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 `Al^(3+)` ions and electrons at 298 k

Options

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

Answer

To solve the problem of how much heat needs to be supplied to convert 13.5 g of aluminum into \( \text{Al}^{3+} \) ions and electrons, we will follow these steps:

### Step 1: Calculate the number of moles of aluminum.
To find the number of moles of aluminum, we use the formula:

\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

Given mass of aluminum = 13.5 g
Molar mass of aluminum = 27 g/mol

\[ \text{Number of moles} = \frac{13.5 \, \text{g}}{27 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 2: Calculate the total energy required for sublimation and ionization.
The total energy required includes the enthalpy of sublimation and the ionization enthalpies.

1. **Enthalpy of sublimation of aluminum** = 330 kJ/mol
2. **Ionization enthalpies**:
- First ionization enthalpy = 580 kJ/mol
- Second ionization enthalpy = 1820 kJ/mol
- Third ionization enthalpy = 2740 kJ/mol

Total ionization energy = \( 580 + 1820 + 2740 = 5140 \, \text{kJ/mol} \)

Total energy required per mole to convert aluminum to \( \text{Al}^{3+} \) ions:

\[ \text{Total energy} = \text{Enthalpy of sublimation} + \text{Total ionization energy} = 330 + 5140 = 5470 \, \text{kJ/mol} \]

### Step 3: Calculate the total energy required for 0.5 moles of aluminum.
Now, we need to find the energy required for 0.5 moles:

\[ \text{Total energy for 0.5 moles} = 0.5 \times 5470 \, \text{kJ/mol} = 2735 \, \text{kJ} \]

### Final Answer:
The total heat that needs to be supplied to convert 13.5 g of aluminum into \( \text{Al}^{3+} \) ions and electrons is **2735 kJ**.

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