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.