- [Voiceover] A one litre reaction vessel
contains 1.2 moles of carbon monoxide,
1.5 moles of hydrogen gas,
and 2.0 moles of methanol gas.
How will the total pressure change
as the system approaches equilibrium
at constant temperature?
So, our carbon monoxide is reacting with our hydrogen
in a one to two ratio
to give us methanol.
And this reaction is reversible.
We also know the equilibrium constant for this reaction
is 14.5 at some temperature.
And we know that the temperature is staying constant.
So, we are going to break this problem up into two parts.
In part one, we're gonna try to figure out,
using the reaction quotient,
whether our system is at equilibrium or not.
So, for this reaction,
our reaction quotient 'Q'
is the product concentration.
"C-H-3-O-H" for methanol.
Divided by the concentration of our hydrogen gas;
to the second power,
because of that stoichiometrical efficient.
And then also in the denominator,
we have our carbon monoxide concentration.
We can calculate 'Q'
by plugging in the concentrations of these,
at this particular moment in time.
And we can calculate the concentrations
using the volume of the vessel,
which is one liter
and the mole quantities.
We know that concentration
is just moles divided by volume.
And since we're dividing everything by one,
the initial concentrations
will be the same as the number of moles.
So, if we write that out,
for carbon monoxide,
the initial concentration is 1.2 molar.
For hydrogen, it's 1.5 molar.
And for methanol,
it is 2.0 molar.
So now we can plug these concentrations into
our expression for 'Q'.
And then we get,
in our numerator, 2.0.
And our denominator, is 1.5
squared times 1.2.
So if we plug this all into our calculators,
what I got is at our 'Q',
for this particular moment in time,
with these concentrations is 0.74.
So this tells us,
first of all, we know that 'Q'
is not equal to 'KC'.
So that means we are not at equilibrium.
"Not at equilibrium".
Which means that our pressures
are indeed going to change
because the system is going to try to reach equilibrium.
The second thing we can do,
using the reaction quotient to,
is figure out how the concentrations will change.
Now that we know our reaction quotient 'Q',
we know that
our reaction quotient 'QC'
is less than 'K'.
We can visualize this on a number line.
So if we look at all possible values of 'Q',
we know that when 'Q' is zero,
we have all reactants.
When 'Q' is infinitely large, we have
all reactants, we have all products.
And then we have all of the possible values in between.
What we're really worried about here,
is just looking at the relative value of 'Q' and 'K'.
And seeing how reaction concentrations are going to shift.
So 'Q', we can put on our number line,
is somewhere around here.
And 'K' is 14.5, so we'll say it's somewhere
around here.
So this is our 'Q' and this is our 'K'.
We can see that 'Q' is less than 'K' on our number line.
So what's gonna happen is,
in order to reach equilibrium,
our concentrations are going to shift to the right
to get 'Q' closer to 'K'.
Which means, what's going to happen is,
the reaction
is going to shift to favor
making more products.
So if we look back at the balance reaction,
what's going to happen here is,
it's going to shift to favor the products.
So I'm making that top arrow a little bit more bold.
And to tie this into
what the problem wants to know,
we can figure out how the shift
to make more products will affect the total pressure.
So the total pressure
for a system that has a bunch of gas molecules in it,
we know that
total pressure is related to
the moles of gas in the system.
So since we're shifting to favor the reactants.
And on the reactants side, we have,
we're making one mole of gas.
And we're starting with three moles of reactant gas.
We're favoring the side that has fewer gas molecules.
So that means as we shift to favor the products,
we're going to reduce number of gas molecules in the system,
and that's gonna reduce our 'P' total.
So the answer is,
that P total is going to decrease
as our reaction approaches equilibrium
and that is because
our reaction quotient 'Q' is less than 'K'.