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If you're driving down the highway and want
to see how fast you're going, you look down

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at your speedometer. If you're in a lab and
want to see how fast a chemical reaction is

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going, it's a little more complicated. In
this video we'll look at a few factors that

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influence the speed of chemical reactions.

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This video is part of the Equilibrium video
series. It is often important to determine

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whether or not a system is at equilibrium,
to do this we must understand how a system's

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equilibrium state is constrained by its boundary
and surroundings.

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Hi, my name is George Zaidan and I'm a graduate
of the chemistry department here at MIT.

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Before watching this video, you should have
a basic understanding of chemical equilibrium.

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After watching this video, you will be able
to:

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Understand how reaction rate is influenced
by reactant concentration

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Explain how reaction rates change as a system
establishes equilibrium, AND

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Predict relative equilibrium concentrations
of reactant and product, based on rates of

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forward and reverse processes.

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Suppose you have this general reaction where
A, B, C, and D are molecules and lowercase

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a, b, c, d are their molar coefficients.
As the reaction progresses, A and B will be

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consumed, and C and D will be formed.

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The speed at which any of these four processes
happen multiplied by the reciprocal of the

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appropriate molar coefficient is called the
rate of this reaction.

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If you want your car to go faster, you step
on the gas, but speeding up a chemical reaction

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is not that simple. There are lots of different
factors that affect reaction rate. Three common

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ones are: the concentration of reactants,
the temperature of the reaction mixture and

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the presence of a catalyst. In this video
we'll focus on the effect of concentration

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on homogenous reactions.

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Most chemical reactions are the result of
two or more molecules colliding with each

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other... but not every collision leads to
a reaction. The molecules must collide in

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the proper physical orientation, and they
must do so with enough energy to break their

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bonds.

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Imagine a beaker with 100 mL of water, in
which you dissolve 106 molecules of reactant

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A and 106 molecules of reactant B. How likely
is the reaction to form C and D? Do a back

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of the envelope calculation to support your
answer. Pause the video.

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So let's see. 100 mL of water is 100 grams
of water, which is 100 over 18.02, about 5.5

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moles of water. That means there are about
3.3x1024 molecules of water in the beaker,

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and only 106 molecules of each reactant...So
in any given volume of this solution, 3.3x1024

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over (3.3x1024 + 106 + 106), or 99.99999999999999994
%, of molecules are nonreactive water!

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Clearly, successful collisions between the
reactants would be very rare in this situation.

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Intuitively, you would expect that the more
concentrated the reactants, the faster the

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reaction, and this is generally true.

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Let's draw a reaction coordinate diagram to
better understand our hypothetical reaction.

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Let's assume that the reaction is exothermic,
in other words, that energy is given off as

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the reaction progresses, so we draw the curve
like this.

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The x-axis is the progress of the reaction,
and the y-axis is potential energy.

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Here are the reactants, and here are the products.

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This hump is the activation energy, the point
of highest potential energy of the reaction.

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Two molecules must collide with at least this
amount of energy in order to successfully

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react.

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Do you notice anything about this diagram?
There's no indication of directionality; in

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other words, there's no reason that the reaction
couldn't proceed backwards just as well as

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forwards.

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You're probably used to thinking of chemical
reactions as processes that happen in only

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one direction (forward), but most reactions
are actually reversible. There are a few exceptions,

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for example combustion. *You can't un-burn
a match. But most reactions that chemists

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carry out in the lab or that happen in our
bodies are reversible.

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Draw your own reaction coordinate diagram
and label the forward and reverse reaction

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paths. Suggest what the relationship might
be between the activation energy and the relative

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rates of the forward and reverse reactions.
Pause the video.

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This is the activation energy for the forward
reaction. The higher this activation energy,

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the slower the forward reaction, because the
number of reactant molecules with sufficient

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energy to react when they collide will decrease.

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But now let's look at this picture in reverse.
The reverse activation energy is not the same

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as the forward activation energy! In this
case, it's higher. So that means that the

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reverse reaction will be slower than the forward
reaction.

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If we were to somehow increase the activation
energy, both the forward and the reverse rates

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would slow down, but the relationship between
them -- namely that the forward rate is faster

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than the reverse rate -- would be preserved.

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Here are four other diagrams, each for a different
hypothetical reaction. Predict the relative

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rates of the forward and reverse reactions
in each of these cases. Pause the video.

