Question

For the gaseous reaction, $K+F\to K^{+}+F^{-}$, $\Delta H=19kcal/mol$ under conditions when cations and anions are prevented by electrostatic separation from combining with each other . The $I{E}_2$ of K is 4.3 eV. Calculate electron affinity of F.

• A. 3.476 eV
• B. 4.523 eV
• C. 2.85 eV
• D. 7.62 eV

Answer 1

The correct option is A 3.476 eV

For $1$ mole,

$△H=19KCal$

$=19\times 2.611\times {10}^{22}=4.96\times {10}^{23}eV/mol$

for $1$ atom,

$△H=\frac{4.96\times {10}^{23}}{6.023\times {10}^{23}}=0.823eV/atom$

So, $e^{-}$ affinity of $F$=$4.3-0.823$

$=3.477eV$