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Topic: Equilibrium with parallel reactions  (Read 6562 times)

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Offline Rutherford

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Equilibrium with parallel reactions
« on: September 19, 2013, 01:41:34 PM »
Serum transferrin (abbreviated: Tf) is a monomeric protein whose main function in the human body is the transport of iron(III). Each transferrin molecule can bind up to two iron(III) ions with stepwise binding constants K1 and K2 at biological conditions except that the temperature is 25 °C corresponding to the react ions:
FeIII + Tf  ::equil:: (FeIII)Tf       K1 = 4.7·1020
FeIII + (FeIII)Tf  ::equil:: (FeIII)2Tf     K2 = 2.4·1019

In the diferric protein, (FeIII)2Tf, the two iron(III) ions are bound at two similar, but non-identical sites, and the two possible monoferric protein products, (FeIII)Tf, can be denoted {FeIII · Tf} and {Tf · FeIII}. Their relative abundance at equilibrium is given by the constant
K = [{Tf · FeIII }] [{FeIII · Tf }]-1 = 5.9.

The questions are:
1)Calculate the values of the two constants K1' = [{FeIII . Tf}] [FeIII]-1 [Tf]-1 and K1" =
[{Tf . FeIII }] [FeIII]-1 [Tf]-1, respectively, corresponding to the formation of each monoferric form of transferrin.
This is not so hard, K1'=6.8·1019 and K1"=4·1020.

2)Calculate the values of the two constants K2' = [(FeIII)2Tf] [FeIII]-1 [{FeIII . Tf }]-1 and
K2" = [(FeIII)2Tf] [FeIII]-1 [{Tf . FeIII}]-1 respectively, corresponding to the formation of diferric transferrin from each of the monoferric forms.
Here I am confused. They used the following expression:
K1'K2'=K1"K2"=K1K2
By what logic did they get this?

Offline Big-Daddy

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Re: Equilibrium with parallel reactions
« Reply #1 on: September 19, 2013, 04:32:44 PM »
Don't know about the method, IChO uses odd methods sometimes. I am assuming this actually is IChO. But this answer is not that hard anyway: (I'm uncomfortable with the notation used so I've gone with [Fe3+ (A) Tf] for {FeIII · Tf} and [Fe3+ (B) Tf] for {Tf · FeIII}, bear with me  :P )

c(total)[Fe3+ Tf] = [Fe3+ (A) Tf] + [Fe3+ (B) Tf]
Kt = [Fe3+ (B) Tf] / [Fe3+ (A) Tf]

So as you already noticed [Fe3+ (A) Tf]*Kt=[Fe3+ (B) Tf] and then

c(total)[Fe3+ Tf] = [Fe3+ (A) Tf] * (1 + Kt)

And c(total)[Fe3+ Tf] = [Fe3+ (B) Tf] * (1 + 1/Kt)

Now we have the desired equilibrium [Fe3+ (A) Tf] + Fe3+  ::equil:: [(Fe3+)2 Tf] and also another equilibrium constant

K2 = [(Fe3+)2 Tf] / ([Fe3+] * c(total)[Fe3+ Tf]

This can be rewritten

K2 = [(Fe3+)2 Tf] / ([Fe3+] * [Fe3+ (A) Tf] * (1 + Kt) = [(Fe3+)2 Tf] / ([Fe3+] * [Fe3+ (A) Tf]) * 1/(1 + Kt)

Whereas for our desired equilibrium [Fe3+ (A) Tf] + Fe3+  ::equil:: [(Fe3+)2 Tf]  the constant is

K2A = [(Fe3+)2 Tf] / ([Fe3+] * [Fe3+ (A) Tf])

And thus it is clear that K2A = K2 * (1 + Kt) and symmetrically K2B = K2 * (1 + 1/Kt). Sub in Kt=5.9 and K2=4.7*1020 and we get K2A = 1.656 * 1020 mol-1dm3 and K2B = 2.8068 * 1019 mol-1dm3.

You may want to check, not completely sure if that's correct but the logic seems ok to me.

Offline Big-Daddy

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Re: Equilibrium with parallel reactions
« Reply #2 on: September 19, 2013, 05:27:38 PM »
I have an interesting question of my own to ponder from this problem: let's say you were given K2A and K2B, but not Kt; how could you work out K2?

Offline Rutherford

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Re: Equilibrium with parallel reactions
« Reply #3 on: September 20, 2013, 12:31:57 PM »
I I had a similar idea but this confuses me: How can the ratio Kt be used for both the first part and the second part of the problem? If the ratio is established after the first reaction, why is it constant after the equilibrium of the second reaction is established, when these are equilibrium constants with different values?

Offline Big-Daddy

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Re: Equilibrium with parallel reactions
« Reply #4 on: September 20, 2013, 01:30:13 PM »
I don't understand what you're asking. Kt is a simple equilibrium constant or ratio between two forms of the one-substituted Tf. Even if each of these transfers out to the same two-substituted Tf form, the same thing applies; this equilibrium constant always has to hold true. If the system is at equilibrium, each equilibrium constant's expression (and this ratio is the same as an equilibrium constant expression here) is valid and always true.

It may not be immediately obvious that Kt is an equilibrium constant as well as a ratio but it is called a "constant" (which is enough for it to hold, in all cases, for this problem) not to mention that it is given the symbol K in a problem flooded with other equilibria. So I think it's clear enough.

Offline Rutherford

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Re: Equilibrium with parallel reactions
« Reply #5 on: September 20, 2013, 03:26:35 PM »
Okay, then it can be easily proven that K1'K2'=K1"K2"=K1K2 (use your method for the first case, too). Which means that this expression is valid only if the ratio of the quantities of the intermediate compounds is constant. Thank you for the help.

Offline Big-Daddy

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Re: Equilibrium with parallel reactions
« Reply #6 on: September 20, 2013, 04:14:42 PM »
Yes, the expression is easy to derive and tells us a lot about concentration relationships if we choose to find them. But in my opinion it is an absolutely bizarre and non-obvious place to start with the solution of either part 1 or part 2.

If they'd asked students to derive the formula from the beginning (before part 1 or 2 as here), I think I would have found it much more difficult than the problem itself, and may have found it impossible due to not knowing where to start (clearly the place to start is trying to express K_1A, K_1B, K_2A, K_2B in terms of K_1, K_2 and K_t which are the three equilibria given; maybe I could have gone from here to the solution). In any case I would put it down to another weird IChO method use, unless you know where this principle comes from. IMO the real way to solve the problem is just by looking at the equilibrium constant expressions outright.

No problem :)

Actually it was a good enough question. You normally don't get them these good (or requiring this kind of thinking) from National Olympiads so I will assume it is IChO; what IChO did it come from?

Offline Rutherford

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Re: Equilibrium with parallel reactions
« Reply #7 on: September 21, 2013, 05:51:21 AM »
32nd.

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