Question 7.FP.5: Aqueous Ni^2+ forms a 1:1 complex with the anion of murexide......
Aqueous \ {\mathrm{Ni}}^{2+} forms a 1:1 complex with the anion of murexide ,\,\mathrm{mu}^{-},
{\mathrm{Ni}}^{2+}(\mathrm{aq})+\mathrm{mu}^{-}(\mathrm{aq}){\overset{k_{1}}{\underset {k_{-1}}{\leftrightarrows}}}\mathrm{~Ni~mu}^{+}(\mathrm{aq})
At 25 °C observed values of the rate constant k_{1} and the equilibrium constant K_{1} depend on pressure as follows
Determine the volume of activation \Delta\,V^{\neq *}: and standard \Delta\,V^{\theta} for the formation of the complex, and comment on the values obtained.
| \frac{p}{\operatorname{atm}} | 1 | 800 | 1500 |
| \frac{k_{1}}{\operatorname*{mol}^{-1}~dm^{3}~\mathrm{s}^{-1}} | 10000 | 6900 | 4800 |
| \frac{K_{1}}{\operatorname*{mol}^{-1}\mathrm{d}\mathrm{m}^{3}} | 1920 | 940 | 510 |
k={\frac{\kappa\,k\,T}{h}}K^{\neq\ast}
\log_{\mathrm{e}}k=\log_{\mathrm{e}}{\frac{\kappa\,k\,T}{h}}+\log_{\mathrm{e}}\,K^{\neq{*}}
=\log_{\mathrm{e}}{\frac{\kappa\,k\,T}{h}}-{\frac{\Delta\,G^{\neq\,{\mathrm{*}}}}{R T}}
\left({\frac{\partial\log_{\mathrm{e}}k}{\partial p}}\right)_{T}=-{\frac{1}{R T}}\left({\frac{\partial\Delta G^{\neq\ast}}{\partial p}}\right)_{T}
=-{\frac{\Delta\ V^{\neq*}}{R T}}
A plot of \log_{e}k versus p should have slope = -\Delta\ V^{\neq *}/R T.
Likewise, for the equilibrium constant,
\left({\frac{\partial\log_{\mathrm{e}}K}{\partial p}}\right)_{T}=-{\frac{\Delta\,V^{\theta}}{R T}}
A plot of \log_{e}k versus p should have slope = -\Delta\ V^{\theta}{}/R T.
The plot of \log_{\mathrm{e}}\,k_{1} versus p is linear, with
\mathrm{slope}=-4.8\times10^{-4}\,\mathrm{atm}^{-1}
∴ \Delta\,V^{\neq{*}}=+4.8\times10^{-4}\,\mathrm{atm}^{-1}\times8.314\,\mathrm{J\,mol}^{-1}\,\mathrm{K}^{-1}\times298\,\mathrm{K}
=+{\frac{4.8\times10^{-4}\times8.314\,\mathrm{J\,mol^{-1}\,K^{-1}\times298\,\mathrm{K}}}{1.013\,25\times10^{5}\,\mathrm{N\,m^{-2}}}}
=+{\frac{4.8\times10^{-4}\times8.314\,\mathrm{N}\,\mathrm{m\,mol^{-1}\,K^{-1}\times298\,\mathrm{K}}}{1.013\,25\times10^{5}\,\mathrm{N}\,\mathrm{m^{-2}}}}
=+1.17\times10^{-5}\,\mathrm{m^{3}}\,\mathrm{mol}^{-1}
=+11.7\,{\mathrm{cm}}^{3}\,{\mathrm{mol}}^{-1}
The plot of \log_{\mathrm{e}}\,K_{1} versus p is linear, with
\mathrm{slope}=-8.94\times10^{-4}\,\mathrm{atm}^{-1}
∴ \Delta\,V^{\theta}=+\left(\frac{8.94\times10^{-4}\,\mathrm{N^{-1}\,m^{2}}}{1.013\,25\times10^{5}}\right)\times8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}\times298\,\mathrm{K}
=+2.19\times10^{-5}\,\mathrm{m^{3}\,m o l}^{-1}
=+21.9~\mathrm{cm^{3}\,m o l^{-1}}
Note: take care with the units as this is probably the most difficult part of the question
\Delta\,V^{\theta} is positive: the product is less highly charged than the reactants, and is therefore less heavily solvated. Solvent is thus released from its ordered and packed structure, therefore the volume increases.
\Delta\,V^{\neq*} is positive, but less so than \Delta\,V^{\theta}: the activated complex has a lower charge than the reactants and so the solvent is less {\mathrm{packed~around~it.~Solvent~is}} thus released from the solvation sheaths, and therefore the volume increases on forming the activated complex.
The magnitude of \Delta\,V^{\neq*} is less than that for \Delta\,V^{\theta} , indicating that the solvation arrangement and charge neutralization is not so fully developed on forming the activated complex as when the final product is formed. The activated complex does resemble the products somewhat, and is probably half-way between reactants and products.
Note how solvation is dominant here: two species ! one activated complex → one product. On this basis a decrease in volume for both processes would be expected. This is not observed.
| \frac{p}{\operatorname{atm}} | 1 | 800 | 1500 |
| \frac{k_{1}}{\operatorname*{mol}^{-1}~dm^{3}~\mathrm{s}^{-1}} | 10000 | 6900 | 4800 |
| \log_{\mathrm{e}}\,k_{1} | 9.21 | 8.84 | 8.48 |
| \frac{K_{1}}{\operatorname*{mol}^{-1}\mathrm{d}\mathrm{m}^{3}} | 1920 | 940 | 510 |
| \log_{\mathrm{e}}\,K_{1} | 7.56 | 6.85 | 6.23 |