Statistical model of ligand substitutionIs 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?
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Statistical model of ligand substitution
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:
$$ceM(OH2)_$N - n$L_n + L <=> M(OH2)_$N - n - 1$L_$n + 1$ + H2O$$
The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to
$$fracN - nn + 1$$
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:
$$ceM(OH2)_$N - n$L_n + L <=> M(OH2)_$N - n - 1$L_$n + 1$ + H2O$$
The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to
$$fracN - nn + 1$$
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
Apr 14 at 20:55
add a comment |
$begingroup$
Recently, I was told that in case of a particular step of a generic ligand substitution reaction:
$$ceM(OH2)_$N - n$L_n + L <=> M(OH2)_$N - n - 1$L_$n + 1$ + H2O$$
The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to
$$fracN - nn + 1$$
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:
$$ceM(OH2)_$N - n$L_n + L <=> M(OH2)_$N - n - 1$L_$n + 1$ + H2O$$
The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to
$$fracN - nn + 1$$
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 Apr 14 at 20:38
andselisk♦
21.4k774143
21.4k774143
asked Apr 14 at 20:35
Shoubhik Raj MaitiShoubhik Raj Maiti
1,405934
1,405934
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
Apr 14 at 20:55
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
Apr 14 at 20:55
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
Apr 14 at 20:55
$begingroup$
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
Apr 14 at 20:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think the answer may just come down to a simple counting of available sites. The equilibrium constant for a reaction is equal to the ratio of the forward and reverse reaction rates. For the forward reaction, there are $N-n$ sites available at which a ligand can replace an $ceH2O$. Conversely, for the reverse reaction, there are $n+1$ ligand sites at which a water molecules can replace it. If we assume that in each case the reaction rate with $m$ sites available is equal to $m$ times the reaction rate with $1$ site available, we obtain an equilibrium constant that is proportional to the ratio of sites available for the forward and reverse reactions.
$$K=frack_f,N-nk_r,n+1approxfrac(N-n)k_f,1(n+1)k_r,1proptofrac(N-n)(n+1)$$
$endgroup$
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 $ceMA_nB_$N-n$$,
$$Omega = N choose n = fracN!n!(N-n)! quad left[ = N choose N-n right]$$
and that for the individual molecules $ceA$ and $ceB$, $Omega = 1$, then the equilibrium constant $K$ for
$$ceMA_nB_$N-n$ + A <=> MA_n + 1B_$N-n-1$ + B$$
is given by
$$beginalign
K &= expleft(frac-Delta_mathrm r GRTright) \
&= expleft(fracDelta_mathrm r SRright) \
&= expleft(fracS_mathrmm(ceMA_n + 1B_$N-n-1$) + S_mathrmm(ceB) - S_mathrmm(ceMA_nB_$N-n$) - S_mathrmm(ceA)Rright) \
&= exp[lnOmega(ceMA_n + 1B_$N-n-1$) + lnOmega(ceB) - lnOmega(ceMA_nB_$N-n$) - lnOmega(ceA)] \
&= expleft[lnleft(fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA)right)right] \
&= fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA) \
&= fracN!(n+1)!(N-n-1)! cdot fracn!(N-n)!N! \
&= fracN-nn+1
endalign$$
The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the answer may just come down to a simple counting of available sites. The equilibrium constant for a reaction is equal to the ratio of the forward and reverse reaction rates. For the forward reaction, there are $N-n$ sites available at which a ligand can replace an $ceH2O$. Conversely, for the reverse reaction, there are $n+1$ ligand sites at which a water molecules can replace it. If we assume that in each case the reaction rate with $m$ sites available is equal to $m$ times the reaction rate with $1$ site available, we obtain an equilibrium constant that is proportional to the ratio of sites available for the forward and reverse reactions.
$$K=frack_f,N-nk_r,n+1approxfrac(N-n)k_f,1(n+1)k_r,1proptofrac(N-n)(n+1)$$
$endgroup$
add a comment |
$begingroup$
I think the answer may just come down to a simple counting of available sites. The equilibrium constant for a reaction is equal to the ratio of the forward and reverse reaction rates. For the forward reaction, there are $N-n$ sites available at which a ligand can replace an $ceH2O$. Conversely, for the reverse reaction, there are $n+1$ ligand sites at which a water molecules can replace it. If we assume that in each case the reaction rate with $m$ sites available is equal to $m$ times the reaction rate with $1$ site available, we obtain an equilibrium constant that is proportional to the ratio of sites available for the forward and reverse reactions.
$$K=frack_f,N-nk_r,n+1approxfrac(N-n)k_f,1(n+1)k_r,1proptofrac(N-n)(n+1)$$
$endgroup$
add a comment |
$begingroup$
I think the answer may just come down to a simple counting of available sites. The equilibrium constant for a reaction is equal to the ratio of the forward and reverse reaction rates. For the forward reaction, there are $N-n$ sites available at which a ligand can replace an $ceH2O$. Conversely, for the reverse reaction, there are $n+1$ ligand sites at which a water molecules can replace it. If we assume that in each case the reaction rate with $m$ sites available is equal to $m$ times the reaction rate with $1$ site available, we obtain an equilibrium constant that is proportional to the ratio of sites available for the forward and reverse reactions.
$$K=frack_f,N-nk_r,n+1approxfrac(N-n)k_f,1(n+1)k_r,1proptofrac(N-n)(n+1)$$
$endgroup$
I think the answer may just come down to a simple counting of available sites. The equilibrium constant for a reaction is equal to the ratio of the forward and reverse reaction rates. For the forward reaction, there are $N-n$ sites available at which a ligand can replace an $ceH2O$. Conversely, for the reverse reaction, there are $n+1$ ligand sites at which a water molecules can replace it. If we assume that in each case the reaction rate with $m$ sites available is equal to $m$ times the reaction rate with $1$ site available, we obtain an equilibrium constant that is proportional to the ratio of sites available for the forward and reverse reactions.
