Question

electron gain enthalpy of fluorine is 333 kj/mol . heat liberated by 1.9 grams of fluorine is ?

Answer 1

Answer:

Answer:.Electron gain enthalpy of F atom is 333 kJ,mole and dissociation energy of F_(2) is 158.8 kJ ,mole . The energy released during the formation of 2g of F^(-) ions form 2g of F_(2).

Explanation:

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Answer 2

Heat liberated by 1.9 grams of fluorine is 33.3 KJ.

We have to find the heat liberated by 1.9 grams of fluorine, if the electron gain enthalpy of fluorine is 333 kj/mol.

Electron gain enthalpy : Electron gain enthalpy is the amount of energy released to add an electron in an isolated gaseous atom.

Let A is an isolated gaseous atom and an electron e⁻ is added to it.

A + e⁻ ⇔ A⁻ + Q, then Q is electron gain enthalpy.

Mass of fluorine = 1.9g

Atomic weight of fluorine = 19 g/mol

∴ no of moles of fluorine = weight of fluorine/molecular weight

= 1.9/19 = 0.1 mol

so, the heat liberated to add electrons in 0.1 mol of fluorine, H = no of moles × electron gain enthalpy

= 0.1 mol × 333 kj/mol = 33.3 kj

Therefore the heat liberated by 1.9 grams of fluorine is 33.3 kj.

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