00:01
Okay, so we have this integral version of butane and isobutane.
00:08
They tell us the k is 2 .5 at 25 degrees celsius, and they tell us the volume is 0 .5 liter.
00:16
So they want us to calculate the concentration of butane and isobutane at equilibrium.
00:22
They also tell us that we start off with, so butane, they tell us we start off with 0 .117 moles.
00:31
So the first thing that we should do is convert this moles into molarity, concentration, right? so the concentration of butane is equal to 0 .117, and they tell us the volume is 0 .5, so we divide it by 0 .5.
00:49
And we should get 0 .034 mole per liter or molarity.
00:59
Right, so what we can do now is we set up an ice box.
01:02
So ice, the initial concentration, change in concentration, equilibrium concentration.
01:10
So we know the initial concentration is 0 .034 molarity.
01:16
And we know the initial concentration of isobutane is zero.
01:20
So what is the change? well, we have change is minus x because the butane is being converted to isobutane.
01:28
That's why it's minus x, but we don't know how much...