Question

K+F→K + + F − , Δ H was calculated to be 19 kcal under conditions where the cations and anions were prevented by electrostatic separation from combining with each other. The ionization potential of K is 4.3 eV. What is the electron affinity of F?

Answer 1

CORRECT QUESTION:

K + F → K⁺ + F⁻ , Δ H was calculated to be 19 kcal/mole under conditions where the anions and cations were prevented by electrostatic separation from combining with each other. The ionization potential of K is 4.3 eV. What is the electron affinity of F?

ANSWER:

• The electron affinity of F = 3.476 eV.

GIVEN:

• K + F → K⁺ + F⁻

• Δ H was calculated to be 19 kcal/mol.

• The ionization potential of K is 4.3 eV.

TO FIND:

• The electron affinity of F.

EXPLANATION:

$\boxed{\bold{\large{\pink{1\ kcal = 2.611 \times 10^{22}\ eV}}}}$

$\sf \leadsto 19\ kcalmol^{-1}= 19\times2.611\times 10^{22}\ eVmol^{-1}$

$\sf \leadsto 19\ kcalmol^{-1}= 49.609\times 10^{22}\ eVmol^{-1}$

$\boxed{\bold{\large{\green{1\ eVmol^{-1} = \dfrac{1}{6.023\times10^{23}}\ eVatom^{-1}}}}}$

$\sf \leadsto 49.609\times 10^{22} \ eVmol^{-1} = \dfrac{ 49.609\times 10^{22}}{6.023 \times {10}^{23}}\ eVatom^{-1}$

$\sf \leadsto 49.609\times 10^{22} \ eVmol^{-1} = \dfrac{ 8.24}{10}\ eVatom^{-1}$

$\sf \leadsto 49.609\times 10^{22} \ eVmol^{-1} = 0.824 \ eVatom^{-1}$

$\leadsto \textsf{Electron affinity of F = 4.3 - 0.824}$

$\leadsto \textsf{Electron affinity of F = 3.476 \ eV}$

Answer 2

$\huge\color{purple}{\underline{\underline{Answer\::}}}$

3.476 eV

$\huge\color{green}{\underline{\underline{explanation\::}}}$

Refers to the attachment....