Worked example: Using the reaction quotient to predict a pressure change | Khan Academy

A one liter 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 going to try to figure out, using the reaction quotient, whether our system is at equilibrium or not.

For this reaction, our reaction quotient ( Q ) is the product concentration ( [CH_3OH] ) or methanol, divided by the concentration of our hydrogen gas squared, because of that stoichiometric coefficient. 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 1 liter, and the mole quantities. We know that concentration is just moles divided by volume. Since we're dividing everything by one, the initial concentrations will be the same as the number of moles.

So, if you write that out for carbon monoxide, the initial concentration is 1.2 M; for hydrogen, it's 1.5 M; and for methanol, it is 2.0 M.

Now we can plug these concentrations into our expression for ( Q ), and then we get in our numerator 2.0, and our denominator is ( (1.5^2) \times 1.2 ). If we plug this all into our calculators, what I got is that 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 ( K_c ). 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 is figure out how the concentrations will change. Now that we know our reaction quotient ( Q_c ) is less than ( K ), we can visualize this on a number line.

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 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 the reaction concentrations are going to shift.

So, ( Q ) we can put on a number line is somewhere around here, and ( K ) is 14.5, so we'll say it's somewhere around here. This is our ( Q ), and this is our ( K ). We can see that ( Q ) is less than ( K ) on our number line, and so what's going to happen is, in order to reach equilibrium, our concentrations are going to shift to the right to get ( Q ) closer to ( K ).

This 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 balanced 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.

Total pressure for a system that has a bunch of gas molecules in it—we know that total pressure is related to the moles, the moles of gas in the system. So since we're shifting to favor the reactants, and on the reactant side we are 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 the number of gas molecules in the system, and that's going to 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 ).