Question

The enthalpy of sublimation of aluminium is \(330\,\text{kJ mol}^{-1}\). Its I\(^\text{st}\), II\(^\text{nd}\) and III\(^\text{rd}\) ionization enthalpies are 580, 1820 and \(2740\,\text{kJ mol}^{-1}\) respectively. How much heat must 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

Hint:

Always multiply molar enthalpy by number of moles.

Answer 1

The Correct Option is C

Step 1: Moles of Al: \[ n=\frac{13.5}{27}=0.5 \]
Step 2: Total energy per mole: \[ 330+580+1820+2740=5470\,\text{kJ mol}^{-1} \]
Step 3: \[ Q=0.5 \times 5470 = 2735\,\text{kJ} \] Including electron removal work gives closest option (C).