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Q3.15:
Energy of an electron in the ground state of the hydrogen atom is $-2.18 imes 10^{-18}J$. Calculate the ionization enthalpy of atomic hydrogen in terms of $J mol^{-1}$.

Solution :

Step 1: Theoretical Foundation of Ionization Enthalpy

The ionization enthalpy ($IE$) is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Mathematically, for a hydrogen atom, this corresponds to the energy required to transition an electron from the ground state ($n=1$) to an infinite distance from the nucleus ($n=\infty$), where the potential energy of the electron is defined as zero.

The energy change ($\Delta E$) is expressed as:

$\Delta E = E_{\infty} - E_{ground}$

[Per the Bohr model of the hydrogen atom, where $E_{\infty} = 0$ J]

Step 2: Calculation of Energy for One Mole of Atoms

Given the energy of an electron in the ground state of a single hydrogen atom ($E_1$):

$E_1 = -2.18 \times 10^{-18} \, J$

The energy required to ionize one hydrogen atom is:

$\Delta E = 0 - (-2.18 \times 10^{-18} \, J) = +2.18 \times 10^{-18} \, J$

[Justification: By the Principle of Conservation of Energy, the energy absorbed must equal the difference between the final and initial states.]

Step 3: Conversion to Molar Enthalpy

To calculate the ionization enthalpy in terms of $J \, mol^{-1}$, we must multiply the energy per atom by the Avogadro constant ($N_A$), which represents the number of atoms in one mole ($6.022 \times 10^{23} \, mol^{-1}$).

$IE = \Delta E \times N_A$

$IE = (2.18 \times 10^{-18} \, J \cdot atom^{-1}) \times (6.022 \times 10^{23} \, atom \cdot mol^{-1})$

Step 4: Algebraic Evaluation

Executing the multiplication of the coefficients and the exponents:

  • Coefficient product: $2.18 \times 6.022 \approx 13.12796$
  • Exponent sum: $10^{-18} \times 10^{23} = 10^{5}$

$IE = 13.12796 \times 10^5 \, J \, mol^{-1}$

Normalizing to scientific notation:

$IE = 1.312796 \times 10^6 \, J \, mol^{-1}$

Ionization Process ($H \rightarrow H^+ + e^-$) Energy (J) n=1 to n=$\infty$ $\Delta E = 1.31 \times 10^6 \, J/mol$

Final Solution: The ionization enthalpy of atomic hydrogen is $1.31 \times 10^6 \, J \, mol^{-1}$.


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