Video Transcript
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 𝐾 𝑝 for this
equilibrium, giving your answer to two decimal places in scientific notation.
We are asked to calculate 𝐾 𝑝. 𝐾 𝑝 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. 𝐾 𝑝 is the ratio of the
concentrations of the products and reactants but expressed as partial pressures.
In this problem, 𝐾 𝑝 is equal to
the partial pressure of the product, ammonia, NH3, raised to the power of two, since
ammonia’s stoichiometric coefficient is two. The denominator will be the partial
pressure of the reactant, N2, nitrogen gas, raised to the power of one, since
nitrogen’s 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 𝐾 𝑝
expression for this reversible reaction. It is not necessary to show powers
of one, and so to simplify, let’s remove it. If we choose, we can simplify
further by removing these parentheses here. To calculate 𝐾 𝑝, 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 𝐾
p.
Let’s 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 𝐾 𝑝. To calculate the mole fraction of a
substance, let’s call it 𝐴, 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. Let’s 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, let’s call it 𝐴, 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 nitrogen’s 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
𝐾 𝑝. We can substitute the partial
pressures for each substance into the 𝐾 𝑝 expression. And we get a 𝐾 𝑝 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
𝐾 𝑝 is normally a dimensionless or unitless quantity.
