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Chemistry · THERMODYNAMICS

हिन्दी

For the ionization process \( \text{M}(g) \rightarrow \text{M}^+(g) + e^- \), the value of \( C_p \) for each species is \( \frac{5}{2}R \). At a temperature \( T = 300\text{ K} \), if the ionization energy \( E_0 \) at \( 0\text{ K} \) is \( 490\text{ kJ mol}^{-1} \), what is the ionization enthalpy \( \Delta_i H^\ominus \) at \( 300\text{ K} \)? (Take \( R = 8.314\text{ J K}^{-1}\text{mol}^{-1} \))

  1. A
    \( 483.76\text{ kJ mol}^{-1} \)
  2. B
    \( 496.24\text{ kJ mol}^{-1} \)
  3. C
    \( 490.00\text{ kJ mol}^{-1} \)
  4. D
    \( 502.48\text{ kJ mol}^{-1} \)
View Answer & Solution

Correct answer · Option B

\( 496.24\text{ kJ mol}^{-1} \)

Explanation

\( \Delta_r C_p^\ominus = C_p(\text{M}^+) + C_p(e^-) - C_p(\text{M}) = \frac{5}{2}R + \frac{5}{2}R - \frac{5}{2}R = +\frac{5}{2}R \). Thus, \( \Delta_i H^\ominus(T) = E_0 + \frac{5}{2}RT = 490\text{ kJ mol}^{-1} + \frac{5}{2}(8.314 \times 10^{-3} \times 300) = 490 + 6.2355 \approx 496.24\text{ kJ mol}^{-1} \).
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