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Chemistry · CHEMICAL BONDING AND MOLECULAR STRUCTURE

हिन्दी

Given the ionization enthalpy of \( \text{Na(g)} \) is \( +495.8\text{ kJ mol}^{-1} \) and the electron gain enthalpy of \( \text{Cl(g)} \) is \( -348.7\text{ kJ mol}^{-1} \), calculate the net enthalpy change for the gaseous ion formation: \( \text{Na(g)} + \text{Cl(g)} \to \text{Na}^+(\text{g}) + \text{Cl}^-(\text{g}) \).

  1. \( -147.1\text{ kJ mol}^{-1} \)
  2. \( +147.1\text{ kJ mol}^{-1} \)
  3. \( +844.5\text{ kJ mol}^{-1} \)
  4. \( -788.0\text{ kJ mol}^{-1} \)

Correct answer · Option 2

\( +147.1\text{ kJ mol}^{-1} \)

Explanation

\( \Delta H = \Delta_i H + \Delta_{eg}H = +495.8 + (-348.7) = +147.1\text{ kJ mol}^{-1} \).
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