^2

now you just find the forward and reverse rates for your fast step (since it's in equilibrium you can set them equal to each other)

Forward rate = k1[F2] Reverse rate = k-1[F]^2 (set them equal to each other and...) k1[F2]=k-1[F]^2

now you wanna solve for [F], so you take the forward rate (left side) and divide it by your reverse rate constant (k-1)
Next you take the square root of both sides in order to achieve your desired [F]=(k1/k-1)^0.5 [F2]^0.5

Now you can go back and plug this in for [F] in your slow step equation (you remember the rate=k2[CHF3][F]^2

One of the most confusing things is all the rate constants you have, good news is since you haven't determined any of them yet, you can just substitute one fat k for all of them. After you plug your [F] into that equation, you should have:

rate=k2(k1/k-1)^0.5 ([F2]^2)^0.5 [CHF3] and like I said just replace that k junk with k and your square root and square cancel out to give you a nice rate equation of

rate = k [F2] [CHF3] Which matches your observed rate law and justifies it for the given mechanism.

Know this is long but I hope it helps!

Alex Chiodo Ortiz 3G
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Re: Pre-Equilibrium Help

Postby Alex Chiodo Ortiz 3G » Sat Mar 12, 2016 10:34 pm

why isn't water included in the rate law?

Posts: 92
Joined: Wed Sep 21, 2016 2:59 pm

Re: Pre-Equilibrium Help

Postby 704709603 » Sat Feb 18, 2017 9:20 am

What if there are more than two steps in the mechanism? Also, how did you go from rate = k1 [CHBr3] ((k2[Br2])/k' 2) to rate = k [CHBr3][Br2]?

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Joined: Wed Sep 21, 2016 2:57 pm

Re: Pre-Equilibrium Help

Postby Ana_Andrade_1A » Mon Feb 20, 2017 11:13 pm

Alex Chiodo Ortiz 3G wrote:why isn't water included in the rate law?

Water does not determine or limit the rate since reactions with water as a solvent have H2O, H3O+, and OH- in the solution throughout the reaction making its presence fixed. The relative concentration of water is also temperature dependent while k is for a constant temp. Since there is such a larger concentration of water present compared to the solute, it is generally omitted.

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