Question

Calculate the enthalpy change when infinity dilute solutions of $CaC{l}_2$ and $N{a}_{2}C{O}_3$ are mixed. $\Delta H_f^{\circ }$ for $C{a}^{2+\left(aq\right),C{O}_3^{2-}}\left(aq\right)$ and $CaC{O}_3\left(s\right)$ are -129.65 and -288.45 kcal mol${}^{-1}$ respectively.

Answer 1

Dilute solution of $CaC{l}_2$ contains $C{a}^{2+}$ and $C{l}^{-}$ ions.
Dilute solution of $N{a}_{2}C{O}_3$ contains $N{a}^{+}$ and $C{O}_3^{2-}$ ions.
On mixing $CaC{l}_2(aq.)$ and $N{a}_{2}C{O}_3$
$CaC{l}_2+N{a}_{2}C{O}_3\to CaC{O}_3+2NaCl$
Solutions are very dilute and thus, $100\%$ dissociation occurs.
$C{a}^{2+}(aq.)+2C{l}^{-}(aq.)+2N{a}^{+}(aq.)+C{O}_3^{2-}(aq.)\to CaC{O}_3\downarrow +2N{a}^{+}(aq.)+2C{l}^{-}(aq.)$

or ,This can also be written as
$C{a}^2+(aq.)+C{O}_3^{2-}(aq.)\to CaC{O}_3\left(s\right)$
$\therefore \Delta H^{=}\sum \Delta H_f^0\left(Products\right)-\Delta H_f^0\left(reactants\right)or\Delta H=\Delta H_{fCaC{O}_3}^0-[\Delta H_{fc{a}^2+}^0+\Delta H_{fC{O}_3^{2-}}^0]$
putting the values, we get;
$=\Delta H=-288.45-(-129.80-161.65)=3$ k cal