yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

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


3m read
·Nov 11, 2024

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 ).

More Articles

View All
Kirsty Nathoo with Shan-Lyn Ma, Founder of Zola
Okay, hi everybody. I’m Kirsty Nathu. I’m one of the partners at Y Combinator, and it is my great honor to introduce Shanna Lynn, MA, who’s the CEO of Zola. Zola has reinvented the wedding gift registry, and they’ve now worked with hundreds of thousands o…
Robinhood Just Got Cancelled - Again
What’s up you guys, it’s Graham here. So historically, they say that on average September is the worst month for the stock market, dating all the way back to 1950. Now whether or not that comes true for this month is yet to be seen, but I have to say the…
Fishing With Dynamite Is Harmful—Why Does It Persist? | National Geographic
[Music] You can come out here on a fine morning and you know there’ll just be ramp and blasting in areas where there may be tuna feeds, or if there aren’t tuna feeds, then they may target the reefs. I would say probably for the last 5 years it’s at least …
Influential points in regression | AP Statistics | Khan Academy
I’m pretty sure I just tore my calf muscle this morning while sprinting with my son. But the math must not stop, so I’m here to help us think about what we could call influential points when we’re thinking about regressions. To help us here, I have this …
Example dividing a whole by a unit fraction
Let’s think about what 3 divided by 1⁄4 is equal to. Pause this video and see if you can figure it out on your own. And I’ll give you a hint: take three holes and divide it into pieces, or sections, that are each one-fourth of a hole. Then think about how…
Embracing Death | Explorer
It’s interesting in our society, and you know how we do things. You know, we plan for so many life celebratory events. We plan for a wedding, we plan for a baby, we plan for a graduation from high school, from college. We plan for our career. But the one…