Question

energy of an electron in the ground state of the hydrogen atom is -2.8 x 10 -18 j. calculate the ionisation enthalpy of atomic hydrogen in terms of kj mol -1 .

Answer 1

Ionization energy can be calculated using the equation=

=$E_{\infty }-E_{1}$

=$0-(-2.18\times 10^{-18 }J)$

= $2.18\times 10^{-18 }J$

Ionization energy per mole of hydrogen atom;

=$(2.18\times 10^{-18 }J\times 6.023\times 10^{23})/1000 \frac{KJ}{mole}$

=$1312.36 \frac{KJ}{mole}$

So, the Ionization energy per mole of hydrogen atom is$1312.36 \frac{KJ}{mole}$.

Answer 2

Answer: The ionization enthalpy of Hydrogen atom is 1686.16 kJ/mol.

Explanation: Ionization enthalpy is defined as the amount of energy required by an isolated gaseous atom to lose an electron from its ground state.

Mathematically,

$I.E.=E_{\infty}-E_1$

We are given:

$E_1=-2.8\times 10^{-18}J/atom$

So, putting values in above equation, we get:

$I.E.=0-(-2.8\times 10^{-10})J/atom=2.8\times 10^{-10}J/atom$

To convert it into kJ/mol, we multiply it by Avogadro's number and divide it by 1000 we get:

$I.E.=\frac{2.8\times 10^{-18}\6.022\times 10^{23}}{1000}kJ/mol=1686.16kJ/mol$