First slide

Enthalpy of hydration

Question

The lattice energy of solid NaCl is 180 K.Cal per mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of 1K.Cal per mol. If the solvation energies of Na⁺ and Cl^– ions are in ratio 6:5, what is the enthalpy of hydration of sodium ion?

Moderate

Solution

$\begin{array}{l} N{a^ + }_{(g)} + C{l^ - }_{(g)} \to NaC{l_{(s)}}\,\,\Delta U = - 180Kcal\\ \underline {NaC{l_{(s)}} + {H_2}O \to N{a^ + }_{(aq.)} + C{l^ - }_{(aq.)}\,\Delta H = + 1} \\ N{a^ + }_{(g)} + C{l^ - }_{(g)} + {H_2}O \to N{a^ + }_{(aq.)} + C{l^ - }_{(aq.)}\,\,\,\,\Delta {H_{hyd}} = - 180 + 1 = - 179K.cal \end{array}$
$\begin{array}{l} \Delta {H_{hyd}}(N{a^ + }) = \frac{6}{{(6 + 5)}} \times \Delta {H_{hyd}}(NaCl)\\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{{11}}( - 179) = - 97.6K.cal/mole \end{array}$

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