Statistical model of ligand substitution Announcing the arrival of Valued Associate #679:...
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Statistical model of ligand substitution
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there a reason for the mathematical form of the equilibrium constant?Why is ligand substitution only partial with copper(II) ions and ammonia?Ligand Binding Paradox?How do we decide which ligand is monodentate, bidentate, etc.?In the reversible reactions of acyl substitution, how do backward reactions happen by being against the forward reaction drives?Why is thiourea a monodentate ligand?Is glycine strong or weak field ligand? If yes, how?What is an uninegative ligandWhy does ligand substitution occur when OH- is a better ligand than NH3?Is there a difference between a chelate ligand and a polydentate ligand?
$begingroup$
Recently, I was told that in case of a particular step of a generic ligand substitution reaction:
$$ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$
The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to
$$frac{N - n}{n + 1}$$
by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?
equilibrium coordination-compounds
$endgroup$
add a comment |
$begingroup$
Recently, I was told that in case of a particular step of a generic ligand substitution reaction:
$$ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$
The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to
$$frac{N - n}{n + 1}$$
by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?
equilibrium coordination-compounds
$endgroup$
1
$begingroup$
I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
$endgroup$
– Tyberius
54 mins ago
add a comment |
$begingroup$
Recently, I was told that in case of a particular step of a generic ligand substitution reaction:
$$ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$
The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to
$$frac{N - n}{n + 1}$$
by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?
equilibrium coordination-compounds
$endgroup$
Recently, I was told that in case of a particular step of a generic ligand substitution reaction:
$$ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$
The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to
$$frac{N - n}{n + 1}$$
by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?
equilibrium coordination-compounds
equilibrium coordination-compounds
edited 1 hour ago
andselisk
19.2k662125
19.2k662125
asked 1 hour ago
Shoubhik Raj MaitiShoubhik Raj Maiti
1,393732
1,393732
1
$begingroup$
I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
$endgroup$
– Tyberius
54 mins ago
add a comment |
1
$begingroup$
I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
$endgroup$
– Tyberius
54 mins ago
1
1
$begingroup$
I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
$endgroup$
– Tyberius
54 mins ago
$begingroup$
I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
$endgroup$
– Tyberius
54 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,
$$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$
and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for
$$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$
is given by
$$begin{align}
K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
&= expleft(frac{Delta_mathrm r S}{R}right) \
&= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
&= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
&= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
&= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
&= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
&= frac{N-n}{n+1}
end{align}$$
The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,
$$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$
and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for
$$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$
is given by
$$begin{align}
K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
&= expleft(frac{Delta_mathrm r S}{R}right) \
&= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
&= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
&= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
&= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
&= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
&= frac{N-n}{n+1}
end{align}$$
The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.
$endgroup$
add a comment |
$begingroup$
I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,
$$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$
and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for
$$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$
is given by
$$begin{align}
K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
&= expleft(frac{Delta_mathrm r S}{R}right) \
&= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
&= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
&= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
&= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
&= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
&= frac{N-n}{n+1}
end{align}$$
The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.
$endgroup$
add a comment |
$begingroup$
I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,
$$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$
and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for
$$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$
is given by
$$begin{align}
K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
&= expleft(frac{Delta_mathrm r S}{R}right) \
&= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
&= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
&= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
&= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
&= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
&= frac{N-n}{n+1}
end{align}$$
The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.
$endgroup$
I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ce{MA_nB_{$N-n$}}$,
$$Omega = {N choose n} = frac{N!}{n!(N-n)!} quad left[ = {N choose N-n} right]$$
and that for the individual molecules $ce{A}$ and $ce{B}$, $Omega = 1$, then the equilibrium constant $K$ for
$$ce{MA_nB_{$N-n$} + A <=> MA_{n+1}B_{$N-n-1$} + B}$$
is given by
$$begin{align}
K &= expleft(frac{-Delta_mathrm r G}{RT}right) \
&= expleft(frac{Delta_mathrm r S}{R}right) \
&= expleft(frac{S_mathrm{m}(ce{MA_{n+1}B_{$N-n-1$}}) + S_mathrm{m}(ce{B}) - S_mathrm{m}(ce{MA_nB_{$N-n$}}) - S_mathrm{m}(ce{A})}{R}right) \
&= exp[lnOmega(ce{MA_{n+1}B_{$N-n-1$}}) + lnOmega(ce{B}) - lnOmega(ce{MA_nB_{$N-n$}}) - lnOmega(ce{A})] \
&= expleft[lnleft(frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})}right)right] \
&= frac{Omega(ce{MA_{n+1}B_{$N-n-1$}})Omega(ce{B})}{Omega(ce{MA_nB_{$N-n$}})Omega(ce{A})} \
&= frac{N!}{(n+1)!(N-n-1)!} cdot frac{n!(N-n)!}{N!} \
&= frac{N-n}{n+1}
end{align}$$
The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.
answered 19 mins ago
orthocresol♦orthocresol
40.3k7117247
40.3k7117247
add a comment |
add a comment |
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$begingroup$
I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. chem.libretexts.org/Bookshelves/Inorganic_Chemistry/…
$endgroup$
– Tyberius
54 mins ago