Principle of chemical equilibrium
• Criterion of spontaneity:
G<0 at constant T and P
reaction mixture tends adjust to the
composition until G is minimum
Equilibrium may be different
composition of the reaction
mixture
G
A. The equilibrium is reached at high
reactant concentration
B. Equilibrium is reached at equal
concentration or reactant and
products
Reactant
Products
C. Equilibrium is reached at high amount
of product
• The reaction Gibbs energy
Let’s consider:
A. a conversion of glucose-6phosphate (G6-P) into fructose-6 phosphate (F-6-P)
B. Formation of ammonia
N2(g) +3H2(g)  2NH3
G = (NH3) 2 n –[(H2) 3n + (H2) n]
rG = 2 (NH3)–[3(H2) + (H2) ]
For transfer
ofBdnJ molecules from
A
nstate
tonstate
J  J A
J J B
dG   JB dnJ   JA dn j  (  JB   JA )dnJ
G = ((G-6-P) - (F-6-P)) n
Reaction Gibbs energy: rG
General expression
aA + bB  cC + dD
rG =[ c(C)+ d(D)]–[a(A) + b(B) ]
rG = G/ n = ((G-6-P) - (F-6-P))
a, b, c, d – stochiometric coefficients
Meaning of the reaction Gibbs energy
A. Reaction Gibbs energy corresponds to the difference of the chemical potential of the
reactant and product at the composition of reaction mixture
B. Reaction Gibbs energy is a slope of the plot of Gibbs energy versus the
compositions of the reaction mixture
Point A: the slope is negative – reaction
proceeds in the direction of products
G
Point B: the slope is zero, rG = zero. The
reaction reaches equilibrium
rG = G/ n
Sponta
-neous
A
Spontaneous
B
Reactant
C
Point C: the slope is positive, the reaction
proceeds in the direction of reactants, the
product decomposis into reactants. The
reverse reaction is spontaneous
Products
rG < 0 for spontaneous reaction - exergonic reaction
rG = 0 for equilibrium
rG >0 for nonspontaneous reaction – endergonic reaction
- Chemical potential is function of system composition
 J   Jo  RT ln aJ
for solute: aJ = c/co
For pure solid and pure liquid : aJ = 1
For gases: aJ = p/po
Lets express the reaction Gibbs energy in terms of standard chemical potential and activities
 r G  [c (C )  d ( D)]  [a ( A)  b ( B)]
 r G  [c o (C )  cRT ln aC  d o ( D)  dRT ln aD ] 
[a o ( A)  aRT ln a A  b o ( B)  bRT ln aB ]
 r G o  [c o (C )  d o ( D)]  [a o ( A)  b o ( B)]
 r G   r G o  RT {[c ln aC  d ln aD ]  [a ln a A  b ln aB ]}
c d
a
a
 r G   r G o  RT ln Ca Db
a A aB
aCc aDd
Q a b
a A aB
Q – reaction quotien
 r G   r G  RT ln Q
At equilibrium
rG  0
c d
a
a
G o   RT ln( Ca Db ) equilibrium
a A aB
aCc aDd
K  ( a b ) equilibrium
a A aB
G o   RT ln K
Q = K only at equilibrium
For K > 1 rGo < 0
products are dominant at equilibrium, the
reaction is thermodynamically feasible
For K< 1
 rG o < 0
Reactants are dominant at equilibrium
Comparison of rGo and rG
If rGo is large negative or large positive, the
sign of rG is determined by rGo and thus the
direction of the reaction is determined by rGo
For reactions with small rGo , the reaction can
go either way, depending on Q term.
rG = RTln(Q/K) and for K>Q the reaction is
spontaneous and for K<Q reaction proceeds
in reverse direction
Thermodynamic criteria of spontaneity
Go
rGo = rHo- TrSo
 r Ho
 r So
 rGo
Conditions
<0
>0
<0
always
>0
<0
>0
always
>0
>0
<0
for T> rHo/rSo
<0
<0
<0
For T< rHo/rSo
 rG o
T  rS o
rHo
0
K<1
Endothermic reaction can have K>1 and
exothermic reaction can have K<1
depending on TrSo
rGo can be determined from rHo and rSo
rGo = rHo- TrSo
 r G o   mG o ( products )   mG o (reac tan ts )
K>1
Temperature
Standard enthalpy of formation: is the change of enthalpy that accompanies the
formation of 1 mole of a substance in its standard state.
rHo = fHo (products)- fHo (reactants)
Standard Gibbs energy of formation
rGo = fGo (products)- fGo (reactants)
fGo = 0 for an element in its more stable state
fGo : determine the stability of the compound
Compound with fGo > 0 is unstable and decomposes into its elements
Such compounds can not be synthesized directly from their elements and different
synthetic roads have to be used
However fGo = + 124 kJ mol-1
Despite it’s thermodynamic instability, benzene does not decompose, decomposition is
kinetically unfavorable
• The fractional saturation
For a oxygen binding to myoglobin, determine the fraction of protein with bound
oxygen molecule knowing that the equilibrium constant is K = 107 M-1 at 1 mM
oxygen concentration.
