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Gibbs Free Energy & Spontaneity

There are 1 m...

Question

There are 1 mole of atoms of an element A. When half of the atoms transfer one electron to another atom, 405 $k J m o l^{- 1}$ of energy was found to be consumed. An additional energy of 745 $k J / m o l$ was further required to convert all the $A^{-}$ ions to $A^{+}$ . Calculate the ionisation energy of atom A in $k J$

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Solution

$(A \to A^{+}); Δ H_{I E} = x k J / m o l$
$(A + e^{-} \to A^{-}); Δ H_{e g} = y k J / m o l$
Let x and y be IE and $Δ H_{e g}$ respectively.
$\frac{1}{2} (x + y) = 405 k J$
$\frac{1}{2} (x - y) = 745 k J$
On solving, we get,
x = 1150 $k J$ and y = -340 $k J$

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