If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Comparing Q vs K example

## Video transcript

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 1 to 2 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 1 we're going to 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 ch3oh or methanol divided by the concentration of our hydrogen gas to the second power because of that stoichiometric coefficient 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 constitution's using the volume of the vessel which is one liter and the mole quantities we know that concentration is just moles divided by volume since we're dividing everything by 1 the initial concentrations will be the same as the the number of moles so if you 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 that our queue 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 Q C 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 0 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 the 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 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 which means what's going to happen is reaction 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 on to know we can figure out how this shift to make more products will affect the total are sure so to a little 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 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 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
AP® is a registered trademark of the College Board, which has not reviewed this resource.