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Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.49,0:00:02.78,Default,,0000,0000,0000,,- [Voiceover] We're gonna talk about
Dialogue: 0,0:00:02.78,0:00:05.16,Default,,0000,0000,0000,,the small x approximation.
Dialogue: 0,0:00:06.49,0:00:10.47,Default,,0000,0000,0000,,The small x approximation\Nis an approximation method
Dialogue: 0,0:00:10.47,0:00:14.70,Default,,0000,0000,0000,,used to solve equilibrium problems.
Dialogue: 0,0:00:14.70,0:00:16.64,Default,,0000,0000,0000,,Maybe the most important thing to remember
Dialogue: 0,0:00:16.64,0:00:20.02,Default,,0000,0000,0000,,that I'll start out with is\Nthat the small x approximation
Dialogue: 0,0:00:20.02,0:00:21.65,Default,,0000,0000,0000,,doesn't always work.
Dialogue: 0,0:00:21.65,0:00:25.73,Default,,0000,0000,0000,,It works best and you should\Ngenerally try to use it
Dialogue: 0,0:00:25.73,0:00:28.58,Default,,0000,0000,0000,,only when your reaction
Dialogue: 0,0:00:29.97,0:00:34.61,Default,,0000,0000,0000,,is strongly product or reactant favored.
Dialogue: 0,0:00:35.59,0:00:38.57,Default,,0000,0000,0000,,So strongly product
Dialogue: 0,0:00:40.71,0:00:43.00,Default,,0000,0000,0000,,or reactant favored.
Dialogue: 0,0:00:43.99,0:00:45.47,Default,,0000,0000,0000,,You might be asking at this point,
Dialogue: 0,0:00:45.47,0:00:46.87,Default,,0000,0000,0000,,okay, well, how do you know
Dialogue: 0,0:00:46.87,0:00:49.51,Default,,0000,0000,0000,,when your reaction is strongly favoring
Dialogue: 0,0:00:49.51,0:00:52.41,Default,,0000,0000,0000,,either products or reactants?
Dialogue: 0,0:00:52.41,0:00:54.85,Default,,0000,0000,0000,,The answer to that is\Nit depends on the value
Dialogue: 0,0:00:54.85,0:00:57.49,Default,,0000,0000,0000,,of your equilibrium constant K.
Dialogue: 0,0:00:57.49,0:01:02.07,Default,,0000,0000,0000,,So if something is\Nstrongly favoring product,
Dialogue: 0,0:01:02.07,0:01:06.11,Default,,0000,0000,0000,,that means K is really\Nbig, and by really big,
Dialogue: 0,0:01:06.11,0:01:09.07,Default,,0000,0000,0000,,we usually mean, or I usually mean
Dialogue: 0,0:01:09.07,0:01:13.78,Default,,0000,0000,0000,,it's on the order of 10\Nto the fourth or larger.
Dialogue: 0,0:01:15.21,0:01:16.87,Default,,0000,0000,0000,,We're talking about Kc here
Dialogue: 0,0:01:16.87,0:01:19.50,Default,,0000,0000,0000,,where our equilibrium constant is interms
Dialogue: 0,0:01:19.50,0:01:21.58,Default,,0000,0000,0000,,of molar concentration,
Dialogue: 0,0:01:21.58,0:01:26.38,Default,,0000,0000,0000,,and similarly, when we say\Nsomething is reactant favored,
Dialogue: 0,0:01:26.38,0:01:30.34,Default,,0000,0000,0000,,that means that Kc is really small.
Dialogue: 0,0:01:30.34,0:01:32.65,Default,,0000,0000,0000,,So we have mostly reactants around it
Dialogue: 0,0:01:32.65,0:01:34.52,Default,,0000,0000,0000,,and not very many products.
Dialogue: 0,0:01:34.52,0:01:39.52,Default,,0000,0000,0000,,And when Kc is very small,\NI mean that Kc is less than
Dialogue: 0,0:01:40.09,0:01:43.90,Default,,0000,0000,0000,,or about equal to 10 to the -4.
