Q15 / 200 · Manipal (MET) 2013 · Chemistry
Calculate second electron affinity of oxygen for the process,
$\mathrm{O}^{-}(\mathrm{g})+e^{-}(\mathrm{g}) \rightarrow \mathrm{O}^{2-}(\mathrm{g})$
by using the following data
(i) Heat of sublimation of
$\mathrm{Mg}(s)=+147.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(ii) Ionisation energy of $\mathrm{Mg}(g)$ to form
$\mathrm{Mg}^{2+}(g)=+2189.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(iii) Bond dissociation energy for
$\mathrm{O}_2=+498.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(iv) First electron affinity of
$\mathrm{O}(\mathrm{g})=-141.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(v) Heat formation of
$\mathrm{MgO}=-601.7 \mathrm{~kJ} \mathrm{~mol}-1$
(vi) Lattice energy of
$\mathrm{MgO}=-3791.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- A$235.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$468,7 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- C$544.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerD$744.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
Stepwise formation of MgO involves
\begin{array}{l|l|l}\hline Step & {|c|}{ Reaction } & \Delta H in \mathrm{kJ} \mathrm{mol}^{-1} \\\hline (i) & \mathrm{Mg}(\mathrm{s}) \rightarrow \mathrm{Mg}(\mathrm{g}) & 147.7 \\(ii) & \mathrm{Mg}(g) \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} & 2189.0 \\(iii) & \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{O}(\mathrm{g}) & 498.4 / 2 \\(iv) & \mathrm{O}(g)+1 e^{-} \rightarrow \mathrm{O}^{-}(g) & -141.0 \\(v) & \mathrm{O}^{-}(g) t e^{-} \rightarrow \mathrm{O}^{2-}(g) & q \\(vi) & \mathrm{Mg}^{2+}(g)+\mathrm{O}^{2-}(g) \rightarrow \mathrm{MgO}(s) & -3791.0 \\\hline\end{array}
$\mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{MgO}(\mathrm{s}) \Delta \mathrm{H}=-1346.1+q$
$\mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{MgO}(\mathrm{s}) \Delta H_2=-601.7$
By Born Herber cycle (based on Hes s law)
$\Delta H_1=\Delta H_2$
$-1346.1+q=-601.7$
$q=744.4 \mathrm{Jmol}^{-1}$
Hence, the correct answer is option: 'D'.