Question

The first IP of lithium is 5.41eV and electron gain enthalpy of Cl is −3.61eV.ΔH in kJmol−1 for the reaction Li(g)+Cl(g)⟶Li+(g)+Cl−(g), is

The first IP of lithium is ( 5.41 mathrm{eV} ) and electron gain enthalpy of ( mathrm{Cl} ) is ( -3.61 mathrm{eV} . Delta mathrm{H} ) in ( mathrm{kJ} mathrm{mol}^{-1} ) for the reaction ( mathrm{Li}(g)+mathrm{Cl}(g) longrightarrow mathrm{Li}^{+}(g)+mathrm{Cl}^{-}(g), ) is
a. ( 173.7 mathrm{J} )
b.( 173.7 mathrm{kJ} )
C. ( 185.4 mathrm{J} )
d. ( 185.4 mathrm{kJ} )

JEE/Engineering Exams

Chemistry

Solution

245

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( Delta mathrm{H} / mathrm{molecule} ) of ( mathrm{Li}^{+} ) and ( mathrm{Cl}^{-}= )
( mathrm{IP}_{1},+mathrm{EA} )
( 1 mathrm{mol}=18 times 6023 times 10^{23} )
( =1.8 times 6.023 times 10^{23} times 1.602 times 10^{-19} )
( =1.8 times 6.023 times 10^{3} times 1.602 times 10^{-19} times 10^{-3} mathrm{k}=173.7 mathrm{k} )