Dialogue: 0,0:01:45.26,0:01:48.77,Default,,0000,0000,0000,,So these are the situations\Nwhere the small x approximation
Dialogue: 0,0:01:48.77,0:01:51.45,Default,,0000,0000,0000,,works really well, and\Nthen, anywhere in between
Dialogue: 0,0:01:51.45,0:01:54.95,Default,,0000,0000,0000,,where Kc is between 10 to\Nthe -4 and 10 to the 4,
Dialogue: 0,0:01:54.95,0:01:56.77,Default,,0000,0000,0000,,it actually doesn't work as well,
Dialogue: 0,0:01:56.77,0:01:58.15,Default,,0000,0000,0000,,and you'll wanna use a different method.
Dialogue: 0,0:01:58.15,0:02:01.52,Default,,0000,0000,0000,,You'll either usually wanna\Nuse the quadratic equation
Dialogue: 0,0:02:01.52,0:02:04.40,Default,,0000,0000,0000,,or you might wanna use\Nsuccessive approximation,
Dialogue: 0,0:02:04.40,0:02:06.06,Default,,0000,0000,0000,,which we're not covering in this video.
Dialogue: 0,0:02:06.06,0:02:08.39,Default,,0000,0000,0000,,We'll be covering those\Nthings in separate videos.
Dialogue: 0,0:02:08.39,0:02:11.18,Default,,0000,0000,0000,,So how does the small\Nx approximation work?
Dialogue: 0,0:02:11.18,0:02:15.55,Default,,0000,0000,0000,,The first step is you assume\Nyour reaction goes fully
Dialogue: 0,0:02:15.55,0:02:20.05,Default,,0000,0000,0000,,to the side you know\Nit's gonna go based on K.
Dialogue: 0,0:02:20.05,0:02:24.27,Default,,0000,0000,0000,,So assume reaction goes 100%
Dialogue: 0,0:02:26.21,0:02:28.72,Default,,0000,0000,0000,,to either products or reactants
Dialogue: 0,0:02:30.20,0:02:34.79,Default,,0000,0000,0000,,depending on whether K\Nis really small or large.
Dialogue: 0,0:02:37.97,0:02:41.59,Default,,0000,0000,0000,,Our second step is to use\Nthis information from step one
Dialogue: 0,0:02:41.59,0:02:43.30,Default,,0000,0000,0000,,to make your ICE table.
Dialogue: 0,0:02:44.88,0:02:48.56,Default,,0000,0000,0000,,So initial concentration,\Nchange in concentration,
Dialogue: 0,0:02:48.56,0:02:50.81,Default,,0000,0000,0000,,and equilibrium concentration.
Dialogue: 0,0:02:50.81,0:02:54.76,Default,,0000,0000,0000,,We'll then solve for\Nthe x in our ICE table,
Dialogue: 0,0:02:57.04,0:02:59.06,Default,,0000,0000,0000,,assuming that x is really small.
Dialogue: 0,0:03:02.13,0:03:05.70,Default,,0000,0000,0000,,Hence the name of the method,\Nthe small x approximation.
Dialogue: 0,0:03:08.29,0:03:10.40,Default,,0000,0000,0000,,And we'll go into a\Nlittle bit more specifics
Dialogue: 0,0:03:10.40,0:03:12.42,Default,,0000,0000,0000,,about exactly what we mean by that
Dialogue: 0,0:03:12.42,0:03:14.06,Default,,0000,0000,0000,,when we do an example.
Dialogue: 0,0:03:14.06,0:03:16.87,Default,,0000,0000,0000,,Then, the last step,\Nwhich is really important,
Dialogue: 0,0:03:16.87,0:03:18.46,Default,,0000,0000,0000,,really, really, really important
Dialogue: 0,0:03:18.46,0:03:20.61,Default,,0000,0000,0000,,is to check your answer.
