MBB 323

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MBB 323

• Lecture 13
• Read p121-139
• Assignment 3 available, due Oct. 20

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Midterm results
3
Gibbs Free Energy

• Consider the ssDNA lt-gt dsDNA problem further.
• Last lecture we wrote
• X 5-ATAGCA and Y 5-TGCTAT
• (with X and Y being the concentrations of each
  strand).
• G2(T,P,X,Y) the state in which the oligos are
  double stranded.
• G1(T,P,X,Y) the state in which the oligos are
  single stranded.
• In particular we did a calculation with X and Y
  equal to 1M (in the standard state).
• This gave us DG0 G2 - G1
• We used this value to show that at equilibrium
  conditions the equilibrium constant (K) was 1416.
  (You could picture two systems that exchange
  particles until equilibrium is reached).
• What does this mean in terms of relative
  probabilities (see overhead)?
• How would we use this information to find DG for
  other concentrations of X and Y?

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Gibbs Free Energyand Chemical Potential

• nX The top strand in moles.
• nY The bottom strand in moles.
• nZ The number of moles of dsDNA
• - nZ is constrained by nX , nY and the
  approximate volume of the system, this is why
  last lecture I wrote G(T,P,X,Y).
• G(T,P, nX, nY, nZ)
• dG -SdT VdP mXdnX mYdnY mZdnZ
• For constant T and P, we must have that dT 0,
  dP 0.
• dG mXdnX mYdnY mZdnZ

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Gibbs Free EnergyThe sum rule of partial molar
quantities

• Consider an open DNA system
• dG mXdnX mYdnY mZdnZ
• If we hold the chemical potentials fixed we can
  integrate this expression, keeping the relative
  proportions of X, Y and Z fixed. You can imagine
  X, Y and Z strands being added to the system from
  the outside incrementally to reach the final
  amounts desired.
• This integration gives
• G(T,P, nX, nY, nZ) nXmX nYmY nZmZ (for
  constant T and P).

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Gibbs Free EnergyChemical potential and
stoichiometry

• Now consider a closed DNA system
• The stoichiometry of this particular example
  requires
• dnX dnY -dnZ
• Spse that the oligo is self complimentary. Then
  only need to consider nX and nZ
• dnX/2 -dnZ (read p123-124 for more detail)
• (in other words it takes two self complimentary
  strands to make one dsDNA)

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Gibbs Free EnergyChemical potential and
stoichiometry

• Now consider a closed DNA system
• XY lt-gt Z
• The stoichiometry of this particular example
  requires
• Case 1 dnX dnY -dnZ
• Spse that the oligo is self complimentary. Then
  only need to consider nX and nZ
• 2X lt-gt Z
• Case 2 dnX/2 -dnZ (read p123-124 for more
  detail)
• (in other words it takes two self complimentary
  strands to make one dsDNA)

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Gibbs Free Energy Chemical potential and
stoichiometry

• Again at constant T and S
• Case 1 dnX dnY -dnZ -da
• dG mXdnX mYdnY mZdnZ
• dG -mX da -mY da mZ da
• (mZ - mX - mY )da
• dG/da mZ - mX - mY (Change in G per mole of
  reaction p 124).
• Case 2 dnX/2 -dnZ -da
• dG mXdnX mZdnZ
• dG -2mX da mZ da
• (mZ - 2mX)da
• dG/da mZ - 2mX
• Either case must be zero at equilibrium meaning
  (DG 0).

9
Gibbs Free Energy Standard state to arbitrary
states activity

• Let concentrate on dilute solutions (the texts
  starts with ideal gases).
• How does the chemical potential depend on
  concentration?
• mB m0B RTln(aB)
• where the unitless activity aB is defined as
• aB gBCB
• where CB is the concentration in molar.
• and gB is in inverse molar and is a function of
  c.
• m0B is the chemical potential in the extrapolated
  standard state 1M (read carefully p136-137).
• Rember aB is a number!

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Gibbs Free Energy Standard state to arbitrary
states activity

• Take our DNA example (Case 1) then if we add a
  certain amount of X and Y (thus defining a
  particular initial and final state)
• DG mZ - mX - mY
• DG m0Z - m0X - m0Y RTln(aZ)
  -RTln(aX)-RTln(aY)
• DG DG0RTln(aZ/(aXaY))
• DG DG0RTln(Q))
• This gives us the change in free energy of a
  particular state relate to the standard state (1M
  X and 1M Y in this case) with a factor that is
  determined by the actual activities of the
  reactants. See p131.
• The only complication is that it could be hard to
  relate activities to concentrations.
  Fortunately for dilute solutes g normally is
  close to 1/M.