Q vs. K

Jules Bruno
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Hey, guys. So here we're dealing with a new concept. The reaction quotient, which is represented by Q. Now we're gonna say we've learned thus far that any given chemical reaction of equilibrium has an equilibrium constant k associated with it. Now we're looking at Kew, which is our reaction quotient. It's used to determine if our chemical reaction is at equilibrium. Now we're going to say, if the reaction question Q is equal to the great Khan, the equilibrium, constant K and our reaction is at equilibrium. But what if our Q and care not equal to each other? What happens then? Well, let's say that our Q is greater than K here. We're gonna make up Ah, fictitious equation. So a plus B gives me C plus deep, and once they hear that K is equal to 10 but Q was equal to 150 to deal with this. All we do is we do a number line. Que is where we wanna be Equilibrium que, however, is larger. Our equation will shift to try to get back to equilibrium. So which way does Q have toe walk to get to K? Well, he was gonna have to shift this way to get to catch and the same direction. Accused shifts To get two K is the same direction that our equation will shift to get back to equilibrium. So our equation will shift in the same direction. Whichever where we shift that side will be increasing. So this side here is increasing. And if this side is increasing, what's happening to the opposite side? The opposite side would have to be decreasing. Now let's look at the opposite. What if Q is smaller than K? Come up with the same exact idea? So I will say again, a plus B gives me C plus deep now K is still 10. But now we're gonna say Que was equal 2. do a number line. Okay, again is in the middle where we wanna be. The queue is smaller. Which way does Q have to shift to get to K? Que has to shift this way now. Whichever way it shifts on the number line is the same direction shifts in our equation. So we'll shift that way and as a result, wherever we're shifting increases and then this side here is decreasing and That's the approach you need to take. If Cuba is equal to K, they're both on the same exact spot on the number line, so there was no shifting that occurs. Knowing this, let's take a look at this question. It says here for the reaction to H two s gives us hte to plus s to the equilibrium. Constants. Okay, is equal to 1.60 times 10 to the negative, too, at 400 Kelvin temperatures just given. But it's not necessary here. If the reaction question, which is Q has given us 4.18 times tension negative four. Which Wilfong statements is are not true. So do a number line case in the middle what we wanna be. It's to the negative, too, whereas Cuba's to the negative four. So cue was definitely smaller. Say to yourself, which way does Q have to shift to get to equilibrium? Q. S to shift this way to get to equilibrium, Therefore, our equation shift that same way. Remember, wherever we shift will be increasing and therefore the other side will be decreasing. So here, if we take a look, the pressure of s to increase all right. So these are all gasses. When the amount of a gas increases, its partial pressure or pressure will increase. If it's amount decreases, then the gas is pressure will decrease. So we're shifting to the right, which is the site of S two. So s two will be increasing. This is true. The pressure of H two will decrease. No, we're shifting to the side of H two, so it's going to increase, not decrease. Or remember, we're looking for what is not true. The equilibrium constant will not change. That is true. Remember the only thing that can change our equilibrium constant is temperature temperatures not being affected here. So the equilibrium, constant K will stay the same. And here the pressure off H two s will decrease. Yes, because we're shifting away from the react inside, so the react inside would decrease. So based on the options given Onley option, B would be the correct choice. So remember, que was just used as a way of determining. Are we at equilibrium? We're only equilibrium When Q equals K. If they're different, que will have to shift either to the four direction, the reverse direction in order to get back to equilibrium which is kept