Practice English Speaking&Listening with: Multistep reaction energy profiles | Kinetics | AP Chemistry | Khan Academy

- [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.