- Kinetics questions
- Introduction to reaction rates
- Rate law and reaction order
- Worked example: Determining a rate law using initial rates data
- First-order reaction (with calculus)
- Plotting data for a first-order reaction
- Half-life of a first-order reaction
- Worked example: Using the first-order integrated rate law and half-life equations
- Second-order reaction (with calculus)
- Half-life of a second-order reaction
- Zero-order reaction (with calculus)
- Collision theory
- The Arrhenius equation
- Forms of the Arrhenius equation
- Using the Arrhenius equation
- Elementary reactions
- Reaction mechanism and rate law
- Kinetic and thermodynamic enolates
An elementary reaction is a reaction that occurs in a single step. The rate law for an elementary reaction can be derived from the coefficients of the reactants in the balanced equation. For example, the rate law for the elementary reaction 2A + B → products is rate = k[A]²[B]. Created by Jay.
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- On previous videos, I thought you couldn't write the rate law with the coefficient. Why can you do it in this case?(61 votes)
- The individual steps in a mechanism are elementary reactions. For such reactions, the exponent in the rate law is the coefficient.
Many reactions consist of a number of elementary steps. Since we can't "see" what the steps are, we don't know what exponents to use.
We have to do experiments to determine the rate law, and then we can make guesses about the elementary steps.(66 votes)
- How do we know these reactions, or any reactions in general, are elementary?(14 votes)
- they are one step reactions and mostly can be identified when two or more reactants(maximum 3 as 4 is quite unlikely) participate to give a single product. Also in most cases it will certainly be mentioned in the question.(9 votes)
- how do we know if your reaction is an elementary one step or not?(5 votes)
- You need to run experiments in order to determine if a reaction is elementary. It's reasonably easy to experimentally prove that a reaction is not elementary, but it's harder to prove that it is elementary. We can predict the rate law for an elementary reaction easily, so we know that if a reaction doesn't match this predicted rate law experimentally, it must not be elementary. Also, if we see any experimental evidence of an intermediate (e.g. we find some chemical in our reaction that isn't our reactant or product), then we also know it must occur over at least two steps. We also know that elementary reactions involving 3 reactants are rare and those involving >3 are all but impossible. So, if you have >3 reactants, you can assume the reaction is not elementary.
We can support that a reaction is elementary if:
-The elementary rate law matches the rate law we measure experimentally
-We have no evidence of intermediates
-There are 3 or fewer (usually 1 or 2) reactants
-Computer-based calculations help us determine a reasonable 1-step process
Of course if you're not the one in the lab doing the experiments, you'll just have to trust what the question tells you. You'll have to be told if a reaction is elementary or not. Or, you'll have to be told the experimental data so that you can compare for yourself whether the elementary rate law and experimental rate law match (=it could be an elementary reaction) or don't (=it must not be an elementary reaction).(15 votes)
- I found this video quite confusing in the context of the AP/College Chem course.
1. I don’t recall there being a formal introduction to the concept of an elementary one-step reaction before this. Even if there had been, a quick review here and emphasis on how critical it is to the remainder of this video would have been helpful to me.
2. For each example where the stoichiometric coefficients are used to build the rate laws directly, it would be good to emphasize (repetitively) that doing so is valid only for elementary one-step reactions, and not generally valid.
3. The examples discussed covered only the coefficients/exponents 1 and 2. It’s unclear in the video if the concept is generalizable to any coefficient/exponent. Seeing something like x A + y B -> products yields R = [A]^x + [B]^y would have been super handy.
Thank you.(4 votes)
- My homework just said the exponents in rate law are not determined by the coefficients in a chemical equation, but this video says they are??(3 votes)
- I think that this only refers to elementary reactions (reactions with one step and no intermediaries). Most reactions aren’t elementary, so we can only use this with elementary reactions.(2 votes)
- Why is it that we can take the coefficients as our exponents in these reactions but not others?(2 votes)
- The reaction happens in one step versus a reaction mechanism involving slow-steps, intermediates, catalysts, etc.(2 votes)
- So for first order reactions the increase in concentration increases the rate of reaction but the half life remains constant?(1 vote)
- That's right.
