You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trend in Ionisation Enthalpy

The first, se...

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

51.40

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2745

No worries! We‘ve got your back. Try BYJU‘S free classes today!

514.0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Similar questions

^{- 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} .

I E

A l (g) ⟶ {A l}^{+} (g) + e

577.5 k J

{m o l}^{- 1}

Δ H

A l (g) ⟶ {A l}^{3 +} (g) + 3 e

5140 k J

{m o l}^{- 1}

I E

2 : 3

{I E}_{2}

{I E}_{3} - 0.5

k J

{m o l}^{- 1}

N a, M g, A l

S i

N a

M g

S i

k J m o l^{- 1}

A l

Join BYJU'S Learning Program