- 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
Collision theory states that molecules must collide to react. For most reactions, however, only a small fraction of collisions produce a reaction. In order for a collision to be successful, the reactant molecules must collide both with sufficient energy to overcome the activation energy barrier and in the proper orientation to form any new bonds in the products. Created by Jay.
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- what decides the correct orientation?(6 votes)
- The molecules can be very large and each have a small portion that can interact. The correct orientation would be the 2 reactive sites exposed to each other. It's based on sterics.(17 votes)
- In the graphs the X-axis is the reaction progress. Does it have some sort of units, e.g. ms, µs or ns?(4 votes)
- The progression of the reaction is basically the same as the time, so it is in some units of time, whether it be minutes, seconds, etc. They aren't included in this graph because it doesn't really matter, since the video is focused on the collision theory and how it affects chemical reactions.(13 votes)
- At6:41doesn't jay mean an exergonic reaction?(5 votes)
- Actually, 'Exergonic' implies that the change in free energy is negative (i.e. energy flows from the system to the surroundings). But here, we can't determine the change in free energy by looking at the energy diagram (the energy diagram is for enthalpy only).
Hence, it is an exothermic reaction (since we can't speculate about the change in free energy).(7 votes)
- In this case if the energy required for the products to form is 40 KJ/mol and the energy of the products is 10 KJ/mol, where did the other 30 KJ/mol go?(5 votes)
- Well, if the system is closed (no particles can come inside or go out) then this energy would be converted into heat or movement.
If we have the inverse process (the product have more internal energy) then it means that the energy from heat, pressure, friction, electricity... put into the system is converted into chemical energy.(4 votes)
- how do the reactants reach the activation energy ?is it by itself or in other case its supplied through heat to the system ?(2 votes)
- They reach the activation energy by means of collisions.
At a given temperature there is always a distribution of speeds.
Some particles are moving fast, others move more slowly.
But there will always be a few collisions with enough speed that the collision energy is enough to get over the activation energy barrier·
As you supply heat and increase the temperature, a greater percentage of the collisions will have enough energy to get over the barrier and react, so the reaction will be faster.(5 votes)
- Whats the difference between activated complex and transition state if any?(2 votes)
- Transition state theory describes a hypothetical transition state that exists between reactants and products during a chemical reaction. The species formed in this hypothetical transition state is called the activated complex. They are practically the same, but note that there is a key difference.(3 votes)
- Why can't activation energy be negative?(1 vote)
- If activation energy were negative, it would completely invalidate the theory of collisions. Let's say that 2 molecules of different elements collide to form 2 molecules of a compound. Let's talk about this, first analytically and then mathematically.
First, analytically. The idea of an activated complex is a highly energetic state where bonds are being broken and formed simultaneously. This makes sense if the activation energy is positive. But if the activation energy is negative, the activated complex is MORE stable than both the reactants and the products! If that were the case, then the reaction should stop at the activated complex, because that is apparently the most stable configuration of the system. If this were the case, no molecules of any element/compound should be stable because every atom is in limbo - breaking and reforming bonds with other atoms every instant.
If activation energy is negative, that means the reaction keeps on proceeding faster and faster with time. (by Arrhenius equation, k = Ae^(-Ea/RT))
But we know for a fact that a reaction either slows or attains a constant rate as it goes to completion.
I hope, by now, it is clear why activation energy being negative is absurd.(5 votes)
- Why is the energy of products less than the energy of the reactants?(1 vote)
- That is because this example is an exothermic reaction where the products are thermodynamically more stable than the reactants. It could just as easily be the other way around. If the products are thermodynamically less stable than the reactants, the reaction is endothermic.
Please note: A reaction may proceed on its own even if the products are less thermodynamically stable than the reactants.
