The Reaction Quotient

So our reaction quotient Q is just a tool that we use to see is our chemical reaction at equilibrium. We're gonna say here, if the reaction quotient Q is equal to the equilibrium constant K. Then our reaction is at equilibrium. Q is just like K. It equals products. Overreact ints and just like K. It ignores solids and liquids. Here in this example we have a gas giving us be gas so neither one is a solider liquid. So the equilibrium expression or cue expression in this case would just be be divided by a You'd get values for your Q. You plug them in to find out what your Q value is. If you figured out what your Q value is and you saw that your equilibrium constant equal the same exact number. We would be at equilibrium. In this case, I've just come up with the value of being equal to 50. So again, when Q equals K, we are at equilibrium. Now when Q does not equal K, we can use Q to determine which direction will our reaction shift to get to equilibrium. Now we're gonna say here, the direction our reaction shift, determine whether our reactions or products are increasing or decreasing. Remember K. Your equilibrium constant, K. Is your equilibrium constant? It is where we want to be. So whatever Q is Q will always shift to get to K. So for example, let's say that we did Q equals products. Overreact ints and we determined Q was equal to 10, R. K is still equal to the value of 50. So in this case Q is less than K. What, wherever que is que will shift there in order to reach equilibrium. So here Q would shift in the four direction to reach K so that it can get to equilibrium. The direction it shifts on the number line is the same direction it shifts in the equation. So here we have a gas, gives us b gas again in the number line we shifted in the four direction to get to K. So we're gonna shift in the four directions in our reaction, wherever we're shifting is increasing in amount. And if that side is increasing an amount that means the other side has to be decreasing in amount. Remember there's a balance involved in a chemical reaction. If one side is increasing its at the detriment of the other side. Okay, so one side increases so that the other side decreases. They both can't be increasing or decreasing together. Okay, it's kind of like a balance. Now. In the other example here I've calculated Q. Again I did Q equals products. Overreact ints I used given values and I figured out Q was equal to 1 40. In this case Q is larger than K but we still do the same thing. Q will always shift to get two K. It'll always move in the direction to get to K. So they can reach equilibrium here on the number line, we shift in the reverse direction to get to K to get to equilibrium. And if I shift in that direction on the number line, I shift in the same direction in my equation again, wherever I'm shifting will always be increasing, so I'm shifting to the reacting side, so the reacting side is increasing since the reacting side is increasing, that means the product side is decreasing. So remember Q is just a tool that we use to see if we're at equilibrium, if Q equals K, we're at equilibrium. There will be no shifting. If Q is less than a greater decay, then we're gonna shift in some direction to get to equilibrium. Remember the fundamental steps we see here so that we can answer any question asking us um which side is increasing or decreasing? Which direction will our reaction shift to get to equilibrium? Now that we've established these fundamentals will attempt to do the example question on the bottom. So click on to the next video and see how I approach this same exact question that we have here on the bottom

consider the hypothetical reaction below. So we have three moles of a gas reacts with one mole of gas to produce two moles of C. Gas. Here we're told the equilibrium constant equals 1.5 times 10 to the negative three were asked which one of the following statements is correct. If the initial concentrations are a equals 10.85 moller B equals 0.36 moller and C equals 0.5 moller. Alright, they're giving us our K value then they're asking are we at equilibrium? Um is such and such increasing or decreasing? These are clues that you need to figure out what Q is. Once we find Q we can compare it to K. So again they'll give you your K value. They'll give you some other new numbers which will help you figure out what Q is. Once you do that you'll know which direction your reaction is shifting and therefore what side is increasing versus what side is decreasing. So Q is just like K equals products. Overreact ints we ignore solids and liquids. So here we have C squared divided by a cubed because remember the coefficients become powers times B. All we do now is that we insert the values so CS .005 which will be squared divided by A. Which is 0.85 cubed times 0.36 When we plug all that in. That gives us 1.13 times 10 to the -4. Now that we have Q. We can compare it to K. Set up your number line. So K is where we wanna be, It's to the -3. Q. is to the -4. So Q. is smaller. So we have to shift in the four direction to get to K. If we shift in that direction on the number line we shift in the same exact direction in our reaction, we're moving towards the product side. So the product side will be increasing in amount. Mhm. And my reaction side will be decreasing in amount. Alright so here at equilibrium the amount of C will increase. Um Yes that's true because we're moving towards products so the products would increase. So it is true. Um At equilibrium the amount of A will increase. Now we're shifting away from react ints. So it should be decreasing at equilibrium. The amount of B will increase in the amount of C will increase. We're moving away from visa decreases. Plus both sides can't increase and both sides can decrease. So that's not a possibility. He had the amount of A. And B. We said would both decrease and then the reaction could only be at equilibrium if Q equals K. Here, K is 1.5 times 10 to the negative three. Q. Was definitely not that number since they did not equal each other. We are not at equilibrium. So out of all the choices. Only option A is the correct choice

