Order and mechanism

[chiddler]

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So consider a reaction that is second order with two first order reagents:

Rate = k[A]

2nd order overall. Generally, first order is a decomposition of the reagent (dissociative, as BR calls it) and second order is a collision reaction (associative).

This mean that the rate limiting step is some collision in this reaction even though both reagents are first order, right?

 

[chiddler]

What I was trying to get at was understanding the mechanistic differences between Rate=k[A] and Rate=k[C]^2. Answering myself. Correct me if i'm wrong.

The difference is if it's Rate=k[A], then the rate limiting step is a decomposition of A and B. There is a collision step, but it is not rate limiting. Comparing to Rate=k[C]^2, then the rate limiting step is a collision. The mechanism may include some decomposition, but it is not rate limiting.

New question: What about rate = k [A]^2?

 

[MedPR]

What I was trying to get at was understanding the mechanistic differences between Rate=k[A] and Rate=k[C]^2. Answering myself. Correct me if i'm wrong.

The difference is if it's Rate=k[A], then the rate limiting step is a decomposition of A and B. There is a collision step, but it is not rate limiting. Comparing to Rate=k[C]^2, then the rate limiting step is a collision. The mechanism may include some decomposition, but it is not rate limiting.

New question: What about rate = k [A]^2?

I think you have it backwards. [A] would be limited by collisions, [C]^2 would be limited by decomposition.

 

[chiddler]

I think you have it backwards. [A] would be limited by collisions, [C]^2 would be limited by decomposition.

Yes, came back to correct myself regarding [A]. But [C]^2 limited by decomposition? That doesn't sound right.

Consider X + Y => A + B

X + X = Z (slow)
Z + Y = A + B (fast)

Rate = k[X]^2

a collision is rate limiting and it makes it [X]^2. This is a TBR example with variables instead of compounds.

So both of the reactions I listed are limited by some collision I think.

 

[typicalindian]

So consider a reaction that is second order with two first order reagents:

Rate = k[A]

2nd order overall. Generally, first order is a decomposition of the reagent (dissociative, as BR calls it) and second order is a collision reaction (associative).

This mean that the rate limiting step is some collision in this reaction even though both reagents are first order, right?

I dont thin you can make a conclusion about the rate limiting step unless you are provided with data or are told which step is slower. Also I think the decomposition / collision only applies to the overall rate law and not individual steps right? Like even though each reaction is first order, which are decomposition, the overall reactio is collision right? Not sure if I even answered your question or if I rambled.. Let me know haha

 

[MedPR]

If your rate order equation is [C]^2, then in order for the reaction to proceed backwards, C would have to decompose.. Right?

Now I'm confused about what we're trying to figure out.

 

[chiddler]

If your rate order equation is [C]^2, then in order for the reaction to proceed backwards, C would have to decompose.. Right?

Now I'm confused about what we're trying to figure out.

Yes. So the forward must be a collision it seems.

My question is what is the difference in mechanism between a [A] and a [C]^2 reaction? The C^2 must be a bimolecular collision. What can be said about [A]?

 

[chiddler]

I dont thin you can make a conclusion about the rate limiting step unless you are provided with data or are told which step is slower. Also I think the decomposition / collision only applies to the overall rate law and not individual steps right? Like even though each reaction is first order, which are decomposition, the overall reactio is collision right? Not sure if I even answered your question or if I rambled.. Let me know haha

I think you can make a conclusion about rate limiting without data, such as Rate = k[C]^2.

Decomposition/collision can apply to only overall rate law? I don't know. This is what i'm trying to figure out. Your question is exactly what i'd like to know:

"Like even though each reaction is first order, which are decomposition, the overall reactio is collision right?"

 

[typicalindian]

I think you can make a conclusion about rate limiting without data, such as Rate = k[C]^2.

Decomposition/collision can apply to only overall rate law? I don't know. This is what i'm trying to figure out. Your question is exactly what i'd like to know:

"Like even though each reaction is first order, which are decomposition, the overall reactio is collision right?"

Haha that quote makes me sound like a ditz. Guess that's what I get for sdn-ing in class haha

 

[MedPR]

Yes. So the forward must be a collision it seems.

My question is what is the difference in mechanism between a [A] and a [C]^2 reaction? The C^2 must be a bimolecular collision. What can be said about [A]?

Oh, I had myself turned around. rate law isn't the same as equilibrium equation, lol. Ok so if your rate law is rate=k[A] that means it depends on two reactants, as you stated. Obviously the rate will increase as you increase both A and B as long as the container can support the reaction.

So more A and B = more collisions between A and B = faster reaction rate. Regardless of the mechanism, A and B must collide for the reaction to go forward. If you are saying the rate equation is [A], then A and B are both involved in the slow step. If there is a slower step that doesn't involve A and B, then rate=k[A] is the incorrect rate law equation.

The [C]^2 doesn't have to be a collision at all. It could be a simple decomposition of [C] under say acidic conditions. You can collide [C] with a ton of stuff at a really high rate, but if those things don't cause [C] to decompose, the reaction won't proceed.

 

[chiddler]

Oh, I had myself turned around. rate law isn't the same as equilibrium equation, lol. Ok so if your rate law is rate=k[A] that means it depends on two reactants, as you stated. Obviously the rate will increase as you increase both A and B as long as the container can support the reaction.

So more A and B = more collisions between A and B = faster reaction rate. Regardless of the mechanism, A and B must collide for the reaction to go forward. If you are saying the rate equation is [A], then A and B are both involved in the slow step. If there is a slower step that doesn't involve A and B, then rate=k[A] is the incorrect rate law equation.

The [C]^2 doesn't have to be a collision at all. It could be a simple decomposition of [C] under say acidic conditions. You can collide [C] with a ton of stuff at a really high rate, but if those things don't cause [C] to decompose, the reaction won't proceed.

Ok. So [A] must be a collision somewhere in the mechanism. This is understandable and explains why increasing either A or B increases total rate.

But C^2 being a decomposition? What comes to mind is radioactive decay that is 1st order. How can it be second order?

I guess it can be

C + C => A (slow)
A => X + Y (fast)

yes this looks right. Slowest step must be forming a bond in a second order. Slowest step is decay in a first order. This says nothing about overall mechanism but only the rate limiting.

 

[MedPR]

Ok. So [A] must be a collision somewhere in the mechanism. This is understandable and explains why increasing either A or B increases total rate.

But C^2 being a decomposition? What comes to mind is radioactive decay that is 1st order. How can it be second order?

I guess it can be

C + Acid => A (slow)
A => X + Y (fast)

yes this looks right. Slowest step must be forming a bond in a second order. Slowest step is decay in a first order. This says nothing about overall mechanism but only the rate limiting.

Yea, rate law equations are only about the slowest step as far as I know.