Reaction mechanics depend on both thermodynamics and kinetics.(3 votes)
- When explaining activation energy Ea at2:45- isn't Ea a property of the reaction itself , and therefore independent of the energy of the reactants? Shouldn't Ea be constant for each reaction, and in this case be 60kJ/mol?(2 votes)
- My understanding is that activation energy is always in reference to the energy of the reactants. Remember the reactant themselves contain energy in the form of their chemical bonds. It takes energy to break bonds (in the form of activation energy), but then energy is released when a bond is broken. I think you are right that Ea is unique to the reaction (not sure if we can call it a property), but it's not independent energetically.(1 vote)
- In that hill-ball analogy , when the ball is at the top of the hill , it's velocity is zero since that enough energy is used. Ball could roll down in both direction. I mean reaction could have not taken place?(1 vote)
- I think you're confunding the analogy with the real concept... When you give enough energy to the system, the mollecules will collide enoughly so that the probability of happening good collisions - that ones which break the B-C bonds, get to the activated complex and then criate the A-B ones - is greater, and so the reaction happens. I hope this can clarify you...(2 votes)
- To think about collision theory, let's consider the following reaction. Here we have atom A reacting with a diatomic molecule B C to form a new diatomic molecule A B and C. According to collision theory, molecules must collide to react. So for this example, atom A has to collide with molecule B C, in order for the reaction to occur. Next, the collisions must have the correct orientation in space to be an effective collision. For example, let's say for this reaction, we have our molecule B-C. Our molecule B-C approaches A in this orientation, and since we're forming a bond between A and B, let's say this is the proper orientation. So this is the way our collision has to occur in order for the reaction to occur. If the diatomic molecule B-C approaches in the opposite direction, so let's say we have our atom A here, and then we have C-B, so the atom C approaches the atom A, here this is not the proper orientation for the reaction to occur. So this would be no. So, there has to be a collision, but the collision has to be in the proper orientation. And finally, collisions must have enough energy. So, if the collision doesn't have enough energy, the molecules, or in this case the atom and the molecule, will just bounce off of each other. If you do have enough energy, the colliding molecules will vibrate strongly enough to break bonds. So let's go ahead and draw this in. We're starting with a certain energy for our reactants. So right here we're going to draw in the energy for our reactants. So we have our atom A, and we have our molecule B-C at this point, and let's say our total energy is 20 kilojoules per mol. When the atom and the molecule collides, they need enough energy. They need enough energy to break this bond between B and C. So we're trying to break this bond in here. And so, we can find that energy on our diagram here. So we're starting with 20 kilojoules per mol, and we need to get up to here, to 60. So this is how much energy we need for the reaction to occur. And we call this the activation energy, which is symbolized by E sub-a here. So this is the activation energy. And the activation energy is important, because this is the minimum amount of energy that's required to initiate a chemical reaction, and for this reaction, we can see we need to get to 60 kilojoules per mol. So this point right here is at 60 kilojoules per mol. We're starting out with 20. So 60 minus 20 would, of course, be 40. So the activation energy for this reaction, so Ea, according to our diagram, is positive 40 kilojoules per mol. So the energy of the collision must be greater than or equal to the activation energy, and at the top right here, we're going to get a transitional structure. So let me go ahead and draw in a possible transitional structure for this reaction. So we have a bond forming between A and B. At the same time, we have a bond breaking between B and C, and we call this transitional structure right here, we call this the transition state. So our structure right here is called the transition state. You might also see this called the activated complex. So, the transition state or the activated complex. And you can see I've drawn in partial bonds here. So the bond between B and C is breaking. At the same time, we have the bond forming between A and B. And so, let's think about an analogy here. Let's say we have a hill. So here's my hill right here, and if we have a ball. Let's say we have a ball right here at this end of the hill. Well, it takes energy to push the ball up the hill. And let's say we have enough energy to get the ball to right here. Well, in that case, it's not enough for the ball to roll down the other side of the hill. Here the ball's going to roll back to the starting position. So that's like thinking about having not enough energy and the molecules just bouncing off of each other. But if you have enough energy, so if you're starting out with the ball right here, and you have enough energy to bring the ball to the top of the hill, so just barely to the top here, the ball can now roll down. And so, the ball's going to end up at the bottom of the hill right here. And that's thinking about formation of your products. So for our example, our products would be our new diatomic molecule A-B. So let me draw that in here. So here we have A-B, and we also have plus C at this point. So this would be our products, and this represents the energy of our products. So let's go here to find that on our diagram. So we have our products here. So what energy is that? So we go over to here, and we find the energy, and let's say that's at 10. Let's say this is 10 right here. So the energy of our products is equal to 10 kilojoules per mol, and the energy of our products for this example is lower than the energy of our reactants. We started out with 20 kilojoules per mol, and we ended up with 10 kilojoules per mol. So to find that difference in energy, to find that change in energy, that would be the energy of the products minus the energy of the reactants, and for this example, the energy of the products is 10 kilojoules per mol, so we have 10, minus the energy of the reactants. We started out with 20 kilojoules per mol. So 10 minus 20 gives us a change in energy equal to negative 10 kilojoules per mol. So on our diagram, we can see this change in energy right here. So it's the change in energy. This right here is a negative change in energy. So I'm writing delta E is equal to, is negative. So this is an exothermic reaction. So heat is given off here. Remember, it takes energy to break bonds, and energy is given off when bonds form, and for this reaction, we're giving off heat. So this represents an energy diagram for an exothermic reaction. Now let's think about another reaction. So down here we have reaction progress. So we're starting out with the energy of our reactants right here. So this is the energy of our reactants at 20 kilojoules per mol, let's say. And we know that to get to this point right here, this represents the energy of the activated complex. So right there will be our activated complex, and that's at 80 kilojoules per mol. So this difference in energy, this is our activation energy. So this is Ea. That represents our activation energy, which for this reaction would be 80 minus 20, which would be equal to 60 kilojoules per mol. So we get to here. We get to the transition state or the activated complex, and then this would represent the energy of our products. So this our energy of our products here, which for this reaction, you can see the energy of our products is greater than the energy of our reactants. And so, if we go over to here, let's say it's at 40. So let's say this level is at 40 kilojoules per mol. So what's the change in energy here? So the change in energy for this reaction, once again, is the energy of the products minus the energy of the reactants, so that would be 40 kilojoules per mol. So, 40 kilojoules per mol minus we started out with 20. So, 40 minus 20 gives us a change in energy equal to positive 20 kilojoules per mol. So, positive 20 kilojoules per mol. So that would be this difference right here. So, our change in energy for this reaction is positive. And so, this represents an endothermic reaction. So the previous example was an exothermic reaction, where heat is given off, and this is an endothermic reaction. This represents an endothermic reaction, where heat is absorbed.