>> narrator: In this video,
we'll take a closer look at
Equilibrium Calculations.
Equilibrium constants
can be used to calculate
the concentrations of reactants
and products that are present
at equilibrium.
The equilibrium constant
is important because
its magnitude tells you
which substances predominate
at equilibrium.
For example, for equilibrium
constants larger than one,
products will predominate
at equilibrium.
When an equilibrium constant
has a value that is
less than one,
reactants will predominate
the equilibrium mixture.
The law of mass action
provides a general description
of the equilibrium condition.
For a chemical reaction mixture
that is in equilibrium,
the ratio between
the concentration of reactants
and products is constant.
Using this hypothetical
equation, where substances
A, B, C and D represent
chemical species
and the lower case letters
e, f, g and h
are their coefficients
in the balanced equation,
the law of mass action
is represented by the following
equilibrium expression:
K sub c is known as
the equilibrium constant,
and it is found by taking
the product of
the equilibrium concentrations
of the products
raised to their coefficients,
over the product
of the equilibrium
concentrations
of the reactants
raised to their coefficients.
This problem asks us
to calculate the value
of the equilibrium constant.
The following equilibrium
concentrations were observed
for the Haber process,
which is the synthesis
of ammonia.
This reaction took place
at 128 degrees Celsius.
The equilibrium concentrations
of the substances
are as follows:
for ammonia, 3.2 times 10
to the negative second molar,
for nitrogen,
8.4 times 10
to the negative first molar,
and for hydrogen, 3.2 times
10 to the negative third molar.
Here is the balanced equation
for the reaction at equilibrium,
which we can tell
by the double arrow.
These are equilibrium
concentrations
that are given to us--
that is very important
because an equilibrium
calculation has to have
equilibrium concentrations
plugged into the equilibrium
expression before we can
solve for K sub c.
The equilibrium expression
for this problem would be
K sub c is equal to
the equilibrium concentration
of ammonia,
raised to the second power,
over the equilibrium
concentration of nitrogen
times the equilibrium
concentration of hydrogen
raised to the third power.
And again, this expression
is found by taking
the equilibrium concentration
of the products
raised to their coefficients
over the equilibrium
concentration of the reactants
raised to their coefficients.
This is an equilibrium
expression, so before
we plug in any values,
we need to make sure
they are equilibrium
concentrations.
Taking the equilibrium
concentrations that were
given to us in the problem
and plugging those in
to the equilibrium expression,
we get this expression here.
When we plug in the values
and solve for K sub c,
we find that the equilibrium
constant for this reaction
at this temperature
is 37,000, which is rounded
to two significant figures
and tells us that the mixture
reaching equilibrium
will contain mostly products.
Notice there aren't
any units listed
for the equilibrium constant--
equilibrium constants
are typically considered
to be dimensionless quantities.
Here's another problem
that asks us to calculate
the value of K sub c,
the equilibrium constant.
A system initially containing
2.00 times 10
to the negative third molar
hydrogen and 4.00 times
10 to the negative third
molar iodine at 388° Celsius
is allowed to reach equilibrium.
Analysis of
the equilibrium mixture
shows that the concentration
of the hydrogen iodide
is 2.81 times 10
to the negative third molar.
Calculate K sub c,
the equilibrium constant,
at 388° Celsius
for the reaction taking place,
where the balanced equation
shows for every one mole
of hydrogen that reacts
with one mole of iodine,
two moles of hydrogen iodide
are produced.
Don't forget that this system
is at equilibrium.
We can use an "ICE" chart here,
which is a tool for tabulating
data from equilibrium problems.
"I" stands the initial
conditions, "C" stands for
the change in conditions
and "E" represents
the equilibrium concentrations.
And remember, to solve
equilibrium constant expressions
for the value of K sub c,
we need to know these
equilibrium concentrations,
that's what we plug
into the expression.
We know that initially,
there was 2.00 times 10
to the negative third
molar hydrogen,
4.00 times 10
to the negative third
molar iodine, and we also know
that at equilibrium
there's 2.81 times
10 to the negative third molar
hydrogen iodide.
So we know that the hydrogen
iodide must have increased
by 2.81 times 10
to the negative third.
And the stoichiometry
from the equation tells us
that the hydrogen
and the iodine
must have decreased
by half as much as the rate
of appearance
of the hydrogen iodide.
And this coefficient of 2
tells us that 2 moles
of the hydrogen iodide
are being produced
for every 1 mole of hydrogen
that is consumed
or every 1 mole of iodine
that is consumed,
and so that's how we came up
with the values
for the change
in the concentration
of the hydrogen and the iodine.
