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## Factors that affect chemical equilibrium

Current time:0:00Total duration:5:53

# Comparing Q vs K example

## Video transcript

- [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'.