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Chemistry library
Course: Chemistry library > Unit 17
Lesson 3: Arrhenius equation and reaction mechanisms- Collision theory
- The Arrhenius equation
- Forms of the Arrhenius equation
- Using the Arrhenius equation
- Collision theory and the Maxwell–Boltzmann distribution
- Elementary reactions
- Reaction mechanism and rate law
- Reaction mechanism and rate law
- The pre-equilibrium approximation
- Multistep reaction energy profiles
- Catalysts
- Types of catalysts
- Types of catalysts
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The pre-equilibrium approximation
The pre-equilibrium approximation is used to find the rate law for a reaction with a fast and reversible initial step. In this method, we first write the rate law based on the slow (rate-determining) step. Then, to eliminate any intermediates from the rate law, we use the fast initial step to solve for the intermediate concentration(s) in terms of reactant and/or product concentrations. Created by Jay.
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At2:13he says, "...we can do that by assuming that the first elementary step in our mechanism comes to a fast equilibrium." How do we know that's a valid assumption to make?(6 votes)
If the reaction is fast, then it'll reach its equilibrium concentrations quickly.
Hope that helps.(7 votes)
- Hello, why do k_2 * k_1 / k_-1 = k ?(5 votes)
(k^2 * k^1)/k-1 would lead t another constant. In the video, he gave the other constant k. The k still stands for (k^2 * k^1)/k-1(7 votes)
Why is k-1 written that way? It's just a subscript/marker and it's not literally subtracting 1 from k, right?(4 votes)- Yeah it's a subscript and it's just there to label the rate constant. In this instance if the forward rate reaction constant is k1, then the reverse will be its opposite; k-1.
Hope that helps.(5 votes)
- I don't understand why it is necessary to use this approximation. Isn't the overall rate just equal to the rate determining step, which in this example would be step 2? And if step 2 is an elementary reaction, why can't we just say rate = k [NOBr2][NO]?(3 votes)
- The slow elementary step here, the rate determining step, includes an intermediate, NOBr2. Rate laws are generally not written with intermediates. Jay explains this at1:50. Conceivably this is done because we are able to change the rate the reaction by varying the concentrations of the reactants, but are unable to directly control the intermediate. So it's more useful to us to have the rate law in terms of the reactants than the intermediates.
Hope that helps.(5 votes)
- can a reaction not have a reverse reaction? Which might be equivalent to say k_{-1}=0. If that is possible, how can we get the overall rate law?(2 votes)
- Technically all chemical reactions are reversible (if they remain in a mixture) and therefore have rate laws. In reality there are certain reactions which favor one side of the reaction so overwhelming that they are practically irreversible. A classic example of this is strong acid/base reactions. But for the purposes of constructing an overall rate law, all elementary reactions have an reverse rate law we can utilize.
Hope that helps.(4 votes)
- Isn't the overall reaction an elementary step already? Wouldn't it be a termolecular reaction? A + A + B -> products or 2A + B -> products?(1 vote)
- So it looks like you're confusing the order of the reaction with molecularity. The order of the reaction just tells us the sum of the chemical species' exponents in a rate law, but molecularity applies to elementary reactions and tells us how many atoms/molecules are colliding in that one step.
So we would say that this reaction is third order overall because of the exponents in the final rate law. But the overall reaction is composed of two elementary reactions. Both steps are bimolecular because they involve the collision of two reactant molecules.
