Download Chem 452 – Homework # 1A

Autumn 2012
452B
hw 7 (A)
CHEM 452B – Homework 7A KEY
Q1) Consider the dissociation of a gas e.g. N 2O4 which is a dimer of NO2 , which is the brown
agent in smog. The gas can form a dimer through the relatively weak N-N interaction. The
reaction is:
Dimer (gas)  2 Monomer (gas)
1-
2
On a per mole basis, the reaction can be thought of as starting with one mole of dimer
so that nD  1   , and nm  2 , as the relative amount of dimer and monomer, the mole
fractions then are based on nTot  nD  nm  1   
The degree of dissociation, , is the fraction of dimer molecules dissociated.
The partial pressures in terms of the degree of dissociation and the overall pressure, P, are:
n
Pm   m
 nTot

 2 
P 
P
1





 n
Pd   d
 nTot

 1 
P 
P
1





a. Find an expression for the reaction quotient QP in terms of  and P only.
2
2
 Pm 
  2  P 
 

 
P0 
4 2
P
 1    P0 


K P  QP 


PD
 1  P
1   1    P0


P0
 1    P0
n 
 2 
 m P 
P
 1  
 nTot 
4 2
Q P
Q  P o 
P 1   2 
2 
 n 
 1 
Pd   d  P  
P
 1  
 nTot 
Qx
Qx  4
b. At 25˚C, suppose the equilibrium constant for the dissociation is 0.14 at all pressures.
Calculate the fraction of dimer molecules dissociated at a total pressure of 1 atm and then
at 5 atm, assuming equilibrium at these pressures. (Answers: 1 atm = 0.18, 5 atm = 0.08)
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Autumn 2012
452B
K   Q 
QP Po
hw 7 (A)
4 2

P 1   2 
0.14
0.14
 0.14
K 
 0.03
P
1
5
Kx
Kx
0.14
0.03
2 

 0.0338  2 

 0.007
K x  4 4.14
K x  4 4.03
K 
K P Po

  0.184
  0.083
c. Find Kx in the two scenarios. Explain why Kx depends on the pressure but Kp does not.
Kp is defined at 1 Atm, so you can change pressure all you like but Kp cannot be affected. But
Kx contains the P in it, so as P goes up Kx goes down, and so the dissocation goes down, as one
sees with these two examples.
0
d. Find the standard molar free energy, G , of this reaction from
0
G 0 rxn  G   RT ln K P .
0
G   RT ln K P  2.47 ln 0.14  4.86 kJ
mol  rxn
e. Find G (using the values in QP) if the system is out of equilibrium such that
Pm  Pd  0.5 Atm . (Answer: 3.2 kJ/mol)
G  RT ln
QP
.52
 2.47 ln
 3.15 kJ
mol  rxn
KP
.5  .14
0
f. For the case of N 2O4 dissociating find the actual G and the equilibrium constant Kp.
(Ans: 2.8x103 J/mole, 0.33 )
o
o
o
Grxn
 H rxn
 T Srxn
 55.2  0.298 176  2.75 kJ
mol  rxn
g. What is the equilibrium constant at 500K (as a result of the heat of the car in which this is
generated). (Ans: 2.2x103)
0
o
o
H rxn
 T Srxn
G
55.2  0.5 176
  ln K P 

 7.9
RT
RT
8.3*.5
K P  2.7 103
0
0
h. For the reaction above, explain the sign of  H and  S .
Both the enthalpy and the entropy are positive because for the reaction an N-N bond is broken,
that costs energy, so the reaction is endothermic, and the entropy is increased because there are
more molecules that have more ways to rearrange.
Q2) If one mole of dimer is present initially in a cylinder at a volume such that its pressure
is1Atm, and the equilibrium constant is Kp=0.14, what is the degree of dissociation if the
volume is fixed (and not the total pressure).
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Autumn 2012
452B
 Pm 
 
P
0.14  K P  QP   0 
PD
P0
hw 7 (A)
2
PD0V  nRT
V  RT  0.082* 298  24.4
PM  2 nD0
 2 

RT
 2
V
PD  1  
2
 KP
1
4 2  K P  K P  0
QP

K P 2  16 K P  K P
 0.174
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b) Compare your answer from this problem with question Q1) where the pressure was held fixed.
The answer is remarkably close to the above (0.174 here vs 0.184 above). The reason the
dissociation is smaller is because the cylinder could no longer relax (expand) to accommodate
more gas molecules. So holding the volume fixed depresses the degree of dissociation. The
pressure is larger here than in the constant pressure system above.
Q3) Many biological macromolecules undergo a transition called denaturation. Denaturation is a
process whereby a structured, biological active molecule, called the native form, unfolds or
becomes unstructured and biologically inactive. The reaction expression is:
native  folded   denatured  unfolded 
For a protein at pH = 2, the enthalpy change associated with denaturation is H o  418 kJ
and the entropy change is S  1.3 kJ
o
molK
mol
at T=298.15 K.
a. Calculate the Gibbs energy change for the denaturation of the protein at pH = 2 and T =
303 K. Assume the enthalpy and entropy are temperature independent between 298.15
and 303 K. (Ans: ~2.41x104 J/mole )
o
o
o
Grxn
 H rxn
 T Srxn
 418  303 1.3  24.1 kJ
mol
This is the free energy per mole of product, given the simplicity of the stoichiometry of the
reaction.
b. Calculate the equilibrium constant for the denaturation of protein at pH = 2 and T = 303
K. (Ans: ~6x10-5)
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Autumn 2012
452B
hw 7 (A)
0
G
24.1
 ln K P 

 9.6
RT
2.52
K P  e9.6  6.8 105
c. Based on your answers for parts (a) and (b), is protein structurally stable at pH = 2 and T
= 303 K?
Because DG0 is a very large positive number, K is a very small number, so Q=K at equilibrium,
and therefore the product must be very small, so the native, folded form is quite stable under
these conditions. The Enthalpy holds it together, and the entropy is not strong enough to break it
apart. At a bit higher temperature the protein will denature.
EXTRA PRACTICE PROBLEMS:
Read the following article and answer the questions below:
Article: Bustamante, Carlos et. al. “Ten years of Tension: Single Molecule DNA Mechanics.”
Nature. 421: (Jan 2003) 423-426.
1.
According to the article the energy for folding a DNA molecule is dependent on three
variables, temperature (T), length (L), and radius of curvature (R). Assume that the flexural
persistence length is constant for this process.
a. Directly from the article, write the expression for E(T, L, R).
b. Write an expression you would use to test if the change in energy for bending a segment
of DNA was a state function at constant temperature.
c. Determine if the energy of bending a segment of DNA is a state function at constant
temperature.
d. Write an expression for the change in energy (dE) required to bend a segment of DNA in
terms of its partial derivatives. Note: does this relationship depend on dE being a state
function?
e. Use your equation from part d. to write an expression for the change in energy (dE)
required to bend a segment of DNA without the partial derivatives. (Remember the
conditions for when this is a state function, constant temperature.)
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