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Question

The first, second and third ionisation energies of $A l$ are $578, 1817$ and $2745$ kJ $m o l^{- 1}$ respectively. Calculate the energy required to convert all the atoms of $A l$ to $A l^{+ 3}$ present in $270$ mg of $A l$ vapours.

5140

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51.40

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2745

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514.0

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Solution

The correct option is C $51.40$ kJ
Given $A l ⟶ {A l}^{3 +} + {3 e}^{-}$

$I E (I + I I + I I I) =$ ( $578 + 1817 + 2745$ )

For $(\frac{270 \times 10^{- 3}}{27})$ mole of $A l (g)$

$∴$ Energy $=$ ( $578 + 1817 + 2745$ ) x ( $0.01$ ) = $51.40 k J$

Suggest Corrections

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^{- 1}

^{3 -}

A l t o A l^{3 +} = 5137 k J m o l^{- 1} △ H_{h y d r a t i o n} o f : A l^{3 +} = - 4665 k J m o l^{- 1} a n d f o r C l^{-} = - 381 k J m o l^{- 1} .

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