Question 24 of 48intermediate🔧 ApplyNumerical3 marks

Energy of an electron in the ground state of the hydrogen atom is –2.18×102.18 \times 10⁻¹⁸ J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol⁻¹. Hint: Apply the idea of mole concept to derive the answer.

Correct Answer

1.3128×106 J mol−11.3128 \times 10^6 \text{ J mol}^{-1}

Exercise: EXERCISES | Q: 3.15 | (Page 23)
For More Understanding

Explanation

Ionization enthalpy represents the energy required to remove an electron from an isolated gaseous atom in its ground state. Since energy is always required to remove electrons, ionization enthalpies are always positive. The given ground state energy is negative, so its magnitude gives the energy needed to ionize a single hydrogen atom. By applying the mole concept, we multiply this single-atom energy by Avogadro's number to find the energy required for one mole of hydrogen atoms.

Solution Steps

  1. Step 1: The energy of an electron in the ground state of the hydrogen atom is −2.18×10−18 J-2.18 \times 10^{-18} \text{ J}. The energy required to remove this electron from one atom is the positive magnitude of this value: 2.18×10−18 J2.18 \times 10^{-18} \text{ J}.

  2. Step 2: To calculate the ionization enthalpy per mole, apply the mole concept by multiplying the energy for one atom by Avogadro's number (NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23} \text{ mol}^{-1}).

  3. Step 3: Ionization Enthalpy=2.18×10−18 J×6.022×1023 mol−1\text{Ionization Enthalpy} = 2.18 \times 10^{-18} \text{ J} \times 6.022 \times 10^{23} \text{ mol}^{-1}

  4. Step 4: Ionization Enthalpy=13.12796×105 J mol−1=1.3128×106 J mol−1\text{Ionization Enthalpy} = 13.12796 \times 10^5 \text{ J mol}^{-1} = 1.3128 \times 10^6 \text{ J mol}^{-1}

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