here, we're told for the reaction of two moles of carbon dioxide decomposing to form two moles of carbon monoxide and two moles of oxygen gas. The equilibrium constant. Okay, is 7.22 times 10 to the negative four at 400 kelvin, while the reaction quotient. So Q. Is 6.63 times 10 to the negative two. Here, we're told if initially we have 20.20 atmospheres of C. 2.30 atmospheres of C. O. And 00.15 atmospheres of 02. Which of the following statements is not true. All right. So, Q versus K. So, K. Is where we wanna be Into the -4. Q is larger to the -2 year, but Q always shifts to get to K. So we're shifting in the reverse direction, which means in our reaction we shift in the same exact direction. So on this side we would be decreasing. I mean actually increasing because we're shifting towards it. So, So, sorry, we're increasing here and on this side will be decreasing. So, what effect does that have? Well, for gas, the amount of the gas affects its partial or individual pressure. So, if we're moving towards C. 02 and it's increasing, that would mean that it's partial pressure would also increase. So we'd expect its new pressure to be greater than 20.20 on the other side on the products here are decreasing over time. So there's less of them available. So they're partial pressures are expected to decrease over time. So, if we take a look, the pressure of C. 02 will be greater than 20.20 atmospheres. That's true. Remember we're looking for what is not true? The pressure of C. O. will be less than .30. Yes because we're shifting away from it. The pressure votes will be greater than .15 atmospheres. No it'll be less than that. So that's what's not true. The pressure votes will be less than yes it will be less than because we're shifting away from 02 and the reaction will favor reactant. So here we're shifting towards the reactant making more reactant meaning we're seeing them as being favorable. So out of all the choices only option C. Would be the um the incorrect answer. All the others are correct as stated.

for the following reaction, we have three moles of hydrogen gas reacting with one mole of nitrogen gas to produce two moles of ammonia gasses product here, we're told that the equilibrium constant for the reaction equals 25. Now it says at a particular time, the following concentrations are measured for the giving compounds. Here, H two becomes 20.5 moller and two is 20.170 moller and NH three is 3.12 times 10 to the two molars, which the following statements is true. So here they're talking about things increasing or decreasing. The only way we can determine that is to first figure out which direction. Does my reaction shift doesn't move in the four directions. Favorite products or does it move in the reverse direction to favor react. It's to be able to determine which direction it shifts, we need to find out what Q is and then compare it to K. So Q. Is just like K. It equals products over react mints. So that's NH three squared Divided by H two Cubed Times and two. We're gonna plug in these initial concentrations here to help us determine what Q. Is. So that's 3.12 times 10 to the two squared Divided by .005 Cubed times .170. So when we plug all that in, that's gonna give me as an answer. 4.5.8 Times 10 to the 12. For my q value. Now just compare Q two K. So on my number line K is where we wanna be. So it's in the middle. So it's 25. Q is so much larger? It's 4.58 times 10 to the 12. So remember Q always shifts to get to K. So it's going to shift this way to get to K. Which means in my reaction it shifts the same way. So this side here would be increasing over time and this side here would be decreasing over time. Mhm. Alright. So here, which the following statement is true here the concentration of H2 will increase. That is true because we're heading towards products. So we know that's true. The equilibrium constant will increase. So your equilibrium constant is K. The only time that your equilibrium constant can be changed is if we're um adjusting the temperature. So if we increase or decrease the temperature that will adjust my equilibrium constant to a new value Here. We never talk about influence of the temperature in any way. So the equilibrium constant will still be 25. The concentration of NH three will increase. No, we're shifting away from it so it'll be decreasing the concentration of n tool decrease. No, it should be increasing over time since we're shifting towards the reactant. No change will occur. This would only happen if we are at equilibrium we are at equilibrium when Q equals K but Q and K are definitely not the same value. So a change will occur. Out of all the choices here, we see that option A is the correct choice. So remember your reaction quotient is just used as a way of determining are you at equilibrium when Q equals K. We are at equilibrium. And so your reaction will not shift in the forward or reverse direction. If Q is different from K, we have to compare Q two K and see which way does cumin to get to K. This determines if you move in the forward direction to favor the creation of additional product or reverse where you're going towards reactant to create more reactant. So just remember the fundamentals in terms of Q and how to compare decay, and you'll be able to determine which side is increasing or decreasing within any given chemical reaction.