00:01
For this reaction, we see that we produce three times as much c as we have a.
00:05
So if after three minutes, a has dropped from one molar down to 0 .7 molar, then we would see something like this for a, where this right here is, sorry, 0 .7 is going to be more like right here where it levels off.
00:35
And then c is going to increase to 0 .9, why 0 .9? well, a is decreasing from 1 to 0 .7, so that's the decrease of 0 .3, and c is going to increase three times as much as a decreases.
00:54
So that means it'll increase from 0 .3 times 3 or 0 .9 molar.
01:00
So we would see c doing something like this, increasing to 0 .9.
01:08
The expression for the equilibrium constant is going to be concentrated of the product raised to its coefficient of three, divided by the concentration of the reactant.
01:19
Now that we know these concentrations, we can calculate k by plugging in .9 for the concentration of c, cubing it, and then dividing by the concentration of a at equilibrium, and we get 1 .04...