Hope that helps.(2 votes)
- so at5:27with the (k2*k1)/k-1 stuff, did you simplify it to "k" or "K"(1 vote)
I'm pretty sure it's a lowercase k. But the main point is that all three constants combine to form another single constant.(1 vote)
2:13Video transcript
- [Instructor] The
Pre-equilibrium Approximation is used to find the rate law for a mechanism with a fast initial step. As an example, let's look
at the reaction between nitric oxide and bromine. In the first step of the mechanism, nitric oxide combines with
bromine to form NOBr2. And in the second step of the mechanism, NOBr2 reacts with NO to
form our product 2NOBr. NOBr2 is generated from the first elementary
step of the mechanism. And then NOBr2 is used
up in the second step. Since NOBr2 wasn't there in the beginning and it's not there in the end, we call NOBr2 an intermediate. The first step of the mechanism is fast. And the second step of
the mechanism is slow. Since the second step of
the mechanism is slow, this is the rate determining step. And we can write the rate
law for the overall reaction by writing the rate law for
this elementary reaction that makes up step two of our mechanism. So we can write the rate
of reaction is equal to for step two, our rate constant is K2. And we multiply K2, the rate constant, by the concentration of our two reactants, which would be the concentration of NOBr2, and the concentration of NO. Since the coefficients
in our balanced equation are ones for NOBr2 and
one and a one for NO, we can take the coefficients and turn them into
exponents in our rate law. So we can do this because this
is an elementary reaction. However, we can't leave the rate law for the overall reaction, in terms of the concentration
of our intermediate, NOBr2. It's preferable to have rate laws written in terms of the
concentration of our reactants, which were NO and Br2. So we need some way of substituting N for the concentration of NOBr2. And we can do that by assuming that the first elementary step in our mechanism comes to a fast equilibrium. So if we assume the first step
comes to a fast equilibrium, we can use the
Pre-equilibrium Approximation. If we assume that the first step comes to a fast equilibrium,
or a pre equilibrium, we know at equilibrium, the rate of the forward reaction is equal to the rate of
the reverse reaction. So in the forward reaction for step one, NO combines with Br2 to form NOBr2. And in the reverse reaction, NOBr2 breaks apart to form NO and Br2. So if the rate of the forward reaction is equal to the rate of the
reverse reaction at equilibrium, let's go ahead and write the rate laws for the forward and the reverse reaction. The rate constant for the
forward reaction is K1. So we can go ahead and write the rate of the forward
reaction is equal to K1. And our two reactants are NO and Br2. So we have K1 times the
concentration of NO, times the concentration of Br2. Since the coefficients
in our balanced equation are both ones for these two reactants, we can raise the power of
these two concentrations to the first power. Since this is an elementary
reaction, we can do this. And we set this rate
of the forward reaction equal to the rate of the reverse reaction. The reverse reaction has a
rate constant of K minus one, and we have only NOBr2 with
a coefficient of one in it. So we multiply K minus one times the concentration of
NOBr2, to the first power. Next, our goal is to substitute N for the concentration of our intermediate. And so we can divide both sides of the equation by K minus one. So if we decide if we divide both sides of the equation by K minus one, on the right side, K
minus one cancels out. And we get that the concentration
of our intermediate NOBr2, is equal to K one times the concentration
of NO to the first power, times the concentration
of Br2 to the first power, divided by K minus one. Next, we can substitute all of this in for the concentration of our intermediate. That gives us the rate
of reaction is equal to, we still have this K2 in here, so we need to make sure to include it. And we're gonna substitute everything in all of this in for the
concentration of our intermediate. So that would be times K one, times the concentration
of NO to the first power, times the concentration
of Br2 to the first power, divided by K minus one. And then we still have this concentration of NO to the first power. So we have to make sure to
include that in our rate law. Let's think about what we would get if we multiply two constants together and then divide by a third constant. So multiplying K2 times K1, and then we divide by K minus one, that would just give us another constant, which we could just call K. So K is now the rate constant
for the overall reaction. So we have the rate law
for the overall reaction is equal to K, times the concentration of we have NO to the first power
times NO to the first power, which is just the concentration
of NO to the second power. And we still have to include the concentration of
bromine to the first power. So now we have a rate law
for our overall reaction in terms of the concentration
of our two reactants. The rate of reaction is
equal to the rate constant K, times the concentration of NO squared, times the concentration of
bromine to the first power. The experimentally determined rate law matches the rate law that we found using the Pre-equilibrium Approximation. And if you look at the coefficients for the overall equation, there's a two in front of NO
and a one in front of Br2, it might be tempting just to say, "Can't we just take those coefficients and turn them into exponents,
because in this case, they happen to match the
exponent in our rate law?" That's just a coincidence
for this reaction. We can't just take the coefficients
for an overall equation and turn them into
exponents in the rate law. We can only do that for
elementary reactions. Like the elementary reactions in the two steps of our mechanism. It's important to point out that if the rate of the forward reaction is equal to the rate of
the reverse reaction, the concentration of our intermediate, NOBr2 remains constant. And therefore we can use this
Pre-equilibrium Approximation to find the rate law for a reaction with a fast initial step.