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What can one do with the mass action
law? A lot - but it is not always very obvious. Lets ask a few "dumb"
questions and see how far we get. |
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First we look at a really simple
reaction, e.g |
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To keep it easy, we start with equal amounts of
H2 and Cl2 . |
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How much HCl do we get? Notice
that we have the same number of mols
on both sides of the reaction equation. |
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Well, in equilibrium (denoted by [..]) we
have |
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or, with [H2] = [Cl2] =
[equ] |
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One equation with two unknowns; not
sufficient for calculating numbers. |
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But then we also have the condition that the
number of the atoms involved stays constant, i.e.[H2] + 2[HCl]
= constant = e.g. the number of H2 mols before the
reaction. |
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Next, a little bit harder. Lets start
with arbitrary concentrations of something and see what we can say about the
yield of the reaction. For varieties sake
lets look at |
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Again a simple reaction with the same number of
mols on both sides, so we do not have to
worry about the precise form of the mass action law. |
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We start with
n0H2 and
n0CO2 mols of the reacting gases and
define as the yield y the
number of mols of H2O that the reaction will produce at
equilibrium. This leaves us with |
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nH2O |
= |
y |
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nCO |
= |
y |
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nH2 |
= |
n0H2
y |
= |
equilibrium
concentration of H2 |
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nCO2 |
= |
n0CO2 y |
= |
equilibrium
concentration of CO |
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Σ n |
= |
n0 =
n0H2 +
n0CO2 |
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The last equation holds because the mol count
never changes in this example. |
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The mass action law now gives |
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y2
(n0H2 y) ·
(n0CO2 y) |
= |
K |
y |
= |
( |
1
2 |
( |
1 K |
) |
) |
· |
( |
( |
n0 · K ± |
( |
(n0 · K)2 + 4 · (1
K) · n0H2 · (n0
n0H2) · K |
) |
1/2 |
) |
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The starting concentration of
CO2, i.e nCO2, is expressed as
nCO2 = n0
n0H2. |
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Looks extremely messy, but this is
just the standard solution for a second order equation. Whatever this solution
means in detail, it tells us that the yield is a function of the starting
concentrations of the ingredients. |
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What kind of starting concentrations will give us
maximum yield? To find out, we have to form
dy/dn0H2 = 0. |
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Well, go through the math yourself; this is
elementary stuff. The solution is |
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n0H2 |
= |
n0
2 |
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n0CO2 |
= |
n0
2 |
= n0H2 |
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In other words: maximum yield is
achieved if you mix just the right amounts of the starting stuff. This result
is always true, even for more complicated reactions. |
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At this point we stop, again because otherwise we
might turn irreversibly into chemists. |
© H. Föll