Dialogue: 0,0:03:22.52,0:03:25.68,Default,,0000,0000,0000,,I usually do that by plugging
Dialogue: 0,0:03:27.05,0:03:31.62,Default,,0000,0000,0000,,equilibrium concentrations back
Dialogue: 0,0:03:31.62,0:03:36.60,Default,,0000,0000,0000,,into the Kc
Dialogue: 0,0:03:36.60,0:03:38.07,Default,,0000,0000,0000,,or equilibrium constant expression
Dialogue: 0,0:03:38.07,0:03:41.37,Default,,0000,0000,0000,,to make sure the\Nconcentration that we did get
Dialogue: 0,0:03:41.37,0:03:44.49,Default,,0000,0000,0000,,as our answers make sense\Nand we get the value of K
Dialogue: 0,0:03:44.49,0:03:46.11,Default,,0000,0000,0000,,that we thought we should get.
Dialogue: 0,0:03:46.11,0:03:48.05,Default,,0000,0000,0000,,So to make more sense\Nof all of these things,
Dialogue: 0,0:03:48.05,0:03:50.40,Default,,0000,0000,0000,,we are going to go through an example.
Dialogue: 0,0:03:50.40,0:03:52.21,Default,,0000,0000,0000,,We're gonna go through an example
Dialogue: 0,0:03:52.21,0:03:55.31,Default,,0000,0000,0000,,of when we have
Dialogue: 0,0:03:55.31,0:03:58.04,Default,,0000,0000,0000,,a really small equilibrium constant.
Dialogue: 0,0:03:59.45,0:04:03.67,Default,,0000,0000,0000,,That means that kc is less than or equal
Dialogue: 0,0:04:03.67,0:04:08.67,Default,,0000,0000,0000,,to 10 to the -4, and that means
Dialogue: 0,0:04:09.52,0:04:11.57,Default,,0000,0000,0000,,since our equilibrium\Nconstant is really small,
Dialogue: 0,0:04:11.57,0:04:14.62,Default,,0000,0000,0000,,that we have mostly reactants\Nand not very much product
Dialogue: 0,0:04:14.62,0:04:18.07,Default,,0000,0000,0000,,in our reaction pot when\Nwe reach equilibrium.
Dialogue: 0,0:04:18.07,0:04:20.87,Default,,0000,0000,0000,,So the example reaction that\Nwe're gonna talk about today
Dialogue: 0,0:04:20.87,0:04:23.77,Default,,0000,0000,0000,,is iodine gas, I2,
Dialogue: 0,0:04:23.77,0:04:26.30,Default,,0000,0000,0000,,and it's going to be an equilibrium
Dialogue: 0,0:04:26.30,0:04:30.88,Default,,0000,0000,0000,,with 2I-, also gas.
Dialogue: 0,0:04:30.88,0:04:33.17,Default,,0000,0000,0000,,The K value for this particular reaction
Dialogue: 0,0:04:33.17,0:04:35.80,Default,,0000,0000,0000,,at the temperature we're interested in
Dialogue: 0,0:04:35.80,0:04:40.80,Default,,0000,0000,0000,,is 5.6 times 10 to the -12.
Dialogue: 0,0:04:40.90,0:04:43.42,Default,,0000,0000,0000,,So that's a pretty small number,\Nand it's definitely smaller
Dialogue: 0,0:04:43.42,0:04:44.85,Default,,0000,0000,0000,,than 10 to the -4.
Dialogue: 0,0:04:44.85,0:04:48.60,Default,,0000,0000,0000,,So we should be able to use\Nthe small x approximation.
Dialogue: 0,0:04:48.60,0:04:51.64,Default,,0000,0000,0000,,We have a little bit more\Ninformation about our reaction.
Dialogue: 0,0:04:51.64,0:04:54.82,Default,,0000,0000,0000,,We know the initial\Nconcentration of I2 gas
Dialogue: 0,0:04:54.82,0:04:58.84,Default,,0000,0000,0000,,is 0.45 molar, and we know
Dialogue: 0,0:04:58.84,0:05:01.77,Default,,0000,0000,0000,,that the initial\Nconcentration of I- is zero.
Dialogue: 0,0:05:01.77,0:05:04.23,Default,,0000,0000,0000,,So based on this, we're\Ngonna go through the steps
Dialogue: 0,0:05:04.23,0:05:06.31,Default,,0000,0000,0000,,for using the small x approximation.
Dialogue: 0,0:05:06.31,0:05:09.65,Default,,0000,0000,0000,,We're gonna assume reaction goes 100%
Dialogue: 0,0:05:09.65,0:05:12.72,Default,,0000,0000,0000,,to either products or reactants.
Dialogue: 0,0:05:12.72,0:05:16.12,Default,,0000,0000,0000,,Here we know that we\Nhave mostly reactants.
Dialogue: 0,0:05:16.12,0:05:19.66,Default,,0000,0000,0000,,If we start out with all\Nreactant and no product,
Dialogue: 0,0:05:19.66,0:05:22.21,Default,,0000,0000,0000,,we assume that at equilibrium,
Dialogue: 0,0:05:22.21,0:05:24.92,Default,,0000,0000,0000,,we're still gonna have mostly reactant.
Dialogue: 0,0:05:24.92,0:05:26.86,Default,,0000,0000,0000,,It's gonna change a little bit.
Dialogue: 0,0:05:26.86,0:05:30.15,Default,,0000,0000,0000,,So some of this is gonna turn into I-.
Dialogue: 0,0:05:30.15,0:05:34.69,Default,,0000,0000,0000,,So we'll say -x for some\Nsmall concentration of I2
Dialogue: 0,0:05:34.69,0:05:36.13,Default,,0000,0000,0000,,that gets converted.
Dialogue: 0,0:05:36.13,0:05:41.13,Default,,0000,0000,0000,,And that -x will turn into 2 moles of I-,
Dialogue: 0,0:05:42.73,0:05:45.17,Default,,0000,0000,0000,,but we don't expect x to very big,
Dialogue: 0,0:05:45.17,0:05:48.81,Default,,0000,0000,0000,,and that's the key to being\Nable to use this approximation.
Dialogue: 0,0:05:48.81,0:05:52.05,Default,,0000,0000,0000,,If we add these up to get our\Nequilibrium concentrations,
Dialogue: 0,0:05:52.05,0:05:57.02,Default,,0000,0000,0000,,we get 0.45 molar minus x
Dialogue: 0,0:05:57.02,0:06:00.35,Default,,0000,0000,0000,,for I2, and we get 2x
Dialogue: 0,0:06:00.35,0:06:03.05,Default,,0000,0000,0000,,for our I- concentration.
Dialogue: 0,0:06:03.05,0:06:07.67,Default,,0000,0000,0000,,Now we can use our Kc expression
Dialogue: 0,0:06:07.67,0:06:09.92,Default,,0000,0000,0000,,to solve for x
Dialogue: 0,0:06:09.92,0:06:11.68,Default,,0000,0000,0000,,when we assume x is small.
Dialogue: 0,0:06:11.68,0:06:14.56,Default,,0000,0000,0000,,Kc, for this particular problem,
Dialogue: 0,0:06:14.56,0:06:19.52,Default,,0000,0000,0000,,is the concentration of\Nour product, I- squared
Dialogue: 0,0:06:19.52,0:06:22.98,Default,,0000,0000,0000,,since we have a metric\Ncoefficient of two there
Dialogue: 0,0:06:22.98,0:06:25.57,Default,,0000,0000,0000,,divided by the concentration of I2.
Dialogue: 0,0:06:28.48,0:06:31.46,Default,,0000,0000,0000,,And if we plug in our\Nequilibrium concentrations
Dialogue: 0,0:06:31.46,0:06:36.08,Default,,0000,0000,0000,,from our ICE table, we get\Nthis 2x squared divided
Dialogue: 0,0:06:36.08,0:06:40.99,Default,,0000,0000,0000,,by 0.45 molar minus x.
Dialogue: 0,0:06:42.28,0:06:44.58,Default,,0000,0000,0000,,So far, we haven't made\Nany approximations.
