Q6 / 120 · VITEEE 2015 · Chemistry
For the given reaction,
$\begin{array}{l}
\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \\
\Delta \mathrm{G}^{\mathrm{a}}=-262.4 \mathrm{~kJ}
\end{array}$
The value of free energy of formation
$\left(\Delta \mathrm{G}_{f}^{\mathrm{o}}\right)$ for the ion $\mathrm{Cl}^{-1}(\mathrm{aq})$, therefore will be
- Correct answerA$-131.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- B$+131.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- C$-262.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$+262.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\begin{aligned}\left(\Delta \mathrm{G}^{\circ}\right) \text { reaction }=\Delta \mathrm{G}_{f}^{0} & \text { (Products) } \\ &-\Delta \mathrm{G}_{f}^{0} \text { (reactants) } \end{aligned}$
$\begin{array}{ll}
\therefore & 264.4=\left[2 \Delta \mathrm{G}_{f}^{0}\left(\mathrm{H}^{+}\right)+2 \Delta \mathrm{G}_{f}^{0}\left(\mathrm{C}^{\mathrm{l}}\right)\right] \\
\text { or } & 264.4=-\left[\Delta \mathrm{G}_{f}^{0}\left(\mathrm{H}_{2}\right)+\Delta \mathrm{G}_{f}^{0}\left(\mathrm{Cl}_{2}\right)\right] \\
& =\left[0+2 \Delta \mathrm{G}_{f}^{0}\left(\mathrm{Cl}^{-}\right)\right]+[0+0]
\end{array}$
or, $\quad-262.4=2 \Delta \mathrm{G}_{f}^{0}\left(\mathrm{Cl}^{-}\right)$
or, $\Delta \mathrm{G}_{f}^{0}\left(\mathrm{Cl}^{-}\right)=-131.2 \mathrm{kJmol}^{-1}$.
Hence, the correct answer is option: 'A'.