• The response of a equilibrium to the reactin
conditions
• Le Chatelier principle
– How does the equilibrium shift if you upon
addition of reactant or removal of product?
• The effect of temperature
Increase of temperature shifts the equilibrium
towards reactants for the exothermic reaction
and towards products for the endothermic
reaction.
Effect of T on K Van’t Hoff equation
o
 r H 1 1 lnK
ln K ' ln K 
(  )
R T T'
r H
r S
ln K  

RT
R
o
o
1/T
Effect of the pressure
-For gases the equilibrium will shift to the side that produces lower number of
molecule
-Pressure has no impact on the equilibrium position for the condense phase or if
there is no change in the number of moles going from reactants to products
Effect of catalyst
• Homework
1. Calculate the change in entropy for the following processes:
a) mixing of 1 mol of nitrogen and 1 mol of oxygen
b) mixing of 2 moles of argon and 1 mol of helium
(assume ideal behavior).
2. A solution of ethanol and n-propanol behaves ideally. Calculate the chemical
potential of ethanol in solution relative to that of pure ethanol when its mole
fraction is 0.4 at boiling point (78.3 oC).
3. The solubility of N2 in the blood at 37 oC and a partial pressure of 0.8 atm is 5.6x104 mol L-1. A deep sea diver breathes compressed air with a partial pressure of N2
equal of 4.0 atm. Assume that the total volume of blood in the body is 5.0 L.
Calculate the amount of N2 gas release (in liters) when the diver returns to the
surface of water, where the partial pressure of N2 is 0.80 atm.
4. Two beakers, 1 and 2, containing 50 mL of 0.10 M urea and 50 mL of 0.20 M uea,
respectively, are placed under a tightly sealed bell jar at 298 K. Calculate the mole
fraction of urea in the solutions at equilibrium. Assume an ideal behaviour. (Hint:
use the Raoult’s law and note that at equilibrium, the mole fraction of urea is the
same in both solutions).
5. Trees in the cold climates may be subjected to temperatures as low as 60 oC. Estimate the concentration of an aqueous solution in the body of the
tree that would remain unfrozen at this temperature. Is this a resonable
concentration?
6. A common untifreeze for car radiators is ethylene glucol. How amy
mililiters of this substance would you add to 6.5 L of water in the radiator if
the coldest day in winter is – 20 oC? Would you keep this substance in the
radiator in the summer to prevent the water from boiling? (The density and
the boiling point of ethylene glycol are 1.11 g/mL and 470 K, respcetively.
7. The tallest trees known are the redwoods in California. Assuming the
heights of a redwood to be 105 m. Estimate the osmotic pressure to push
the water up to the treetop.
8. Lyzozyme extracted from chicken egg white has a molecular mass of 13,930
g/mol. Exactly 0.1 g of this protein is dissolved in 50 g of water at 298 K.
Calculate the wapor pressure lowering, the depression of freezing point, the
elevation of boiling point and the osmotic pressure of this solution. The vapour
pressure of pure water at 298 K is 23.76 mmHg.
9. A nonvolatile organic compound was used to make up two solutions. Solutions
A contained 5.00 g of Z compound dissolved in 100 g of water. The solution B
contains 2.31 g of B dissolved in 100 g of benzene. Solution A has a vapor pressure
of 745.5 mmHg at the normal boiling point of water and solution B has the same
vapor pressure at the boiling point of benzene. Calculate the molar mass of
solution A and B, and account for the difference.
10. At 85 oC, the vapor pressure of A is 566 torr and that of B is 250 torr. Calculate
the composition of A and B that boils at 85 oC when the pressure is 0.60 atm. Also,
calculate the composition of the vapor mixture. Assure an ideal behavior.
11. A certain dilute solution has an osmotic pressure of 12.2 atm at 20 oC. Calculate
the difference between the chemical potential of the chemical potential of the
solvent in the solution and that of pure water. Assume that the density is the same
as that of water. (Hint: Express the chemical potential in terms of mol fractions, x1,
and rewrite the osmotic pressure equation as V=n2RT, where n2 is the number of
mole of the solute and V=1L)
• From the text book:
Chapter 6:
Problems: 6.1 (6.12); 6.2 (6.13); 6.4 (6.20); 6.7;
6.15 (6.28); 6.9 (6.22); 6.23 (6.32); 6.9 (6.22);
6.16; 6.17; 6.39 (6.14); 6.40 (6.16);
Chapter 7:
7.8 (7.12); 7.9 (7.13); 7.12 (7.16); 7.17 (7.19); 7.18
(7.20); 7.23 (7.25); 7.31 (7.33); 7.32 (7.34) .