$$K=frack_f,N-nk_r,n+1approxfrac(N-n)k_f,1(n+1)k_r,1proptofrac(N-n)(n+1)$$
edited Apr 15 at 18:04
answered Apr 15 at 2:17
TyberiusTyberius
7,55532264
7,55532264
add a comment |
add a comment |
$begingroup$
I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ceMA_nB_$N-n$$,
$$Omega = N choose n = fracN!n!(N-n)! quad left[ = N choose N-n right]$$
and that for the individual molecules $ceA$ and $ceB$, $Omega = 1$, then the equilibrium constant $K$ for
$$ceMA_nB_$N-n$ + A <=> MA_n + 1B_$N-n-1$ + B$$
is given by
$$beginalign
K &= expleft(frac-Delta_mathrm r GRTright) \
&= expleft(fracDelta_mathrm r SRright) \
&= expleft(fracS_mathrmm(ceMA_n + 1B_$N-n-1$) + S_mathrmm(ceB) - S_mathrmm(ceMA_nB_$N-n$) - S_mathrmm(ceA)Rright) \
&= exp[lnOmega(ceMA_n + 1B_$N-n-1$) + lnOmega(ceB) - lnOmega(ceMA_nB_$N-n$) - lnOmega(ceA)] \
&= expleft[lnleft(fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA)right)right] \
&= fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA) \
&= fracN!(n+1)!(N-n-1)! cdot fracn!(N-n)!N! \
&= fracN-nn+1
endalign$$
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 $ceMA_nB_$N-n$$,
$$Omega = N choose n = fracN!n!(N-n)! quad left[ = N choose N-n right]$$
and that for the individual molecules $ceA$ and $ceB$, $Omega = 1$, then the equilibrium constant $K$ for
$$ceMA_nB_$N-n$ + A <=> MA_n + 1B_$N-n-1$ + B$$
is given by
$$beginalign
K &= expleft(frac-Delta_mathrm r GRTright) \
&= expleft(fracDelta_mathrm r SRright) \
&= expleft(fracS_mathrmm(ceMA_n + 1B_$N-n-1$) + S_mathrmm(ceB) - S_mathrmm(ceMA_nB_$N-n$) - S_mathrmm(ceA)Rright) \
&= exp[lnOmega(ceMA_n + 1B_$N-n-1$) + lnOmega(ceB) - lnOmega(ceMA_nB_$N-n$) - lnOmega(ceA)] \
&= expleft[lnleft(fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA)right)right] \
&= fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA) \
&= fracN!(n+1)!(N-n-1)! cdot fracn!(N-n)!N! \
&= fracN-nn+1
endalign$$
The reason for ignoring $Delta_mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.
$endgroup$
add a comment |
$begingroup$
I think the easy way out is to invoke $S_mathrm m = R ln Omega$. If we assume that for a generic complex $ceMA_nB_$N-n$$,
$$Omega = N choose n = fracN!n!(N-n)! quad left[ = N choose N-n right]$$
and that for the individual molecules $ceA$ and $ceB$, $Omega = 1$, then the equilibrium constant $K$ for
$$ceMA_nB_$N-n$ + A <=> MA_n + 1B_$N-n-1$ + B$$
is given by
$$beginalign
K &= expleft(frac-Delta_mathrm r GRTright) \
&= expleft(fracDelta_mathrm r SRright) \
&= expleft(fracS_mathrmm(ceMA_n + 1B_$N-n-1$) + S_mathrmm(ceB) - S_mathrmm(ceMA_nB_$N-n$) - S_mathrmm(ceA)Rright) \
&= exp[lnOmega(ceMA_n + 1B_$N-n-1$) + lnOmega(ceB) - lnOmega(ceMA_nB_$N-n$) - lnOmega(ceA)] \
&= expleft[lnleft(fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA)right)right] \
&= fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA) \
&= fracN!(n+1)!(N-n-1)! cdot fracn!(N-n)!N! \
&= fracN-nn+1
endalign$$
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 $ceMA_nB_$N-n$$,
$$Omega = N choose n = fracN!n!(N-n)! quad left[ = N choose N-n right]$$
and that for the individual molecules $ceA$ and $ceB$, $Omega = 1$, then the equilibrium constant $K$ for
$$ceMA_nB_$N-n$ + A <=> MA_n + 1B_$N-n-1$ + B$$
is given by
$$beginalign
K &= expleft(frac-Delta_mathrm r GRTright) \
&= expleft(fracDelta_mathrm r SRright) \
&= expleft(fracS_mathrmm(ceMA_n + 1B_$N-n-1$) + S_mathrmm(ceB) - S_mathrmm(ceMA_nB_$N-n$) - S_mathrmm(ceA)Rright) \
&= exp[lnOmega(ceMA_n + 1B_$N-n-1$) + lnOmega(ceB) - lnOmega(ceMA_nB_$N-n$) - lnOmega(ceA)] \
&= expleft[lnleft(fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA)right)right] \
&= fracOmega(ceMA_n + 1B_$N-n-1$)Omega(ceB)Omega(ceMA_nB_$N-n$)Omega(ceA) \
&= fracN!(n+1)!(N-n-1)! cdot fracn!(N-n)!N! \
&= fracN-nn+1
endalign$$
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.
edited Apr 15 at 2:05
Tyberius
7,55532264
7,55532264
answered Apr 14 at 21:30
orthocresol♦orthocresol
41.8k7126256
41.8k7126256
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
Apr 14 at 20:55