Worked example: Using the reaction quotient to predict a pressure change
Near the end the narrator says the reaction is 'shifting to favor the reactants' while describing the product side (the one mole of gas, methanol); this is a verbal slip since Q is less than K and the reaction shifts right toward products. The final conclusion that total pressure decreases is correct. Interventions use the correct right/products framing.
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This worked example uses the reaction quotient to predict how total pressure changes for CO + 2H2 forming methanol (K = 14.5). Because the vessel is one liter, concentrations equal the mole amounts. Q is found to be 0.74, which is less than K, so the reaction shifts right toward products. Since the product side has fewer moles of gas, the total pressure decreases.
You'll be able to
- write Q for a reaction with a coefficient greater than one.
- compute concentrations from moles when the volume is one liter.
- compare Q to K to predict the direction of shift.
- relate a shift in moles of gas to the change in total pressure.
Key words
- reaction quotient (Q)
- Products over reactants at a moment in time, with coefficients as exponents.
- shift direction
- Q less than K shifts the reaction right toward products; Q greater than K shifts it left.
- moles of gas and pressure
- Fewer moles of gas in a fixed volume at constant temperature means lower total pressure.
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'.
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