- [Instructor] As an example of a reversible reaction,
let's look at the hypothetical reaction
where diatomic gas X2 turns into its individual atoms X,
and it would turn into two of them.
So X2 goes to 2X.
The forward reaction is X2 turning into 2X.
And the reverse reaction is 2X combining to form X2.
And let's say the X2 is a reddish brown gas.
If we assume that both the forward and the reverse reactions
are elementary reactions,
we can actually write the rate law
from the balanced equation.
So for the forward reaction,
let's go ahead and write the rate of the forward reaction
is equal to the rate constant for the forward reaction,
which we will symbolize as K with the subscript F,
times the concentration of,
if we're going in the forward direction,
the reactants would be X2.
So times the concentration of X2.
And since we have a coefficient of 1 in front of X2,
for this elementary reaction,
this would be raised to the first power.
Next we can write the rate law for the reverse reaction.
So the rate of the reverse reaction
is equal to the rate constant,
and we'll put in a subscript R here.
So that's the rate constant for the reverse reaction.
And in the reverse reaction, 2X combines to form X2.
So this would be times the concentration of X.
And since we have a two as our coefficient,
we need to raise the concentration of X to the second power.
Next, let's look at these particulate diagrams
and think about what happens for the forward reaction.
So we start at time is equal to zero,
and we start with only X2.
So here are five particles of X2.
If we wait 10 seconds,
now we've gone from five particles of X2
to only three particles of X2.
So overall, two of those particles of X2
have turned into X.
And so there are four particles of X
in this second particular diagram.
We wait another 10 seconds
for a total of time is equal to 20 seconds.
And we've gone from three particles of X2
to only two particles of X2,
and we've increased in the particles of X.
So now we're up to six particles of X.
So the concentration of X2 has decreased.
We went from five particles of X2 to three particles of X2
to only two particles of X2.
And if we look at the rate law for the forward reaction,
the rate of the forward reaction is proportional
to the concentration of X2.
So if the concentration of X2 decreases,
the rate of the forward reaction also decreases.
We can see the same concept
if we look at a graph of rate versus time.
So if we look at this line right here,
we're starting on a certain rate for the forward reaction.
And as the concentration of X2 decreases,
we can see the rate of the reaction decrease.
So the rate of the reaction stops decreasing
when we get to time is equal to 20 seconds.
Next, let's think about the rate of the reverse reaction.
Well, when time is equal to zero,
the rate of the reverse reaction is zero.
And that's because when we start out, we have only X2,
we don't have any X present.
So the reverse reaction doesn't happen.
But as soon as some of that X2 turns into X,
it's possible for the reverse reaction to happen.
And as we increase in the amount of X,
and we look at our rate law here for the reverse reaction,
as we increase in the concentration of X,
the rate of the reverse reaction should increase as well.
And so that's why we see,
that's why we see the rate of the reverse reaction increase
as time increases.
So as the forward reaction is happening,
the reverse reaction is also occurring at the same time.
However, we don't really see that
when we look at our particular diagram.
In our particular diagrams,
we see a net conversion of X2 into 2X,
for example, looking from the first particular diagram
to the second,
we see that two particles of X2
have turned into four particles of X.
And going from the second diagram to the third diagram,
we see that another particle of X2
has turned into 2X.
And therefore we have six particles of X
at time is equal to 20 seconds.
So since we see a net conversion of reactants to products
in our particular diagram,
the rate of the forward reaction must be greater
than the rate of the reverse reaction.
And we can see that.
So before time is equal to 20 seconds here,
if we look at our rates, let's just pick,
for example, time is equal to 10 seconds,
for the forward reaction,
there's a higher rate than for the reverse reaction.
So at times equal to 20 seconds,
the rate of the forward reaction
becomes equal to the rate of the reverse reaction.
So here's the line on our graph,
where the rates become equal,
and also notice the rates become constant at this point.
And when the rate of the forward reaction
is equal to the rate of the reverse reaction,
the reaction has reached equilibrium.
So to the right of the dotted line,
the reaction is at equilibrium.
And to the left of the dotted line,
the reaction is not at equilibrium.
And since the rate of the forward reaction
is equal to the rate of the reverse reaction at equilibrium,
X2 is turning into 2X
at the same rate that 2X is turning back into X2.
Therefore, the concentrations of X2
and X at equilibrium remain constant.
And we can see that when we look at the particular diagrams
where time is equal to 20 seconds,
and time is equal to 30 seconds.
So both of these particular diagrams have two X2 particles
and six X particles.
Let's look at a summary of what the rates of the forward
and reverse reaction mean
in terms of reactants and products.
If the rate of the forward reaction
is greater than the rate of the reverse reaction,
that means there's a net conversion
of reactants to products.
So therefore over time,
the amount of reactants would decrease
and the amount of products would increase.
Eventually, the rate of the forward reaction
becomes equal to the rate of the reverse reaction.
And that means the reaction is at equilibrium
and there's no net change
in the amounts of reactants or products.
So reactants are turning into products at the same rate
the products are turning back into reactants.
And then finally, if the rate of the reverse reaction
is greater than the rate of the forward reaction,
there's a net conversion of products to reactants.
So products are turning into reactants
faster than reactants are turning into products.
And the example that we looked at,
the rate of the forward reaction
was greater than the rate of the reverse reaction.
And eventually the rates became equal
and the reaction reached equilibrium.
If we had looked at an example of this third case here
where the rate of the reverse reaction is greater
than the rate of the forward reaction,
eventually the two rates would become equal
and this reaction would reach equilibrium too.