Question

Calculate the free energy change when 1 mol of NaCl is dissolved in water at 298 K. given:
(i) Lattice energy of NaCl = -778 kJ $mo{l}^{-1}$
(ii) Hydration energy of NaCl = -778 kJ $mo{l}^{-1}$
(iii) Entropy change at 298 K = 43 J $mo{l}^{-1}$

• A. -9.11 kJ
• B. -11.9 kJ
• C. -10.9 kJ
• D. None of these

Answer 1

The correct option is A

-9.11 kJ

${△}_{sol}{G}^{\ominus }$= ${△}_{sol}{H}^{\ominus }$ -T${△}_{sol}{S}^{\ominus }$
${△}_{sol}{H}^{\ominus }={△}_{ion}{H}^{\ominus }+{△}_{hyd}{H}^{\ominus }$
$N{a}^{\oplus }\left(g\right)+C{l}^{\ominus }\left(g\right)\to NaCl\left(s\right)$; $△H^{\ominus }$ = -778 kJ $mo{l}^{-1}$
Or
$NaCl\left(s\right)\to N{a}^{\oplus }\left(g\right)+C{l}^{\ominus }\left(g\right)$
$△H_1$ = 778 kJ $mo{l}^{-1}$- - - - - - (i)
$N{a}^{\oplus }\left(g\right)+C{l}^{\ominus }\left(g\right)+aq\to NaCl\left(aq\right)$
Or
$N{a}^{\oplus }\left(aq\right)+C{l}^{\ominus }\left(aq\right)$
$△H_2$ = -774.3 KJ $mo{l}^{-1}$- - - - - - (ii)
To calculate ${△}_{sol}{H}^{\ominus }$
$NaCl\left(s\right)+aq\to N{a}^{\oplus }\left(aq\right)+C{l}^{\ominus }\left(aq\right)$
$△H^{\ominus }=△H_1+△H_2$
= 778 - 774.3 = 3.7 kJ $mo{l}^{-1}$ = 3700 J $mo{l}^{-1}$
${△}_{sol}{S}^{\ominus }$ = 43 J $mo{l}^{-1}$
${△}_{sol}{G}^{\ominus }$= 3700 -298X43 = 9114J =9.11 kJ