They must have disappeared
at half the rate
that the hydrogen iodide
appeared.
So that's where we got
a change in -1.41
times 10 to the negative third
for both substances,
the hydrogen and the iodine.
So now we have everything
we need to determine
the equilibrium concentration
of all three substances.
Since the concentration
of hydrogen is decreasing
as the system approaches
equilibrium, we take
the 2.00 times 10
to the negative third
and subtract 1.41
times 10 to the negative third
and we're left with 5.9 times
10 to the negative fourth
molar-- that is
the equilibrium concentration
of hydrogen.
Since the concentration
of iodine is also decreasing
as the system approaches
equilibrium, we take
the initial concentration, 4.00
times 10 to the negative third
and subtract the change
in the concentration of iodine,
which was, 1.41 times
10 to the negative third
and we find the equilibrium
concentration of the iodine
is 2.59 times 10
to the negative third molar.
And we already have
the equilibrium concentration
of the hydrogen iodide
which is 2.81 times 10
to the negative third molar.
Now we can solve for
the equilibrium constant.
Remember the equilibrium system,
the equation was this:
1 mole of hydrogen
plus 1 mole of iodine
in equilibrium
with 2 moles of hydrogen iodide.
When we set up our
equilibrium expression,
we want the products raised
to their coefficients
in the numerator
over the reactants
raised to their coefficients
in the denominator.
But since both reactants
have an understood 1
as coefficients, we just have
hydrogen concentration
times iodine concentration
in the denominator.
And remember, this is
an equilibrium expression,
so the only numbers
that we can plug in here
are those molar concentrations
that are at equilibrium.
So for the hydrogen,
we take the 5.9 times 10
to the negative fourth
and plug it in here.
For the iodine
we take the 2.59 times 10
to the negative third
and plug it in here.
For the hydrogen iodide,
we would take the 2.81
times 10 to the negative third
and plug that in here.
So when we plug those values
into our equilibrium expression,
and solve, we find that
the value of K sub c,
our equilibrium constant,
is 5.16, which tells us
that the mixture
reaching equilibrium
will contain mostly products.
Here's one for you to try.
At a given temperature,
the initial concentration
of sulfur trioxide
is 4 molar.
The sulfur trioxide decomposes
according to the reaction here.
Two moles of sulfur trioxide
are in equilibrium
with two moles of sulfur dioxide
and one mole of oxygen.
At equilibrium,
the concentration
of the sulfur trioxide
is two molar.
Calculate the equilibrium
constant for this reaction.
Press pause and give it a try
and then press play
to check your work.
Hopefully the first thing
that you set up was
the equilibrium expression,
which again is the products
raised to their coefficients
over the reactants
raised to their coefficients.
So this is the equilibrium
expression for this
equilibrium reaction.
Then if you set up an ICE chart,
the information that was
given to you was that
the sulfur trioxide
was initially at 4 molar
but at equilibrium,
it was 2 molar,
so that means it must have
changed or the concentration
must have decreased
by 2 molar.
And if we look back
at the stoichiometry
from the balanced equation,
we see that the sulfur dioxide
also has a 2, so that means
that the sulfur dioxide
was appearing at the same rate
that the sulfur trioxide
was disappearing, which is
why we put a + 2
for the sulfur dioxide.
The oxygen has an understood 1
as a coefficient, which means
it is appearing at half the rate
that the sulfur trioxide
is disappearing,
which is why we put a + 1 here
for the change in
the concentration of oxygen.
And so now we have
equilibrium concentrations
for all three species
and we can plug these
into our equilibrium expression
to solve for K sub c.
So for the sulfur dioxide,
we took the 2
and plugged it in here
and we squared it
based on the fact
that there was a coefficient
of a 2 in front of
sulfur dioxide,
which is why there was
an exponent of 2
in the equilibrium expression.
We multiply that times
the concentration of the oxygen
at equilibrium, which was 1.
In the denominator,
we took the sulfur trioxide
concentration at equilibrium
which was 2, and we plugged
that in and squared it as well,
and when we did the math,
we see that 2 squared is 4,
times 1 is 4, divided by
2 squared which is just 4,
so what we are left with
is 4 divided by 4,
or in other words
the equilibrium constant
had a value of 1,
which tells us that
neither reactants nor products
predominate at equilibrium.
And that's a closer look
at equilibrium calculations.
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