Suppose a molecule
Y
, carboxylic acid for instance, can form dimers
Y
:
Y
denoted as
Y
2
. The forward reaction between two molecules of
Y
to form a dimer has rate constant
k
f
=
5
×
1
0
−
5
mM
−
2
s
−
1
, and the backward reaction of one dimer breaking into two monomers has rate constant
k
b
=
2
mM
−
1
s
−
1
.
You prepare a solution of 1 0 0 mM Y , what is the equilibrium concentration (in mM ) of the dimer Y 2 ?
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But I think my solution was wrong. [ Y ] is not equal to 1 0 0 m M but ( 1 0 0 − x ) m M .
Let the concentration of dimer at equilibrium be x m M . This implies that concentration of monomer would be ( 1 0 0 − 2 x ) m M , because 2 molecules of monomers give one dimer. Now equate the rates of reactions to get the equation 5 × 1 0 − 5 ( 1 0 0 − 2 x ) 2 = 2 x . Solve the quadratic to get the answer 0 . 2 4 7 5 3 .
Seriously, we have to solve the quadratic? Why can't I approx it and take 1 0 0 − 2 x as 1 0 0 only?
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Simply due to the fact that the extent of dimerization(of the acid) is non-negligible and needs to be. Evaluated For better accuracy
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I have got perfect 0 . 2 5 m M as answer.
The law of mass action gives that for a reversible reaction 2 Y ⇌ Y 2 ,
the forward reaction rate = k f [ Y ] 2
the backward reaction rate = k b [ Y 2 ]
where [ Y ] and [ Y 2 ] are the concentrations of Y and Y 2 respectively.
At equilibrium, the forward and back reaction rates are equal, therefore,
k f [ Y ] 2 = k b [ Y 2 ] ⟹ [ Y 2 ] = k b k f [ Y ] 2 = 2 5 × 1 0 − 5 × 1 0 0 2 = 0 . 2 5