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Topic: Free energy of a biological reaction - Q&A  (Read 2761 times)

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Offline Iamyourfather

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Free energy of a biological reaction - Q&A
« on: November 24, 2012, 05:23:19 PM »
The value of the equilibrium constant of the chemical reaction
A + 2B C + D
is 8.5 x 10-3 mM-1 at 27oC. Find the value of the free energy change for this reaction maintained at a steady state at the following concentrations of the reaction components [A] = 100 μM; [ B] = 0.2 M; [C] = 200 μM; [D] = 170 mM. The value of R is 8.31 J mol-1 K-1.

Keq = 8.5 x 10-3 mM-1 = 8.5 x 10-6 M-1
ΔG'° = -RT In K
ΔG'° = ΔG' = -RT In K/Γ
Γ = 1 M x 1 M/1 M x 2 M = 0.5 M
In (8.5 x 10-6) = - 11.78
R = 8.31 J K-1 mol-1; T = 273 K + 27 K = 300 K
ΔG'° = -8.31 x 300 x (-11.78) = 29368 J mol-1= 29.4 KJ mol-1
ΔG’ = ΔG°' + RT In ([C’] [D’]/[A’]/[B’]2
ΔG’ = (29368 + 8.31) In ([200] [1.7 x 105]/ [100]/ [2 x 105]2)
= (29368 + 2493) In (3.4 x 107/4.0 x 1012)
= 31861 x In (8.5 x 10-6)
ΔG’ = - 37199 J mol-1
ΔG’ = - 37.2 KJ mol-1

Is the end value correct?

A few problems along the way:

I wasn't sure whether I was meant to mention the mass ratio (Γ). If I did, what was I meant to do with it (because it was 0.5 M and not the typical 1 M)?

Was the Keq meant to be converted from mM to M? Was it necessary?

Lastly were all the concentrations within the reaction components meant to be converted to μm?






« Last Edit: November 24, 2012, 05:41:24 PM by Iamyourfather »

Offline Yggdrasil

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Re: Free energy of a biological reaction - Q&A
« Reply #1 on: November 24, 2012, 07:08:05 PM »
For this reaction [tex]\Gamma = \frac{(1M)(1M)}{(1M) (1M)^2} = (1 M)^{-1}[/tex]

For the expression K/Γ to be unitless, you must convert K to units of M-1 in order to match the units of Γ.  Similarly, for calculating ln(Q) in the calculation of ΔG', you must convert the concentrations to units of M.  Basically, anytime you are taking the natural log of an equilibrium constant (K) or reaction quotient (Q), you should always convert the concentrations to units of M (not mM or µM) before proceeding.  This has to do with the fact that 1M is chosen as the reference concentration for calculation of ΔG'°.

Offline Iamyourfather

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Re: Free energy of a biological reaction - Q&A
« Reply #2 on: November 24, 2012, 09:01:55 PM »
For this reaction [tex]\Gamma = \frac{(1M)(1M)}{(1M) (1M)^2} = (1 M)^{-1}[/tex]

For the expression K/Γ to be unitless, you must convert K to units of M-1 in order to match the units of Γ.  Similarly, for calculating ln(Q) in the calculation of ΔG', you must convert the concentrations to units of M.  Basically, anytime you are taking the natural log of an equilibrium constant (K) or reaction quotient (Q), you should always convert the concentrations to units of M (not mM or µM) before proceeding.  This has to do with the fact that 1M is chosen as the reference concentration for calculation of ΔG'°.

Thanks! x

Would the answer be ΔG’ = + 16.9 KJ mol-1 ?


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