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

This problem involves calculating the total energy required to perform successive ionizations on a specific mass of gaseous lithium. Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion. For lithium to become $\text{Li}^{2+}$, it must lose two electrons sequentially.

First, let's determine the total energy required to convert 1 mole of $\text{Li}(g)$ to $\text{Li}^{2+}(g)$. This is the sum of the first ionization enthalpy ($IE_1$) and the second ionization enthalpy ($IE_2$):
$$\text{Total molar energy } (E_{total}) = IE_1 + IE_2$$
$$E_{total} = 520\text{ kJ mol}^{-1} + 7297\text{ kJ mol}^{-1} = 7817\text{ kJ mol}^{-1}$$

Next, we calculate the number of moles of lithium present in $3.5\text{ mg}$. Using the molar mass of lithium ($7\text{ g mol}^{-1}$):
$$\text{Mass of Li} = 3.5\text{ mg} = 3.5 \times 10^{-3}\text{ g}$$
$$\text{Moles of Li } (n) = \frac{\text{Mass}}{\text{Molar mass}} = \frac{3.5 \times 10^{-3}\text{ g}}{7\text{ g mol}^{-1}} = 0.5 \times 10^{-3}\text{ moles}$$
$$n = 5 \times 10^{-4}\text{ moles}$$

Finally, the total energy required for this amount is calculated by multiplying the moles by the total molar energy:
$$\text{Energy Required} = n \times E_{total}$$
$$\text{Energy Required} = (5 \times 10^{-4}\text{ moles}) \times (7817\text{ kJ mol}^{-1}) = 3.9085\text{ kJ}$$

Rounding to the nearest integer, we get $4\text{ kJ}$. Note: While the question blank ends with 'kJ mol⁻¹', it is asking for the energy for the specified mass, which is typically expressed in kJ. Based on standard numerical patterns, the answer is 4.