Dialogue: 0,0:06:44.58,0:06:46.88,Default,,0000,0000,0000,,Now we can make our approximation.
Dialogue: 0,0:06:46.88,0:06:48.92,Default,,0000,0000,0000,,We're assuming x is small,
Dialogue: 0,0:06:48.92,0:06:52.34,Default,,0000,0000,0000,,and specifically, we're\Nassuming that x is a lot smaller
Dialogue: 0,0:06:52.34,0:06:54.72,Default,,0000,0000,0000,,than 0.45 molar.
Dialogue: 0,0:06:54.72,0:06:56.94,Default,,0000,0000,0000,,What that means is we can assume
Dialogue: 0,0:06:56.94,0:07:01.53,Default,,0000,0000,0000,,that 0.45 molar minus x is approximately
Dialogue: 0,0:07:01.53,0:07:04.04,Default,,0000,0000,0000,,equal to 0.45 molar.
Dialogue: 0,0:07:04.04,0:07:07.06,Default,,0000,0000,0000,,So x doesn't really change\Nthe concentration of I2
Dialogue: 0,0:07:07.06,0:07:08.78,Default,,0000,0000,0000,,at equilibrium very much.
Dialogue: 0,0:07:08.78,0:07:12.61,Default,,0000,0000,0000,,So that simplifies our\Nexpression for Kc quite a lot.
Dialogue: 0,0:07:14.05,0:07:16.93,Default,,0000,0000,0000,,We have 4x squared in the numerator
Dialogue: 0,0:07:16.93,0:07:19.62,Default,,0000,0000,0000,,from squaring two and squaring x,
Dialogue: 0,0:07:19.62,0:07:21.88,Default,,0000,0000,0000,,and in the denominator,\Nwe no longer have x.
Dialogue: 0,0:07:21.88,0:07:24.38,Default,,0000,0000,0000,,So this makes a much\Neasier equation to solve,
Dialogue: 0,0:07:24.38,0:07:29.26,Default,,0000,0000,0000,,and this is all equal to\N5.6 times 10 to the -12.
Dialogue: 0,0:07:29.26,0:07:30.45,Default,,0000,0000,0000,,Awesome.
Dialogue: 0,0:07:30.45,0:07:34.13,Default,,0000,0000,0000,,So now what we can do is we\Ncan multiply both sides by .45,
Dialogue: 0,0:07:34.13,0:07:36.88,Default,,0000,0000,0000,,and we can divide both sides by four,
Dialogue: 0,0:07:36.88,0:07:40.14,Default,,0000,0000,0000,,and what that gives us is\Nthat x squared is equal
Dialogue: 0,0:07:40.14,0:07:45.14,Default,,0000,0000,0000,,to 6.3 times 10 to the -13,
Dialogue: 0,0:07:45.45,0:07:47.41,Default,,0000,0000,0000,,and if we take the square root of this,
Dialogue: 0,0:07:47.41,0:07:50.27,Default,,0000,0000,0000,,we get that x is equal to
Dialogue: 0,0:07:50.27,0:07:55.27,Default,,0000,0000,0000,,7.9 times 10 to the -7 molar.
Dialogue: 0,0:07:56.62,0:07:59.30,Default,,0000,0000,0000,,So that is what we get for x.
Dialogue: 0,0:07:59.30,0:08:02.89,Default,,0000,0000,0000,,Here is the second most important step.
Dialogue: 0,0:08:02.89,0:08:05.13,Default,,0000,0000,0000,,Besides making our approximation\Nin the first place,
Dialogue: 0,0:08:05.13,0:08:07.88,Default,,0000,0000,0000,,we need to make sure that our\Napproximation makes sense.
Dialogue: 0,0:08:07.88,0:08:09.32,Default,,0000,0000,0000,,We've made a change somewhere.
