1. Tardigrade
  2. Question
  3. Chemistry
  4. The enthalpy of sublimation of aluminium is 330 kJ/mol. Its Ist, IInd and IIIrd ionization enthalpies are 580, 1820 and 2740 kJ respectively. How much heat has too be supplied (in kJ) to convert 13.5 g of aluminium into Al3+ ions and electrons at 298 k

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Q. The enthalpy of sublimation of aluminium is $330 \,kJ/mol$. Its $I^{st}, II^{nd}$ and $III^{rd}$ ionization enthalpies are $580, 1820$ and $2740 \,kJ$ respectively. How much heat has too be supplied (in $kJ$) to convert $13.5\, g$ of aluminium into $Al^{3+}$ ions and electrons at $298 \,k$

BITSATBITSAT 2014

Solution:

Heat needed too be supplied per mol $= 330 + 580 + 1820 + 2740 $
$= 5470\, kJ$
No. of mols of $Al$ taken $ = \frac{13.5}{27} = 0.5\, mol$
Heat required $= 0.5 \times 5470 \,kJ = 2735 \,kJ$