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Topic: For rate equations, are AB+A vs 2A+B vs A2+B equivalent?  (Read 2673 times)

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K-Feldspar

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For rate equations, are AB+A vs 2A+B vs A2+B equivalent?
« on: August 19, 2016, 04:22:09 PM »
For rate equations, are AB+A vs 2A+B vs A2+B equivalent?

I.e. if the reaction is dependent on the concentrations of A, do all three of the above give rate=k[A]2.

I think they would NOT be equal as A is a concentration.

For example, if A=0.5 moles/L
• 2A=2*0.5=1.0
• A2=A*A (I think?)=0.5*0.5=0.25
• Note sure how to treat AB
.

FYI, for some context, here is the problem I am working on

I can see the slow reaction in the bottom left box has A+2B - so it is NOT of the form rate=k[A]2[ B]

In the bottom right one we have 2A+B -> D+I , so the rate equation iS of the form rate=k[A]2B

However, in the top right one, We have A2+B -> A2B.

I do not know if this A2, where it is 1 molecule containing 2 A atoms is equivalent to the bottom right one where we had 2 moles of A)

Similarly for the top left one, I do not know what to do when it is AB rather than A and B separately.

Thanks for the help.
« Last Edit: August 20, 2016, 05:02:05 AM by Borek »

mjc123

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Re: For rate equations, are AB+A vs 2A+B vs A2+B equivalent?
« Reply #1 on: August 22, 2016, 07:20:46 AM »
Quote
For rate equations, are AB+A vs 2A+B vs A2+B equivalent?
If AB and A2 are different compounds (not just A+B or 2A) then of course they are not. But as they are not starting materials, you have to determine how their concentrations depend on A and B.
e.g. top left rate = k[AB][A]. But what is [AB]?

docnet

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Re: For rate equations, are AB+A vs 2A+B vs A2+B equivalent?
« Reply #2 on: August 23, 2016, 10:13:44 PM »
The rate law for the overall reaction that consists of a few elementary steps is given by the slowest step which is called the rate determining step. The reaction can only progress as fast as the progress of the slowest step.

Two of those proposed reaction mechanisms in the picture you provided are plausible, I think the one in the bottom right square and the one on the top left.

For the top left reaction mechanism with a fast first step, the overall rate law doesn't agree with the rate law of the rate determining elementary step.

rate law for the slow step is rate=k slow [AB][A]

the given overall rate law is rate=k overall [A]^2[B ]

Because AB is not consumed by the second step as fast it is produced by the first step, the concentration of AB builds up and causes step 1 to happen in reverse, and there is an equilibrium between the reactants and the products. Equilibrium is a condition when the formation of the product happens at the same rate as the breakdown of the product into the reactants.

Therefore,

rate = K forward[A][B ] must be equal to
rate = K reverse[AB] in order for there to be an equilibrium.

Set the two equal to each other. Then, divide both sides by the constant (K reverse)

(K forward/K reverse)[A][B ] = [AB]

Substituting (K forward/K reverse)[A][B ] in the place of [AB] in the rate law for the slow step, you get:

rate = K slow (K forward/K reverse) [A][B ][A]

where K slow (K forward/K reverse) is equal to K overall, and [A][B ][A] is equal to [A]^2[B ]

rate = K overall [A]^2[B ]

the now being equal, this allows us to judge the plausibility of a proposed reaction mechanism.

« Last Edit: August 24, 2016, 01:05:26 AM by docnet »

K-Feldspar

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Re: For rate equations, are AB+A vs 2A+B vs A2+B equivalent?
« Reply #3 on: January 20, 2018, 09:57:05 PM »
Hi Docnet,

Thank you, that makes perfect sense.

Sorry about the late response. I for some reason, forgot to re-check for responses after a while. But I came across the same exercise in my book again and found the thread as I was just as confused.