Dialogue: 0,0:08:09.32,0:08:11.33,Default,,0000,0000,0000,,We've decided x is small,\Nand we've ignored it
Dialogue: 0,0:08:11.33,0:08:14.52,Default,,0000,0000,0000,,in our equation, in one\Npart of our equation.
Dialogue: 0,0:08:14.52,0:08:17.32,Default,,0000,0000,0000,,So we have to make sure\Nthat our original assumption
Dialogue: 0,0:08:17.32,0:08:19.89,Default,,0000,0000,0000,,gave us a common sense answer.
Dialogue: 0,0:08:19.89,0:08:23.82,Default,,0000,0000,0000,,We see that x is 7.9 times 10 to the -7.
Dialogue: 0,0:08:23.82,0:08:26.75,Default,,0000,0000,0000,,We assumed that was a\Nlot smaller than .45,
Dialogue: 0,0:08:26.75,0:08:28.78,Default,,0000,0000,0000,,which looks to be about true.
Dialogue: 0,0:08:31.98,0:08:36.46,Default,,0000,0000,0000,,It's about six orders of\Nmagnitudes smaller than .45.
Dialogue: 0,0:08:36.46,0:08:37.90,Default,,0000,0000,0000,,So, so far, so good.
Dialogue: 0,0:08:37.90,0:08:40.55,Default,,0000,0000,0000,,We can make sure that it really is right
Dialogue: 0,0:08:40.55,0:08:42.02,Default,,0000,0000,0000,,by checking our answer.
Dialogue: 0,0:08:42.02,0:08:46.24,Default,,0000,0000,0000,,We can plug in our x\Nand calculate Kc again.
Dialogue: 0,0:08:46.24,0:08:48.82,Default,,0000,0000,0000,,So then, we get that Kc is equal to
Dialogue: 0,0:08:48.82,0:08:51.46,Default,,0000,0000,0000,,two times our x value,
Dialogue: 0,0:08:51.46,0:08:55.94,Default,,0000,0000,0000,,7.9 times 10 to the -7 molar,
Dialogue: 0,0:08:55.94,0:08:57.75,Default,,0000,0000,0000,,and all of this squared
Dialogue: 0,0:08:57.75,0:09:02.72,Default,,0000,0000,0000,,divided by 0.45 molar
Dialogue: 0,0:09:03.59,0:09:05.65,Default,,0000,0000,0000,,minus our x value,
Dialogue: 0,0:09:05.65,0:09:09.74,Default,,0000,0000,0000,,7.9 times 10 to the -7 molar.
Dialogue: 0,0:09:09.74,0:09:11.74,Default,,0000,0000,0000,,If you multiply this all out,
Dialogue: 0,0:09:11.74,0:09:15.16,Default,,0000,0000,0000,,what you get for Kc, or what I got for Kc
Dialogue: 0,0:09:15.16,0:09:19.70,Default,,0000,0000,0000,,is 5.6 times 10 to the -12.
Dialogue: 0,0:09:19.70,0:09:22.62,Default,,0000,0000,0000,,That matches what we had\Noriginally in our problem.
Dialogue: 0,0:09:22.62,0:09:25.24,Default,,0000,0000,0000,,So this tells us that
Dialogue: 0,0:09:25.24,0:09:27.71,Default,,0000,0000,0000,,A, our approximation was pretty good.
Dialogue: 0,0:09:27.71,0:09:30.25,Default,,0000,0000,0000,,It gave us an x value that\Ngave us concentrations
Dialogue: 0,0:09:30.25,0:09:31.49,Default,,0000,0000,0000,,that make sense,
Dialogue: 0,0:09:31.49,0:09:33.23,Default,,0000,0000,0000,,and B, which is another reason
Dialogue: 0,0:09:33.23,0:09:34.45,Default,,0000,0000,0000,,why we wanna check our answer.
Dialogue: 0,0:09:34.45,0:09:37.16,Default,,0000,0000,0000,,B, it tells is that we didn't\Nmake any silly math errors,
Dialogue: 0,0:09:37.16,0:09:39.04,Default,,0000,0000,0000,,which is really easy to do\Nin this kind of problem.