Na(g)+Cl(g)→Na+(g)+Cl−(g), ΔH=?
2017-10-09 9:07 pm
The ionization energy for sodium is 496 kJ/mol. The electron affinity for chlorine is -349 kJ/mol. Use these values and Hess's law to calculate the change in enthalpy for the following reaction per mole of reagent:
回答 (1)
2017-10-09 10:05 pm
Na(g) → Na⁺(g) + e⁻ …… ΔH = +496 kJ/mol
Cl(g) + e⁻ → Cl⁻(g) …… ΔH = -349 kJ/mol
Add the above two thermochemical equations, and cancel e⁻ on the both sides, i.e.
Na(g) + Cl(g) → Na⁺(g) + Cl⁻(g) …… ΔH = (+496 - 349) kJ/mol = +147 kJ/mol
Cl(g) + e⁻ → Cl⁻(g) …… ΔH = -349 kJ/mol
Add the above two thermochemical equations, and cancel e⁻ on the both sides, i.e.
Na(g) + Cl(g) → Na⁺(g) + Cl⁻(g) …… ΔH = (+496 - 349) kJ/mol = +147 kJ/mol
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