The first ionisation potential of (Na) is (5.1, eV). The val... | Chemistry | SolveeIt

Question

The first ionisation potential of $Na$ is $5.1\, eV$ . The value of electron gain enthalpy of $Na^+$ will be

$- 2.55\, eV$

$- 5.1\, eV$

$- 10.2\, eV$

$+ 2.55\, eV$

Answer

$- 5.1\, eV$

Explanation

Solution

$\, \, \, \, \, \, Na \longrightarrow Na^+ + e^- \, First \, IE$
$Na^+ + e^- \longrightarrow Na$
Electron gain enthalpy of Na+ is reverse of (IE) Because reaction is reverse so
$\, \, \, \, \, ? H (eq) = - 5.1 \, eV$

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