The lattice energy of NaCl is -786 kJ/mol, and the enthalpy of hydration of one mole of gaseous Na+ and one mole of gaseous Cl- ions is -783 kJ/mol. Calculate the enthalpy of solution per mole of solid NaCl. ____ kJ/mol NaCl
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This is an endothermic process, so the energy is positive (+786 kJ/mol). Show more…
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The lattice energy of KCl is -715 kJ/mol, and the enthalpy of hydration of one mole of gaseous K+ and one mole of gaseous Cl- ions is -684 kJ/mol. Calculate the enthalpy of solution per mole of solid KCl. _____ kJ/mol KCl
Hitendra S.
Calculate the lattice energy of NaCl(s) and draw a Born–Haber cycle. Data: Enthalpy of sublimation of Na(s) = 107.5 kJ/mol 1st ionization energy of Na(g) = 496 kJ/mol Bond dissociation energy of Cl2(g) = 242.58 kJ/mol 1st electron affinity of Cl(g) = –348.57 kJ/mol Standard enthalpy of formation of NaCl(s) = –411.2 kJ/mol (ANSWER: –787.42 kJ/mol)
Shaiju T.
The lattice enthalpy of sodium chloride, $\Delta H^{\circ}$ for $$\mathrm{NaCl}(s) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(g)$$ is $787 \mathrm{~kJ} / \mathrm{mol} ;$ the heat of solution in making up $1 \mathrm{M}$ $\mathrm{NaCl}(a q)$ is $+4.0 \mathrm{~kJ} / \mathrm{mol}$. From these data, obtain the sum of the heats of hydration of $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$. That is, obtain the sum of $\Delta H^{\circ}$ values for $$\begin{aligned} \mathrm{Na}^{+}(g) \longrightarrow & \mathrm{Na}^{+}(a q) \\ \mathrm{Cl}^{-}(g) & \longrightarrow \mathrm{Cl}^{-}(a q) \end{aligned}$$ If the heat of hydration of $\mathrm{Cl}^{-}$ is $-338 \mathrm{~kJ} / \mathrm{mol}$, what is the heat of hydration of $\mathrm{Na}^{+}$ ?
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