If a 1 mol/L solution takes 30 min to reach 0.5 mol/L, a 2 mol/L solution will take 30 min to reach 1 mol/L, and another 30 min to reach 0.5 mol/L.
The half-life remains constant even though the concentration changes.(3 votes)
- Ok, I still dont see why we can now just use the stoichiometric coefficients as the exponents for some reason when making these rate laws.
For any other rxn that is not elementary, we have to use experimental data. And then the exponent for each reactant/product's concentration is the order for the corresponding reacantant/product. But now we can just magically ignore all of this?(2 votes)
- [Voiceover] Let's say we have a simple elementary reaction where we have only one reactant, A, turning into our products. We can classify this reaction according to its molecularity, which refers to the number of participating molecules. So if we think about one molecule of A giving us our products, this would be a unimolecular reaction. We have only one molecule, so we call this a unimolecular reaction. Next let's think about writing the rate law. So we know when we're writing rate laws, we write the rate of our reaction is equal to the rate constant k times the concentration of our reactants. And here we have only one reactant, so we say times the concentration of A. For the exponents, we can actually take the coefficient in our balanced equation and turn that into the exponent. So we have a one here for our coefficient, so we make that a one right here. So you can only do this for an elementary, one-step reaction. You can't do this for an overall equation with a detailed mechanism like we'll see in the next video. But for these elementary reactions, you can do this. So the rate of our reaction is equal to the rate constant k times the concentration of A to the first power. So for this unimolecular reaction, it's first order in A. Next, let's look at another reaction. A, one molecule of A plus one molecule of B gives us our products. Here we have two molecules, two participating molecules. So this is a bimolecular reaction. So I write bimolecular here. So we can think about these two molecules colliding in space. So molecule A is going to collide with molecule B to give us our products. And so it makes sense that the rate of the formation of our products depends on how frequently A and B collide. And that depends on the concentration of A and B. If you increase the concentration of A and B, you increase the frequency of collisions, and therefore you increase the overall rate of your reaction. So when you write your rate law, so the rate of our reaction is equal to the rate constant k times the concentration of A, and since this is an elementary, one-step reaction, we can take the coefficient and turn that into our exponent. So times the concentration of A to the first power, times the concentration of B, and once again, we can take our coefficient, which is a one, and turn that into our exponent. And so now we have the rate law for this bimolecular reaction. Let's look at another bimolecular reaction. This time we have two molecules of A reacting to give us our products. So we could say that this is A plus A gives us our products or we could say that this is two A gives us our products. So either one. If we stick with the first version, we have a one for our coefficient here, and a one for our coefficient here, and so we write the rate law for this bimolecular reaction, the rate is equal to the rate constant k times the concentration of A and we look at our coefficient here which is a one so we make that to the first power, and then times the concentration of A again, and once again we look at our coefficient and we turn that into our exponent. And so that of course will become, this will just be the rate is equal to the rate constant k times the concentration of A, this will be to the second power. A to the first times A to the first is equal to A squared. Or we could have looked at our other version of writing it, a two A, and once again our coefficient would become our exponent. So this is another example of a bimolecular reaction. Finally, let's look at a reaction where we have three participating molecules. So one A plus one B plus one C gives us our products. So one molecule of A plus one molecule of B plus one molecule of C. There are three participating molecules here so we call this a termolecular reaction. And for this to occur in one step, these would all have to collide at the same time. So if we had A, B, and C, they would all have to collide at this point in space at the same time. And this is rare if you think about it, trying to get three molecules to collide at once is pretty difficult to do. So these termolecular reactions are rare, but we can write the rate law. So the rate of our reaction is equal to the rate constant k times the concentration of A and we have a coefficient of one here so this is to the first power, times the concentration of B and once again this would be to the first power, and times the concentration of C and this would also be to the first power. So for these elementary rate laws, for these elementary reactions, we can take the coefficients and turn them into the exponents in our rate laws. Once again, in this next video, you'll see that we can't do that, we can't look at an overall balanced equation with a detailed mechanism and just take the exponents and figure out the rate law. The rate law needs to be determined experimentally.