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Welcome back; we are looking at reaction equilibria, and we are working out some examples to make
sure that we are all in the same plane, in terms of comfort in dealing with the quantities
related to reactions. For this, we have looked at some of the concepts that were learnt in
the earlier classes. And we are in the middle of working out a set of three examples; we
worked out one in the last class, we will work out two or you want to work it out I
am going to give you enough time to work it out in this class.
And so in this class, we will just look at these two examples. Example 6.2 reads as,
the standard Gibbs free energy changes at pH 7 delta G dash and recall that it is a
different from delta G naught, because the physiological pH is pH 7 and delta G naught
is defined under conditions of pH 0, which is irrelevant. So, we use delta G dash; this
is what we again recalled in the previous solution. So, the delta G dash values for
the following reactions are given next to each; this is again ATP hydrolysis ATP plus
water giving you ADP and inorganic phosphate; and as we had seen delta G dash the standard
value was minus 7.7 kilo calories per mole. And glucose 6 phosphate plus water, you know
the hydrolysis of glucose 6 phosphate. To give you glucose plus inorganic phosphate,
the delta G dash value for that happens to be minus 3.14 kilo calories per mole; somebody
has very carefully determined the values of delta G dash for these reactions; thousands
of reactions. Now, the question here is, about or the need
here is calculate the standard Gibbs free energy change at pH 7 and the equilibrium
constant for the phosphorylation of glucose, which is coupled with ATP hydrolysis in the
cell. It is a coupled reaction; the phosphorylation of glucose and ATP. If you recall that is
one of the first reactions in glycolysis and is catalyzed by the enzyme hexokinase. For
our purposes it can be considered, as the reaction between glucose and ATP, just take
it as a reaction between glucose and ATP. And so, before I present the solution, let
me give you about 20 minutes to work this out. It is going to involve a few things and
only thing is that, we have seen these reactions in the context of biological systems. Go ahead
please 20 minutes
Hopefully, you experience some deja vu movements moments and clarified some of the concepts
that you had learnt earlier. The solution as is follows, the coupled process that we
are interested in first the reaction that we are interested in the process. As we were
suggested in the problem statement itself; glucose plus ATP giving you glucose 6 phosphate
plus ADP, if you recall we are looking at the phosphorylation of glucose, which is coupled
to ATP and, we have written that reaction as glucose plus ATP giving you glucose 6 phosphate
plus ADP. This particular reaction can be obtained by
the two reactions, or from the two reactions given in the problem statement itself. Just
by adding them appropriately, you know the glucose plus phosphate giving you glucose
6 phosphate plus water is reaction a this is of course given; this was not given as
it is written here in the problem statement; and ATP plus water the hydrolysis of ATP to
give you ADP plus p i. If you add these two, it can be easy to see that you get glucose
plus A T P because water water here will cancel out; p i p i here will cancel out; and what
we are left with is, glucose plus ATP giving you a glucose 6 phosphate plus ADP right here.
So, this addition of these two will be fine. And, this directly comes from, what is known
in the earlier classes that, if the delta G dash values are that the delta G dash values
are additive. Therefore, the needed delta G dash value is nothing but the delta G dash
of a reaction a added to the delta G dash value of the reaction b.
So, if we have delta G dash of a and delta G dash of b, we are done. We already have
delta G dash of b in the problem statement but delta G dash of a we do not have, what
was given in the problem statement is, the reverse of the reaction that we need, or reaction
a is the reverse of the second reaction that was given in the problem statement.
So, from the delta G dash values that were given, we know that delta G dash a value for
the above reaction is minus of minus 3.14 which is plus 3.14 kilo calories per mole,
you know it is a reverse. Therefore, we had put in a minus there and delta G b dash value
was minus 7.7 kilo calories per mole. And therefore, delta G dash of the reaction that
we are looking for, which is the ATP catalyzed phosphorylation of glucose and we have already
established that is being equal to delta G a dash plus delta G b dash; and just adding
these 2 together we get, minus 4.56 kilo calories per mole. Also from delta G dash equals minus
RT log k equilibrium dash and this is, this again we derived in this particular course,
or we saw in the earlier one of the earlier lectures this relationship. We had used this
extension, we had used this equation extensively, subsequently. But this equation itself was
known to you from earlier classes. From this, we can get the equilibrium constant that we
are looking for, is exponential of minus minus delta G dash by RT; substituting the values,
we get exponential of minus delta G dash is nothing but 4.56 into 10 power 3 calories
per mole and multiplied by 4.18 to convert it into s i units divided by 8.31 which is
r in SI units times 292 Kelvin 25 degree c; and that works out to be 2.2 into 10 power
3.
