Question

First and second ionization enthalpies of lithium are $520\text{ kJ mol}^{-1}$ and $7297\text{ kJ mol}^{-1}$ respectively. Energy required to convert $3.5\text{ mg}$ lithium (g) into $\text{Li}^{2+}(g)$ $[\text{Li}(g) \rightarrow \text{Li}^{2+}(g)]$ is ________ $\text{kJ mol}^{-1}$. (nearest integer)
[Molar mass of $\text{Li} = 7\text{ g mol}^{-1}$]

Hint:

Sum the first and second ionization energies to get the energy per mole, then multiply by the number of moles in 3.5 mg of Lithium.

Answer 1

Correct Answer: 4

To find the total energy for the conversion of gaseous lithium atoms to doubly charged lithium ions, we must sum the sequential ionization energies and then scale by the amount of substance provided.

Step 1: Calculate the total energy per mole.
The process is $\text{Li}(g) \xrightarrow{IE_1} \text{Li}^{+}(g) \xrightarrow{IE_2} \text{Li}^{2+}(g)$.
Total Energy ($E_m$) $= 520 + 7297 = 7817\text{ kJ mol}^{-1}$.

Step 2: Convert the mass of lithium to moles.
The given mass is $3.5\text{ mg}$. Since $1\text{ g} = 1000\text{ mg}$, the mass in grams is $0.0035\text{ g}$.
Given atomic weight of $\text{Li} = 7\text{ g/mol}$.
Moles $= \frac{0.0035}{7} = 0.0005\text{ moles}$.

Step 3: Calculate the energy for $0.0005$ moles.
Energy $= \text{moles} \times E_m$
Energy $= 0.0005 \times 7817 = 3.9085\text{ kJ}$.

The question asks for the nearest integer. $3.9085$ rounds to $4$.
Answer: 4.