Introduction to reaction quotient Qc
Introduction to the reaction quotient Qc, and comparing the reaction quotient with the equilibrium constant to predict how concentrations will change.
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- At around5:01, the speaker states that having no reactants gives us a value of zero in the denominator, which makes the reaction quotient equal to infinity. Does that not violate a principle of mathematics: division by zero is undefined? Is this also how it is defined in chemistry, or was the infinity definition invented to make such a calculation possible in a chemistry context ?(11 votes)
- From the author:Great point, I think I misspoke in the video. We generally assume there is always at least some tiny bit (a molecule or two, perhaps) of the reactants left. That means the denominator is never equal to zero, it just gets very very very tiny so that the quotient approaches infinity but isn't undefined. I hope that helps!(28 votes)
- were did you get the concentrations from to get 4.3(13 votes)
- How did you get 4.3 for the first reaction?(5 votes)
- That value would've been determined through experimentation using varying concentrations of reactants.(1 vote)
- Apart from acidic or basic medium, can we balance a redox reaction in a neutral medium?(4 votes)
- I really didn't get the difference between the formula for Kc and Qc
And how did you get Kc = 4.3?(2 votes)
- Consider the reaction: A + B <--> C.
Kc is the equilibrium constant and it equals [C]/[A][B] at equilibrium. But when you first mix A with B, the reaction won't be at equilibrium because there will be a lot of A and B, but very little C. If at this time you measure the concentrations of A, B and C and work out the value of [C]/[A][B] it won't equal Kc because the reaction is not at equilibrium - if fact, it will be smaller than Kc because the numerator will be small and the denominator will be big. We call the result of this calculation Qc.
As the reaction progresses towards equilibrium, the concentrations of A and B will decrease as more C is formed. If we keep measuring the concentrations of A, B and C, and calculating [C]/[A][B] (which we are calling Qc) then we will find that Qc gets closer and closer to Kc until, at equilibrium, Qc equals Kc.(3 votes)
- What will happen if inert gas is added at constant pressure, but increasing volume?(2 votes)
- Just to clarify, what if only one of the reactants gets completely used up?
Would the Keq be infinity? Or would we calculate it without taking the used up reactant into consideration?(1 vote)
- If a reactant gets completely used up, then you don't have an equilibrium reaction!
This section is only relevant to reactions that don't go to "completion" — i.e. they reach a balance point where reactants and products both exist.
Does that help?(3 votes)
- At3:50, why did Kc and Qc get different outcomes? What was the difference between the two?(1 vote)
- If Kc and Qc have different values, this means that the reaction has not yet reached equilibrium. Kc only equals Qc at equilibrium.(2 votes)
- Is water included in reaction quotient expressions?(1 vote)
- Wouldn't Q also be 0 if it was all one reactant with the product without the other reactant?(1 vote)
- [Voiceover] Today we are going to be talking about the Reaction Quotient, Q. In this video, I'm going to go over, how you calculate Q and how you use it. We're gonna start with and example reaction between sulfur dioxide, S02 gas, which will react with oxygen gas, and this is a reversible reaction that makes sulfur trioxide or SO3. We should make sure this is a balanced reaction. We have two sulfur dioxdes reacting with one O2 to give 2S03. At equilibrium we can calculate the equilibrium constant, Kc. So, at equilibrium we know the concentrations should be constant because the rate of the forward and backward reactions are the same. And if we plug those concentrations in to this expression, we will get Kc. So, Kc is the product concentration raised to the second power so that's from this stoichiometric coefficient. And then our reactant concentrations, so S02 squared and the concentration of O2. So we know at some temperature, if you plug in the equilibrium concentrations, Kc is equal to 4.3. But what if we're interested in looking at the reaction and it's not at equilibrium yet or maybe we just don't know if it's at equilibrium. In that case, when you're not sure it's at equilibrium or really at any point in your reaction or any time, we can calculate the Reaction Quotient, Q. So Qc is equal to the concentration of our product squared, so the concentration of the product raised to the stoichiometric coefficient times the reactant concentrations, also raised to their stoichiometric coefficients. So S02 squared and O2. So you may be wondering at this point, what's the difference? The equation for Qc and Kc will always look exactly the same and the main difference is when you use them. The equilibrium constant K you calculate only with the equilibrium concentrations. So the c means everything is in terms of the molar concentration. And for the reaction quotient, Q, again everything is in terms of molar concentration but we can calculate it with any concentrations and we don't have to be at equilibrium. Molar concentration... So let's calculate this for a set of example concentrations. At some point in our reaction we have the following concentrations. We have 0.10 molar S02, 0.30 molar O2, and 3.5 molar of our product. So if we plug these numbers into our expression for Qc, we get 3.5 molar squared in the numerator and 0.10 squared times 0.3 in the denominator. So if I plug this into my calculator, I get that Qc with this set of concentrations is 4,083. So now we know how to calculate Qc. So next we're going to talk about what it tells you. So there are three possible scenarios. So when Q is equal to K, that tells us we're at equilibrium. So, if at any point you're not sure if your concentrations are the equilibrium concentrations, you can calculate Q and check if it's equal to K. And in this case, it's not. So the other two possibilities are that Q is greater than K, which is the case here. Or Q can be less than K. So let's go through both of those possibilities. We can draw all of the possible values of Q on a number line or a Q line. So Q can have values anywhere from zero to infinity. When you have no product your numerator is zero and Q is equal to zero. So that tells us Q equals zero when you have all reactants and no products. And then if you have no reactants left and all products, we have zero in the denominator and that gives us a Q value of infinity. So that means at Q equals infinity, we have all products. Then we have a bunch of values in between. And I'm gonna just write some intermediate values in here but the actual intermediate values here aren't super important. We're mostly gonna wanna compare the relative values of our Q and K. So Q here is equal to 4,083, which I will place right around here. So that's Qc and our K in yellow is 4.3. So we'll place that right around here. So we can see that our Q is larger than K and it's closer to having all products. At the concentrations we have up here, we have way more products than we should at equilibrium. So our reaction is gonna try to adjust the concentrations to get to equilibrium. And what that means in terms of our number line is that our concentrations are gonna shift so that Q can get closer to K. Since our shift is to the right, and it's moving towards all reactants, our reaction is going to favor reactants to get to equilibrium. So when Q is greater than K, like here, we're going to favor reactants... reactants. And then the last scenario when Q is less than K our reaction will favor products. And we can show that also on our number line. If we had a different set of concentrations, where Q was less than K, which I will show using this color here. If we had, say, a Q value around here then our shift would be to the right towards making more products and therefore that would mean our reaction is gonna try to reach equilibrium by favoring the forward reaction. So that's how you calculate Q and how you use it to see how the reaction concentrations will shift to get to equilibrium. In our next video, we'll go over an example problem using Q and trying to figure out how the reactant concentrations will shift for another reaction.