We will work out one more problem to become better comfortable with these concepts, which
is this example, 6.3 NADH nicotinamide adenine dinucleotide the hydrogenated form of that
NADH, is a molecule that has energetic significance. You can recall this, you know NADH molecules
result from the glycolysis pathway the TCA cycle, and they are the electron carriers
between these reactions and the oxidative phosphorylation pathway in the cell. If you
recall the metabolism of the cell, the primary metabolism and the from your biochemistry
course; you would have done this quiet in detail. One of the reaction pathways in the
cell of importance is rather of importance in lactic acid production is the lactic acid
dehydrogenase catalyzed conversion of pyruvate lactate. Remember, the end point of glycolysis
is considered to be pyruvate, and that goes to lactate by this lactate dehydrogenase reaction,
that is how you get lactate and lactococcus lactis essentially has this pathway completely
channelized and therefore, such organisms are used for production of lactic acid, and
that is essentially what produces your curd from milk.
So, that is represented as or let me complete this; of importance in lactic acid production
is the lactic acid lactic dehydrogenase catalyzed conversion of pyruvate to lactate, coupled
with NADH breakdown. This NADH breakdown also happens there and also lactate could be replaced
with ethanol in some cases, in some organisms and there also you have an NADH breakdown.
This is represented as, pyruvate plus NADH plus H plus giving you lactate plus ***. What
is needed is, to estimate the free energy change for the above reaction, when the ratios
of concentrations of lactate to pyruvate; as well as *** plus to NADH. You know NADH,
when it gives out a hydrogen ion, becomes *** plus and the ratios of lactate to pyruvate;
as well as NADH to *** plus to *** H are in the first instance one and the second instance
thousand. The standard reduction potentials for the
half reactions; recall what half reactions are given next to them pyruvate plus 2 H plus
2 e minus gives you lactate and this is one half reaction, where the standard reduction
potential E naught dash; again the dash refers to a pH 7 condition equals minus 0.19 volt
and NADH. I am sorry, this is a standard reduction potential here; NADH *** plus plus 2 H plus
plus 2 e minus giving you NADH plus H plus; and here the standard reduction potential
of this half reaction is, minus 0.32 volts. Take about 25 minutes and work this out. Go
ahead please.
You hopefully, would have recalled something to do with half reactions, reduction potentials
and so on.
If you did not, I will show you how we go about it. And also in this case, we are looking
at two concentration ratios. So let us go through the solution; let us number the reactions
as follows, pyruvate plus 2 H plus plus 2 e minus giving you lactate, let us call that
as 1; and *** plus plus 2 H plus plus 2 e minus giving you NADH plus H plus, let us
call that equation 2. And the needed reaction that, we require in the problem statement
is obtained as, 1 minus 2; and therefore, the delta E naught dash that we require is
nothing but E naught 1 dash minus E naught 2 dash; this you, why do not you work it out;
1 minus 2, you would have done this already. Check whether, 1 minus 2 gives you the reaction
that we are looking for. So, delta E naught dash equals E naught 1
dash minus E naught 2 dash; which is taking the values from the problem statement E naught
1 dash was minus 0.19 volts; and E naught 2 dash was minus 0.32 volts minus 0.19 minus
of minus 0.32 plus 0.13 volts. And, we know from earlier classes the relationship between
delta G and the difference in the reduction potentials that is, delta G dash equals minus
nF, the faraday's constant times delta E naught dash; and n in this case, the number of electrons
transferred happens to be 2; therefore, minus 2 into 92500 is a faraday's constant in the
required units times 0.13 volts, we get 25.1 kilo joules per mole. Recall that faraday's
constant is essentially a conversion factor between energy units in normal energy kilo
joules and in electrical energy units.
Now, the above value is the standard free energy change at pH 7; that is for part a,
when the ratios of lactate to pyruvate as well as *** plus to NADH are equal to 1. Why
is that? To appreciate this, the second ratio is given; for non standard ratios, we need
to go back to the half reactions itself. To do that, we need to look at the Nernst equation
that says, E equals E naught dash plus RT by nF, natural log of the oxidized form concentration
divided by the reduce form concentration. Therefore, E 1 turns out to be minus 0.19
E naught dash plus RT by nF substituted here, log of 1 by 1000 the ratio was 1000 of lactate
to pyruvate; therefore, pyruvate to lactate turns out to be 1 by 1000, which is minus
0.279 volts and E 2 in the similar fashion, happens to be minus 0.231 volts.
And therefore, delta E is E 1 minus E 2, which is minus 0.279 minus of minus 0.321, which
is minus 0.048; and delta G dash, which is minus nF delta E naught dash is minus n, which
is 2 F 96500 delta E naught dash, we have calculated here as minus 0.048, and the value
turns out to be plus 9.264 kilo joules per mole. Now, that the delta G value is positive,
and you know from high secondary school that for a spontaneous reaction, the delta G has
to be negative, and so what this essentially means is that the reaction as written cannot
proceed spontaneously. In fact, the reverse reaction would be the one that is spontaneous
under the given concentration ratios of 1000 to 1, whereas the first part was spontaneous
as written. So, that was interesting as suppose. When we are out of time, so when we come back,
we will begin with something new, which is a formalism to consider electrolytes in biological
systems. See you then.