Statistical model of ligand substitution
$begingroup$
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
3 hours 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 3 hours ago
andselisk
19.2k662125
19.2k662125
asked 3 hours ago
Shoubhik Raj MaitiShoubhik Raj Maiti
1,398732
1,398732
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
3 hours 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
3 hours 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
3 hours 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
3 hours 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 |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "431"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f112749%2fstatistical-model-of-ligand-substitution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 |
$begingroup$
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 2 hours ago
orthocresol♦orthocresol
40.3k7117247
40.3k7117247
add a comment |
add a comment |
Thanks for contributing an answer to Chemistry Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f112749%2fstatistical-model-of-ligand-substitution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
3 hours ago