00:01
This question is an extension of some of the concepts that you have learned in this chapter, namely how the equilibrium will shift based upon a volume change.
00:14
We also need to use avogadro's number and a couple other things.
00:19
We have two containers, well, one container under two separate set of conditions.
00:26
Temperature is the same, but when we have a large container, we have eight, molecules of nitrogen, or sorry, hydrogen, four molecules of nitrogen, and just two molecules of ammonia.
00:41
Then we decrease the volume, and the reaction shifts such that we have six moles of ammonia, two moles of nitrogen, and two moles of hydrogen in one liter.
00:54
The reaction being, one mole nitrogen, reacts to a three moles hydrogen to produce two moles of ammonia.
01:01
So the equilibrium constant will be equal to the concentration of ammonia divided by squared, divided by the concentration of nitrogen, divided by the concentration of hydrogen cubed because of the three right here.
01:21
So if we want to calculate the k value, we need to know these concentrations in moles per liter.
01:27
We know molecules per liter, so we need to carry out a conversion.
01:33
If we have six molecules in a liter, then we can convert the molecules to moles by dividing by avagadro's number.
01:41
We will then have moles per one liter, which is the ammonia concentration, which we need to square.
01:50
Then we'll divide by the nitrogen concentration.
01:53
We have two molecules of nitrogen, where one mole is 6 .022 times 10 to the 23rd molecules.
02:01
So this gives us the moles of nitrogen, which we divide by one to get a concentration.
02:10
Raise it to the first power.
02:12
Then we have two molecules of hydrogen, one, two, which we can then convert to moles by dividing by avagadro's number.
02:24
Then with moles, we divide by the volume 1 liter to get concentration, and then we cube it...