Question

If you are given Avogadro’s number of atoms of a gas ‘X’. If half of the atoms are converted into $X{\left( g \right)^ + }$ by energy $\Delta H$. The IE of X is?
A. $\dfrac{{2\Delta H}}{{{N_A}}}$
B. $\dfrac{{2{N_A}}}{{\Delta H}}$
C. $\dfrac{{\Delta H}}{{2{N_A}}}$
D. $\dfrac{{{N_A}}}{{\Delta H}}$

Answer 1

Since it is given that $\Delta H$ amount of energy is required by half mole of the gas (since, half of the Avogadro’s number of atoms, i.e. 0.5 mole of the gas ‘X’) for ionisation, hence 1 mole of the gas would require $2\Delta H$ amount of energy. Therefore, we can calculate the ionisation energy of gas X.

Complete step by step answer:
Avogadro’s number is an absolute number i.e. there are $6.022 \times {10^{23}}$ elementary entities in 1 mole. We can also write it as $6.022 \times {10^{23\;}}mo{l^{ - 1}}$.
Since, Avogadro number $= \;6.023 \times {10^{23}} = {N_A}$
It is given in the question that $\Delta H$ amount of energy is required by half mole of the gas or half of Avogadro’s number of atoms, i.e. 0.5 mole of gas for ionisation. Each atom gives one electron out. Among half of ${N_A}$ atoms are ionized.
So, 1 mole of the gas would require $2\Delta H$ amount of energy
Thus, the ionisation energy (IE) of gas $X = \dfrac{{\Delta H}}{{2{N_A}}}\dfrac{{kJ}}{{mole}} = \dfrac{{2\Delta H}}{{{N_A}}}\dfrac{{kJ}}{{mole}}$
Thus, if half of the atoms are converted into $X{\left( g \right)^ + }$ by energy $\Delta H$, then the IE of X is $\dfrac{{2\Delta H}}{{{N_A}}}$

Therefore, the correct answer is option (A).

Note: The ionisation energy (IE) is the amount of energy which is required to remove the outermost electron from an isolated, gaseous atom in its ground state to form a cation in the gaseous state. We express the Ionization energy in kJ/mol, i.e. the amount of energy which is required by 1 mole of gaseous atom so that each of them is converted into a cation.