00:01
In this problem, we're asked about the following reaction, where we start with one molar concentration of hydrogen and two molar concentration of iodine with a kc of 49 .7 at this temperature, where all reactants are gases.
00:15
So let's start by writing an equilibrium expression, because we're going to need it later.
00:21
The equilibrium expression is going to be products over reactants.
00:23
The products are hydriotic acid.
00:25
It's going to be squared because it forms two moles.
00:28
And the reactants are going to be hydrogen gas and iodine gas.
00:35
So our next step is going to be making an ice table that is initial change equilibrium.
00:41
The initial is one molar of hydrogen, one molar of iodine, and zero molar of hydrolyotic acid.
00:49
We're going to lose some amount of hydrogen, lose some amount of iodine, and we're going to gain twice as much in hydrolyotic acid.
00:58
So we'll have 1 minus x, or sorry, this starts at 2.
01:04
2 minus x and 2x.
01:07
Although we'll notice that this equilibrium constant is extremely large.
01:13
So we might want to consider instead of this reaction, we might want to consider the reverse, essentially.
01:19
So we'll convert everything we can in hydrolytic acid and then go backwards.
01:23
This will save us some time later when we're testing values.
01:27
Use.
01:28
So instead, we'll start with zero molar of hydrogen because we'll be converting everything we can to hydrolytic acid.
01:35
We'll start with one molar of iodine because we're consuming one molar of each reactant.
01:42
And that'll produce two molar of hydrolyotic acid.
01:46
And so now we'll lose some amount to x of hydrolyotic acid.
01:50
We'll gain x iodine and gain x hydrogen.
01:55
So we'll have x hydrogen 1 plus x iodine and 2 minus 2x hydraedic acid.
02:03
So we can plug these into our equilibrium expression.
02:06
So we can say that kc of 49 .7 equals 2 minus 2x squared over x times 1 plus x.
02:20
Well, note that 49 .7 is a pretty large number.
02:23
And because of that, we can assume that this minus 2x up here is pretty negligible.
02:29
For our first estimate.
02:31
So we'll actually simplify this to 4, which is just 2 squared, ignoring the minus 2 x over x, we'll call it x sub 1 times 1 plus x sub 1.
02:42
This is a test value.
02:44
And we'll make multiple test values until we get a negligible change.
02:48
So now we can say that 49 .7 times x times 1 plus x, sub 1 on a wolf of those, equals 4, and so we'll end up with a quadratic function of x sub 1 squared, 49 .7...
