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Chemistry · THERMODYNAMICS

हिन्दी

At temperature \( T \), the relation between the ionization enthalpy (\( \Delta_i H^\ominus \)) and ionization energy (\( E_0 \)) for an atom in gaseous state is:

  1. \( \Delta_i H^\ominus = E_0 - RT \)
  2. \( \Delta_i H^\ominus = E_0 - \frac{5}{2}RT \)
  3. \( \Delta_i H^\ominus = E_0 + \frac{3}{2}RT \)
  4. \( \Delta_i H^\ominus = E_0 + \frac{5}{2}RT \)

Correct answer · Option 4

\( \Delta_i H^\ominus = E_0 + \frac{5}{2}RT \)

Explanation

From \( \Delta_i H^\ominus(T) = \Delta_i H^\ominus(0) + \int_0^T \Delta_r C_p^\ominus dT \), since \( \Delta_r C_p^\ominus = +\frac{5}{2}R \), \( \Delta_i H^\ominus = E_0 + \frac{5}{2}RT \).
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