### Video Transcript

In this video, we will learn how to
construct and calculate the equilibrium constant for partial pressures.

Letβs start by briefly recapping
equilibrium constants. When a reaction, such as this one,
is reversible, in equilibrium, in a closed system, then the rate of the forward
reaction is the same as the rate of the reverse reaction. If these four substances are
solutions or gases in equilibrium, then we can calculate πΎ c, the equilibrium
constant.

πΎ c is equal to the concentration
of the products multiplied with each other in the numerator divided by the
concentrations of the reactants multiplied with each other in the denominator. And the concentration of each
species is raised to the power of the stoichiometric coefficient in front of it in
the balanced equation. Sometimes, we will see πΎ c written
as πΎ or πΎ eq.

Square brackets refer to molar
concentration or molarity, which is moles per liter or moles per decimeter
cubed. We are perhaps more familiar with
this unit used for solution concentrations. But we can use molar concentration
for gases as long as we have the moles of each gas and the volume or capacity of the
vessel.

Besides molar concentration, there
is another unit we can use instead for gases in an equilibrium constant
expression. We know that in a closed system all
the gases in a gaseous reaction mixture exert pressure on the walls of the
container. We say each gas exerts a partial
pressure towards the total pressure within the vessel. The particles of gas are in
constant motion, and each gas exerts a fraction of the total pressure.

We can therefore express the
equilibrium constant in terms of partial pressures, πΎ π. And in the numerator, we have the
partial pressure of gas C raised to the power π multiplied by the partial pressure
exerted by gas D raised to the power small π divided by, in the denominator, the
partial pressure of gas A raised to the power small π multiplied by the partial
pressure of gas B raised to the power π.

We still have all the products in
the numerator and all the reactants in the denominator. But this time, they are expressed
with units for pressure. There are many possible units for
gas pressure. But two common ones are
atmospheres, atm, and kilopascals, kPa

We now know how to write the πΎ π
expression for a gas reaction mixture in equilibrium. This is the general expression. Letβs look at a real-life example
now.

Here is an example of a gaseous
reaction at equilibrium. 2SO2 gas plus O2 gas reacting
reversibly to give 2SO3 gas. The equilibrium constant for
partial pressures for this equilibrium would be πΎ π is equal to the partial
pressure of the product SO3 raised to the power of two, because there is a
coefficient of two in front of SO3, divided by the partial pressures of the
reactants SO2 and O2. And the partial pressure of SO2
will be raised to the power of two because SO2 has a coefficient of two in the
balanced equation.

We could raise the partial pressure
of O2 to the power of one if we choose because its stoichiometric coefficient is
one. Or we can just leave that power
out. And therefore, for oxygen, it is
not necessary to show the parentheses. So I will erase these too. This is the πΎ π expression for
this gas equilibrium reaction.

Note that sometimes partial
pressure is written with a capital P with the subscript indicating the
substance. But sometimes, we will find partial
pressure written with a lowercase p. If we knew the partial pressures of
these substances in terms of atmosphere units, we would have atmosphere squared
divided by atmosphere squared timesed by atmospheres, which would give us a final
answer with a unit of per atmospheres.

So, depending on the reaction
equation and the stoichiometric coefficients, we might get a different unit for πΎ
p. However, πΎ π is usually considered
to be dimensionless or unitless, just like other equilibrium constants, which means
that in a calculation we can express the numerical answer without units. The size or magnitude of a πΎ π
value for a particular reaction at equilibrium has meaning.

We can interpret the magnitude of
πΎ π. If πΎ π is larger than one, we know
that the numerator value in the πΎ π expression is larger than the denominator,
telling us that the products are favored at equilibrium, or the equilibrium sits to
the right. If πΎ π is less than one, we know
that the numerator value in the πΎ π expression is small relative to the denominator
value, telling us that the reactants are favored. In other words, the equilibrium
lies to the left-hand side.

Like other equilibrium constants,
πΎ π is temperature dependent. Its magnitude is affected by or
influenced by temperature. Heating or cooling will shift the
equilibrium to the right or to the left, depending on whether the forward reaction
is endothermic or exothermic. So, for a specific temperature, we
can calculate a πΎ π value using the partial pressures of all the species in the
system. However, sometimes we are not given
the partial pressures of each gas, but rather the number of moles of each gas at
equilibrium. But we can convert from moles to
partial pressure and then calculate πΎ π.

If we know the number of moles of
each gas of a reaction mixture present at equilibrium in a closed system, then we
can work out the total number of moles of gas particles by summing their values, in
a similar way, and we have seen before, that the total pressure is equal to the sum
of the partial pressures. From these two expressions, we
could derive two handy formulas.

We wonβt do the derivations. But the first expression is mole
fraction, given the Greek symbol π, of substance A is equal to the number of moles
of substance A divided by the total number of moles. And the other expression is the
partial pressure of substance A is equal to the mole fraction of A timesed by the
total pressure. In other words, the pressure
exerted by gas A is a fraction of the total pressure. We could do the same calculation to
work out the partial pressure of gas B, C, and D.

In this first formula, because the
units of the numerator and denominator are both moles, they will cancel, giving a
unitless or dimensionless answer. This answer will be a fraction less
than one. These two equations are based on
two assumptions that all the particles are the same size and all the particles
behave in the same manner.

Now itβs time to practice using
these two equations.

6.00 moles of N2 gas and 20.00
moles of H2 were allowed to react at 650 kelvin and 50 atmospheres of pressure. At equilibrium, 4.00 moles of N2
gas had been converted into ammonia according to the following equation: N2 gas plus
3H2 gas reacting reversibly to give 2NH3 gas. Calculate πΎ π for this
equilibrium, giving your answer to two decimal places in scientific notation.

