Question

The energy changes for many unusual reactions can be determined using Hess’s law.

(a) Calculate ΔE for the conversion of F⁻(g) into F⁺(g).
(b) Calculate ΔE for the conversion of Na⁺(g) into Na⁻(g).

Answer 1

(a) F⁻(g) ⇒ F⁺(g) ΔE=2009 KJ/mol

(b) Na⁺(g) ⇒ Na⁻(g) ΔE=-548.6 KJ/mol

Step-by-step explanation:

Hess's Law

"Energy changes for a reaction is independent of steps or route involved."

(a) F⁻(g) ⇒ F⁺(g)

Now the conversion F⁻ to F⁺ is a two step reaction. In first step F⁻ loses an electron e⁻ to become neutral atom.

F⁻ ⇒ F + e⁻ ΔH=328 KJ/mol (i)

Kindly note this energy is derived from electron affinity, which is the energy changes while adding an electron to neutral atom and its same while removing an electron from ion.

In second step,

Another electron is removed from F.

F ⇒ F⁺ + e⁻ ΔH= 1681 KJ/mol (ii)

This energy is 1st ionization energy for F, which is the energy required to remove first electron from outer most shell of neutral atom.

Now the total energy change can be found by applying Hess's law and adding equations (i) and (ii);

F⁻ ⇒ F + e⁻ ΔH=328 KJ/mol

F ⇒ F⁺ + e⁻ ΔH= 1681 KJ/mol

F⁻ ⇒ F⁺ + 2e⁻ ΔH= 2009 KJ/mol

(b) Na⁺(g) ⇒ Na⁻(g)

This reaction involves two steps in first Na⁺(g) gains electron and become neutral atom while in second step Na(g) accepts another electron to become Na⁻(g).

Na⁺(g) + e⁻⇒ Na(g) ΔH=-495.8 KJ/mol (i)

Na(g) + e⁻ ⇒ Na⁻(g) ΔH=-52.8 KJ/mol (ii) (By Applying Hess's Law)

Na⁺(g) + 2e⁻ ⇒ Na⁻(g) ΔH=-548.6 KJ/mol

Kindly note as there is no molar change in above mentioned reactions therefore enthalpy changes(ΔH) and internal energy changes(ΔE) are same.