1
|
Av. |
\(\ce{Sn}\) |
\(+\) |
\(\ce{Cu^2+}\) |
\(\ce{<-->}\) |
\(\ce{Sn^2+}\) |
\(\ce{Cu}\) |
| État initial |
\(x = 0\) |
\(n_0(\ce{Sn})\) |
|
\(n_0(\ce{Cu^2+})\) |
|
\(0\) |
\(0\) |
| État final |
\(x = x_f\) |
\(n_f(\ce{Sn}) = n_0(\ce{Sn}) - x_f\) |
|
\(n_f(\ce{Cu^2+}) = n_0(\ce{Cu^2+}) - x_f\) |
|
\(n_f(\ce{Sn^2+}) = x_f\) |
\(n_f(\ce{Cu}) = x_f\) |
| État maximal |
\(x = x_m\) |
\(n_m(\ce{Sn}) = n_0(\ce{Sn}) - x_m\) |
|
\(n_m(\ce{Cu^2+}) = n_0(\ce{Cu^2+}) - x_m\) |
|
\(n_m(\ce{Sn^2+}) = x_m\) |
\(n_m(\ce{Cu}) = x_m\) |
2
\(K = \dfrac{\left(\dfrac{[\ce{Sn^2+}]_f}{c^0}\right)^1 × 1^1}{1^1 × \left(\dfrac{[\ce{Cu^2+}]_f}{c^0}\right)^1}\)
Soit, avec des concentrations en \(\pu{mol*L-1}\) : \(K(T) = \dfrac{[\ce{Sn^2+}]_f}{[\ce{Cu^2+}]_f}\)
Donc \(K = \dfrac{\left( \dfrac{n_f(\ce{Sn^2+})}{V_{tot}} \right)}{\left( \dfrac{n_f(\ce{Cu^2+})}{V_{tot}} \right)}\)
Soit \(K = \dfrac{x_f}{n_0(\ce{Cu^2+})-x_f} \)
3
D'après la relation précédente : \(K×\left(n_0(\ce{Cu^2+})-x_f\right) = x_f\)
Soit : \(K×n_0(\ce{Cu^2+})-K×x_f = x_f\)
Soit : \( x_f + K×x_f = K×n_0(\ce{Cu^2+})\)
Soit : \( x_f×(1+K) = K×n_0(\ce{Cu^2+}) \)
Soit :
\(\begin{align}
x_f &= \dfrac{K×n_0(\ce{Cu^2+})}{1+K} \\
&= \dfrac{K×[\ce{Cu^2+}]_0×V_0}{1+K} \\
&= \dfrac{10^{16}×\pu{0,10}×\pu{0,100}}{1+10^{16}} \\
&= \pu{0,010 mol}
\end{align} \)
Remarque :
Si on calculait l'avancement maximal on trouverait \(x_m = \pu{0,010 mol}\) avec les ions \(\ce{Cu^2+}\) comme réactif en défaut.
Donc la transformation est totale.
Cela est cohérent avec le fait que \(K > 10^4\).
