00:02
Pep, which is fossil in all pyruvate, and it's one of the major phosphoryl group donors to atp during glycolysis.
00:13
It's asking about with given the certain amount of molarity for atp, atp, the pyruvate.
00:24
What would be the calculation for the concentration of the pep? and this is in sort of like driving the question is, are all reactions at equilibrium? and this is especially as we're looking at metabolic reactions within the cell.
00:43
So the first thing that we need to do is look at pep, which is an intermediate factor involved in glycolysis.
00:49
And it aids the adp and to make an atp.
00:51
The first thing is writing out the equation.
00:54
We've got pep plus h2o yields pyruvate plus a phosphate, the delta g, the amount of energy is negative 61 .9 kilojoules per mole.
01:04
And then the second reaction that's happening, of course, the whole point is that the phosphate is being donated and given to adp.
01:13
So adp plus the phosphate yields atp plus h2o, and the delta g in that situation is 30 .5 kilojoules per mole.
01:21
When we do a summation between the two reactions we have and we cross out the ones that are in common, we're going to have pep plus adp yields parruvate plus.
01:31
Atp and yielding as a summary negative 31 .4 kilojoules per mole.
01:37
Now assuming that we have 25 degrees celsius we're told to calculate the concentration of pep.
01:45
In order to calculate the concentration of pep first, sorry, calculate the k sub eq, in other words the equilibrium value.
01:56
So the equation that we're going to use is delta g prime equals negative rt, natural log case vq.
02:04
Then moving the things around so that we can put our knowns on one side and put our unknowns on the other side, just like basic algebra.
02:11
We're going to have delta g on the left side, divided by rt, because rt is multiplying to the natural log.
02:20
So if we divide both sides by that, we can bring it over to where the delta g is located.
02:25
So what we end up having then is a negative delta g divided by rt is equal to the natural log of, of, you know, k -e -q.
02:35
And then once we do that, then we just simply plug and play to put the numbers in terms of the various concentrations provided that that's in the problem as well as the one that we figured for the summation of the reductions.
02:46
So the top number is the delta g.
02:48
What's the negative of the delta g...