We are asked to calculate πΎ π. πΎ π is a special equilibrium
constant, which we can use when the substances in a reaction are gases, when the
reaction is reversible and at equilibrium, and this can only take place in a closed
system. πΎ π is the ratio of the
concentrations of the products and reactants but expressed as partial pressures.

In this problem, πΎ π is equal to
the partial pressure of the product, ammonia, NH3, raised to the power of two, since
ammoniaβs stoichiometric coefficient is two. The denominator will be the partial
pressure of the reactant, N2, nitrogen gas, raised to the power of one, since
nitrogenβs stoichiometric coefficient is one, multiplied by the partial pressure of
the other reactant, H2, which is hydrogen gas, raised to the power of three, since
this substance has a stoichiometric coefficient of three.

We have now formulated the πΎ π
expression for this reversible reaction. It is not necessary to show powers
of one, and so to simplify, letβs remove it. If we choose, we can simplify
further by removing these parentheses here. To calculate πΎ π, all we need are
the partial pressures of ammonia, nitrogen, and hydrogen. But we are not given these
values. So we have to use data that we have
been given to first calculate these three partial pressures and then calculate πΎ
p.

Letβs clear some space to do the
calculation. We can draw a handy table to see
what data we have and what is missing. We are told that six moles of
nitrogen and 20 moles of hydrogen react. We can assume that initially there
is no ammonia. We are also told that at
equilibrium four moles of nitrogen have been converted or changed into ammonia.

Because we know the starting number
of moles of the reactants and the moles of one of the reactants that has been
converted or changed, we can use an ICE table, where I is the initial or starting
number of moles, C the changed or converted moles, and E the equilibrium number of
moles for each species. We can fill in these initial
values. Changed or converted moles are the
number of moles of each substance that reacted or were produced. We know that 4.00 moles of nitrogen
are converted to ammonia. But how many moles of hydrogen
reacted and how many moles of ammonia were produced overall? We need to use the mole ratio one
as to three as to two to calculate these values.

Since the ratio of nitrogen to
ammonia is one as to two, then if 4.00 moles of nitrogen reacted or converted or
changed, then 8.00 moles of ammonia must have been produced. In a similar way, if the moles of
nitrogen to hydrogen is one as to three, then the moles of hydrogen that changed
must have been three times that of the moles of nitrogen which were changed, which
gives 12.00 moles of hydrogen that reacted.

The third line on the ICE table,
equilibrium moles, is the number of moles of each species present at
equilibrium. Initially, there were 6.00 moles of
nitrogen. 4.00 moles of this were changed or
reacted. And so the difference, 2.00 moles,
must be the moles of nitrogen present at equilibrium. And this is the number of moles of
nitrogen that did not react. Doing the same calculation for the
other reactant, we get 8.00 moles of hydrogen present at equilibrium or
unreacted. Initially, there was no ammonia in
the system. At the end of the reaction, 8.00
moles of ammonia had been produced. This means at equilibrium there
were 8.00 moles of ammonia present. Can you see that for the product we
are doing an addition and not a subtraction?

We are now going to use the
equilibrium number of moles of each species to calculate the partial pressure for
each gas. But this is a two-step process. First, we need to calculate the
mole fraction of each substance and, from this, the partial pressure of each
substance. Then, we can calculate πΎ π. To calculate the mole fraction of a
substance, letβs call it π΄, we need to know the number of moles of that substance
at equilibrium divided by the total number of moles of all the substances.

If we take the sum of these three
values, we will calculate the total number of moles of particles in the system at
equilibrium. Performing this sum, we get a total
of 18.00 moles. Therefore, the denominator value
for the mole fraction calculation for each substance is 18.00 moles. For nitrogen, the numerator value
is 2.00 moles, and for hydrogen and ammonia, 8.00 moles. Solving for each, we get 0.111
recurring for nitrogen, 0.444 recurring for hydrogen, and 0.444 recurring for
ammonia. Letβs keep too three significant
figures for simplicity.

Note that mole units will cancel,
and mole fraction is a unitless or dimensionless quantity. These values are the fraction of
the total number of moles present at equilibrium. There is about 11.1 percent of
nitrogen present at equilibrium and about 44.4 percent hydrogen and 44.4 percent
ammonia.

Now we can use the equation the
partial pressure of a substance, letβs call it π΄, is equal to the mole fraction of
that substance multiplied by the total pressure. We know the mole fraction of each
substance at equilibrium. And we are told the total pressure
is 50 atmospheres. So nitrogenβs partial pressure is
equal to 0.111, the mole fraction, multiplied by 50 atmospheres, giving us 5.55
atmospheres, the partial pressure of nitrogen. This value means that of the total
pressure of 50 atmospheres, 5.55 atmospheres comes from nitrogen. We can do a similar calculation for
hydrogen and ammonia. And we get a partial pressure for
each of 22.2 atmospheres.

Finally, we are ready to calculate
πΎ π. We can substitute the partial
pressures for each substance into the πΎ π expression. And we get a πΎ π value for this
reaction in scientific notation to two decimal places, which is what we were asked,
of 8.12 times 10 to the negative three. We have not included units because
πΎ π is normally a dimensionless or unitless quantity.

Letβs wrap up what weβve learnt
about πΎ π. For a gas equilibrium reaction in a
closed system, the equilibrium constant for partial pressures is given by this
expression. πΎ π is equal to the partial
pressures of the products C and D multiplied by each other divided by the partial
pressures of the reactants A and B multiplied by each other and each value raised to
the power of the stoichiometric coefficient of that species from the chemical
equation.

We learnt that πΎ π is usually
dimensionless and is temperature dependent. We learnt that the mole fraction of
a substance at equilibrium is equal to the number of moles of that substance divided
by the total number of moles and that mole fraction can be used with total pressure
to work out the partial pressure of a particular substance.