Rate Laws Question

[dw88]

Hi guys,

 

I don't know the policy on posting individual questions here, or posting Kaplan questions anywhere buuuut I'm going to do it anyways.

 

Let me qualify this by saying I've taken 4th year mathematical biology courses which teach the crap out of this stuff and I can't decode Kaplan's answer lol.

 

The question is determine the rate law for:

Co(NH3)6 + H20 (double arrow) Co(NH3)6 + OH- + H+ (fast step)

OH- + Co(NH3)6 (single arrow -->) Co(NH3)6OH (slow step)

 

I get that h20 doesn't show up in rate laws, but my thinking is the 1st step is fast so it should reach equilibrium ([Co(NH3)6 = const*[OH]*[H]) which could be used to simplify the mass-action rate for the 2nd part.

 

However the Kaplan answer is rate = k[Co(NH3)6]^2 with no mention of [H+] as the answer takes the first step to be one directional.

 

I call shananigans =P, unless somebody can explain this one for me? =)

 

Thanks.

[estairella]

The [H+] does not participate in the reaction (it's a byproduct) so I'm not sure why you would consider it important in the rate law.

 

You're right that [Co(NH3)6] is proportional to [-OH] at equilibrium.

 

So your rate is

constant*[Co(NH3)6][-OH] (second, slow reaction) = constant*[Co(NH3)6]^2

 

/Disclaimer: hahah I'm just making stuff up, seriously it's been about 4 years since I last saw rate law, but it makes sense doesn't it?

[nosuperman]

I haven't done this stuff since the Leafs made the playoffs but shouldn't the overall rate just be the rate of the slow step? Which explains why H+ isn't in there.

 

Now why the double impact of the Co(NH3)6? My de-educated guess would be the rate of formation of [OH]- (as a reactant in the slow step) is the same as the rate of formation of Co(NH3)6 (as both are 1:1 products of the first step) so really its OH + Co(NH3)6 >> Co(NH3)6OH... but as we know from the first step, the rate of formation of OH is the same as Co(NH3)6... so why not just use 2 [Co(NH3)6OH]'s for simplicity?

 

That's my take on it. Could be wrong.

[Purplehaze]

In a reaction: aA + bB (arrow) cC + dD, rate = K [A]^a^b.

 

also, when determining rate laws, you always look at the slow step b/c thats the step that sets the overall rate of the reaction.....

 

So in this question:

Co(NH3)6 + H20 (double arrow) Co(NH3)6 + OH- + H+ (fast step)..ignore this

OH- + Co(NH3)6 (single arrow -->) Co(NH3)6OH (slow step)

 

I'd expect rate = k[Co(NH3)6]^1....not k[Co(NH3)6]^2!...can anyone explain the 2?

[Apixaban85]

In a reaction: aA + bB (arrow) cC + dD, rate = K [A]^a[b ]^b.[/b]

 

also, when determining rate laws, you always look at the slow step b/c thats the step that sets the overall rate of the reaction.....

 

So in this question:

Co(NH3)6 + H20 (double arrow) Co(NH3)6 + OH- + H+ (fast step)..ignore this

OH- + Co(NH3)6 (single arrow -->) Co(NH3)6OH (slow step)

 

I'd expect rate = k[Co(NH3)6]^1....not k[Co(NH3)6]^2!...can anyone explain the 2?

 

I believe the bolded statement is only true if aA + bB --> cC + dD is the slow step of the reaction or the ONLY step of the reaction.

 

I'm guessing the ^2 portion of the rate law comes from the 1st reaction being in equilibrium.

[dw88]

While I do agree that

 

rate = k[Co(NH3)6]*[OH-] is a answer (as aA+bB --> stuff would indicate on the slow step), it cannot be the correct answer since [OH-] is a intermediate in the reaction and thus shouldn't exist in the rate.

 

Therefore you need to go to the first reaction to get [OH-]. To me this would be k1[Co(NH3)6]/(k2*[H+]) since h2o plays no role in these equations since [H2O] is roughly constant. However, [H+] does change so it *should* appear.

 

Furthermore - the Kaplan site uses k[Co(NH3)6] as the [OH-] which would give units of (some measure of concentration / s^2) which is no longer a rate but rather an acceleration which adds to the nonsense that the answer seems to be.... I'm going to consider this answer void unless somebody can give me a reason that [OH-] doesn't depend on [H+] lol.

[estairella]

You know after thinking about this further and taking some time to review basics, I'm inclined to possibly agree with you.

 

rate = [Co(NH3)6]^2 / [H+]

 

The only thing is, what is the rate:

 

H2O <-> H + OH

 

See, ignoring the H2O, there is no rate. In the original equation, Co(NH3)6 is merely catalyzing that reaction... which intuitively would make sense that the reaction rate would depend on Co(NH3)6.

[cnb88]

I haven't done this stuff since the Leafs made the playoffs

 

BAHAHAHA.....

 

Awesome:p.... I have nothing to contribute... I just thought this was hilarious. That is all.