Given the lattice energy of NaCl = 787 kJ/mol, the ionization energy of Na = 496 kJ/mol, and the electron affinity of Cl = -349 kJ/mol, calculate the ΔH° for the reaction: Na(g) + Cl(g) → NaCl(s)
Added by Juan Carlos V.
Step 1
The ionization energy of Na is the energy required to remove an electron from a gaseous sodium atom: Na(g) → Na+(g) + e−. This process is endothermic, so ΔH₁ = +496 kJ/mol. Show more…
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The lattice energy of $\mathrm{NaCl}$ is $-786 \mathrm{kJ} / \mathrm{mol}$, and the enthalpy of hydration of 1 mole of gaseous $\mathrm{Na}^{+}$ and 1 mole of gaseous $\mathrm{Cl}^{-}$ ions is $-783 \mathrm{kJ} / \mathrm{mol} .$ Calculate the enthalpy of solution per mole of solid NaCl.
The lattice energy of $\mathrm{NaCl}$ is $-786 \mathrm{kJ} / \mathrm{mol},$ and the enthalpy of hydration of 1 mole of gaseous Na' and 1 mole of gaseous $\mathrm{Cl}^{-}$ ions is $-783 \mathrm{kJ} / \mathrm{mol}$ . Calculate the enthalpy of solution per mole of solid NaCl.
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