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

Question

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 (A) 5470 (B) 2735 (C) 4105 (D) 3765

Tardigrade Admin · Accepted Answer

Heat needed too be supplied per mol = 330 + 580 + 1820 + 2740 = 5470 kJ No. of mols of Al taken = (13.5/27) = 0.5 mol Heat required = 0.5 × 5470 kJ = 2735 kJ

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

5470

2735

4105

3765

Heat needed too be supplied per mol =330+580+1820+2740 =5470kJ No. of mols of Al taken =2713.5​=0.5mol Heat required =0.5×5470kJ=2735kJ

Heat needed too be supplied per mol $= 330 + 580 + 1820 + 2740$
$= 5470 k J$
No. of mols of $A l$ taken $= \frac{13.5}{27} = 0.5 m o l$
Heat required $= 0.5 \times 5470 k J = 2735 k J$