Consider the relationship between Ionization Enthalpy () and Electron Affinity (). If the first ioni — Periodic Table and Periodicity Chemistry Question
Question
Consider the relationship between Ionization Enthalpy ($IE$) and Electron Affinity ($EA$). If the first ionization enthalpy ($IE_1$) of a neutral Sodium ($Na$) atom is $5.1 \text{ eV}$, what is the exact value of the electron gain enthalpy ($\Delta_{eg}H$) of the isolated $Na^+$ ion?
Answer: B
💡 Solution & Explanation
The ionization of $Na$ is given by $Na \rightarrow Na^+ + e^- \ (\Delta H = +5.1 \text{ eV})$. The electron gain of $Na^+$ is the exact reverse process: $Na^+ + e^- \rightarrow Na$. Therefore, by Hess's Law, the enthalpy change for the reverse process is equal in magnitude but opposite in sign, yielding $-5.1 \text{ eV}$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes