- 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
- Types of catalysts
- Types of catalysts
Many chemical reactions have mechanisms that consist of multiple elementary steps. The energy profile for a multistep reaction can be used to compare the activation energies of the different steps and identify the rate-determining step. The energy profile can also be used to determine the overall change in energy for the reaction. Created by Jay.
- [Instructor] Let's consider a reaction with the following multi-step mechanism. In step one, A reacts with BC to form AC plus B. And in step two, AC reacts with D to form A plus CD. If we add the two steps of our mechanism together we can find the balanced equation for this hypothetical reaction. So we're gonna put all of our reactants on the left side here, and we're gonna have all of our products on the right side. And we can see that AC is on the left and it's on the right side, so we can cancel that out. A is also in the left and the right side, so we can cancel that out. So the overall equation would be BC plus D goes to B plus CD. We've just seen that BC and D are our reactants and B and CD are the products for this hypothetical reaction. If we look at the mechanism, A is there in the beginning and A is there in the end. But A is not a reactant or a product, therefore A must be a catalyst. Something else that's not a reactant or a product is AC. You notice how AC was generated in the first step of our mechanism, and then AC is used up in the second step of the mechanism. Therefore AC must be the intermediate for this reaction. Next, let's look at the energy profile for this multi-step reaction. Energy profiles usually have potential energy in the y-axis and then reaction progress on the x-axis. So as we move to the right on the x-axis, the reaction is occurring. This first line on our energy profile represents the energy level of our reactants, which are BC and D. So let's go ahead and show the bond between B and C. And then we also have D present. Our catalyst is also present at the very beginning of our reactions. So I'll go ahead and draw in A above our two reactants. We can see in our energy profile that we have two hills. The first Hill corresponds to the first step of the mechanism and the second hill corresponds to the second step. So the peak of the first hill is the transition state for the first step of the mechanism. And we can see in the first step that the catalyst A, is colliding with BC or reacting with BC to form our intermediate AC. So A must collide with BC and at the transition state, the bond between B and C is breaking, and at the same time, the bond between A and C is forming. We would still have reactant D present at the top of this hill too. So I'll go ahead and draw in D here. When a collides with BC, the collision has to have enough kinetic energy to overcome the activation energy necessary for this reaction to occur. And on this energy profile, the activation energy is the difference in energy between the reactants and the transition state, so the very peak of the hill. So this difference in energy, corresponds to the activation energy for the first step of the mechanism which we will call Ea1. If we assume that the collision has enough kinetic energy to overcome the activation energy, we'll form our intermediate AC, and we'd also form B. So let's go ahead and show the bond between A and C has now been formed. So this valley here between our two hills represents the energy level of the intermediate. We would also have be present, so I can go ahead and I'll just write in B here. And then we still have some D present, D still hasn't reacted yet. So I'll go ahead and draw in D as well. Next we're ready for the second hill or the second step of our mechanism. In the second step, AC the intermediate AC reacts with D to form A and CD. So the top of this second hill would be the transition state for this second step. So we can show the bond between A and C braking, and at the same time the bond between C and D is forming. The difference in energy between the energy of the intermediate and the energy of the transition state represents the activation energy for the second step of the mechanism, which we will call Ea2. So AC and D must collide with enough kinetic energy to overcome the activation energy for this second step. If AC and D collide with enough kinetic energy, we would produce A and CD. So this line at the end here represents the energy level of our products. So CD is one of our products, so we'll write that in here. And remember B is our other products, which we formed from the first step of the mechanisms. So let's go ahead and write in here B plus CD. And we also reformed our catalyst, so A would be present here as well. Next let's compare the first activation energy Ea1 with the second activation energy Ea2. Looking at the energy profile we can see that Ea1 has a much greater activation energy than Ea2. So let's go ahead and write Ea1 is greater than Ea2. The smaller the activation energy, the faster the reaction, and since there's a smaller activation energy for the second step, the second step must be the faster of the two. Since the first step has the higher activation energy, the first step must be slow compared to the second step. Since the first step of the mechanism is the slow step, the first step is the rate determining step. Finally, let's Find the overall change in energy for our reaction. So to find the overall change in energy, that's Delta E, which is final minus initial. So that would be the energy of the products minus the energy of the reactants. So the energy level of the products is right here and then the energy level of the reactants is at the beginning. So let me just extend this dashed line here so we can better compare the two. Representing Delta E on a graph, it would be the difference in energy between these two lines. And since the energy of the products is greater than the energy of the reactants, we would be subtracting a smaller number from a larger number and therefore Delta E would be positive for this hypothetical reaction. And since Delta E is positive, we know that this reaction is an endothermic reaction.