Question

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 $298 \,k$

• A. 5470
• B. 2735
• C. 4105
• D. 3765

Answer 1

Answer: (B)

2735

Answer 2

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 $= 0.5 \times 5470 \,kJ = 2735 \,kJ$