Question

The first ionisation energy of Li is $5.4$ eV and electron affinity of Cl is $3.61$ eV. What is the value of $\Delta$ H(in kJ/mol) for the following reaction?
$Li\left(g\right)+Cl\left(g\right)\to L{i}^{+}\left(g\right)+C{l}^{-}\left(g\right)$
(resulting ions do not combine) ($1$eV $=1.6\times {10}^{-19}$J)

• A. $70$
• B. $100$
• C. $270$
• D. $172$

Answer 1

The correct option is D $172$

First I.P. of $Li=5.4eV$

E.A. of $Cl=3.61eV$

$\Delta H=I.P.-E.A.$

$=5.4-3.61=1.80eV=2.86\times {10}^{-22}KJ$ ($1eV=1.6\times {10}^{-19}J$)

$\Delta H=2.86\times {10}^{-22}\times 6.02\times {10}^{23}=172KJmo{l}^{-1}$ (Since, $1$ mol of Avogadro number $=6.02\times {10}^{23}$ )