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In

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real life chemical processes, the rates of
the forward and reverse reactions are often

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very different. Many of the reactions that
you may have previously thought of as "irreversible"actually

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just have wildly different forward and reverse
rates. For example, you've probably seen the

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dissociation of a strong acid in water written
like this.

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But really, it's this. In this case, the forward
reaction is many orders of magnitude faster

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than the reverse reaction, so we write it
as just the forward reaction.

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So far we've seen that concentration and activation
energy can each independently affect the rate

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of a chemical reaction. The higher the concentration
of a reactant, the faster the reaction; and

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the lower the activation energy, the faster
the reaction. But there's a twist... Thinking

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about the hypothetical reaction A and B yields
C and D. Suggest what that twist might be.

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Pause the video.

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There's no reason that the activation energy
of this reaction would change as the reaction

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progresses. But that's not true for the concentrations:
as A and B are converted to C and D, the concentration

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of all four species change! And as the concentrations
change, the reaction rate also changes.

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To see if we can understand how the reaction
rate changes as the reaction progresses, let's

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go back to the reaction coordinate diagram.
Let's start by considering both the forward

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and reverse reactions as if they're completely
separate.

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Here are A and B reacting to form C and D.
Let's assume this forward reaction is relatively

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fast. As more and more A and B get converted
to products, their concentrations decrease,

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and the initially fast reaction rate slows
over time. Here are C and D reacting to form

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A and B. This reaction would be slower than
this one, but as C and D react, their concentrations

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decrease with time and the reaction becomes
even slower.

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Except that in reality, these aren't two separate
reactions. The products of one reaction are

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the reactants for the other.
Let's now consider the forward and reverse

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reactions at the same time. At the very very
beginning, only A and B are present -- no

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C or D. So the initial rate of the forward
reaction will be relatively high, since the

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concentrations of A and B are high. The reverse
reaction can't happen at all yet, because

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there is no C and D present, so its initial
rate is zero.

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As the reaction proceeds, two things happen:
the concentrations of A and B decrease, and

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the concentrations of C and D increase. So
the forward reaction rate slows, and the reverse

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reaction rate speeds up.
Eventually, we reach a point where the rates

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of the forward and reverse reactions are the
same: this is the definition of a dynamic

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chemical equilibrium. When this happens, the
concentrations of A, B, C, and D stop changing

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with time. Be careful not to confuse equilibrium
with "no reaction."A and B are still reacting

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to form C and D; and C and D are still reacting
to form A and B. But the rate of formation

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of products equals the rate of disappearance
of reactants and vice versa. That means that

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the concentrations of A, B, C, and D don't
change with time.

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Let's go back to our reaction coordinate diagrams
for our four hypothetical reactions. We've

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already worked out the relative rates of the
forward and reverse reactions in each case.

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Using these relative rates, see if you can
predict the relative concentrations of the

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reactants and products at equilibrium. Pause
the video.

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-Endo, high Ea: lots of reactant, not much
product

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-Exo, high Ea: lots of product, not much reactant

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-(In both of these cases, the forward rate
is the same as the reverse rate...)Same energy,

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high Ea: ratio of product/reactant will depend
on specific reaction Same energy, low Ea:

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ratio of product/reactant will depend on specific
reaction

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Now what's the difference between the last
two scenarios? In each case, the forward reaction

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rate will be the same as the reverse reaction
rate. But the rates for this scenario will

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be much faster than for this one... so equilibrium
will be reached much more quickly here than

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here!
The most important thing to understand about

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reaction rates and equilibrium is that just
because the forward and reverse rates are

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the same DOES NOT mean that the concentrations
of the reactants and products are the same.

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Some equilibria, like the dissociation of
a strong acid in water, strongly favor the

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products. Others strongly favor the reactants.
Many of the reactions that keep us alive are

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equilibria, and our body goes to a great deal
of effort to make sure that the position of

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equilibrium heavily favors one side or the
other.

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We hope this video has helped you understand
the relationship between reaction rates and

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equilibrium. We saw that, in general, the
rate of a reaction decreases as reactant concentration

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decreases.

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And while we may think of reactions as irreversible,
most are actually reversible, it's just that

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we "see"the faster of the two.

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As the forward reaction rate decreases, the
reverse reaction rate increases until equilibrium

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is established. At equilibrium, the rate of
the forward reaction equals the rate of the

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reverse reaction, but this does not imply
anything about the equilibrium concentrations

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of the products and reactants.