If the Gibbs free energy change when 1 mole of $N a C l$ is dissolved in water at 298K is $x$ kJ, then - $1000 x$ is
Given that,
(a) Lattice energy of $N a C l$ = $778 k J {m o l}^{- 1}$
(b) Hydration energy of $N a C l = - 774.3 k J {m o l}^{- 1}$
(c) Entropy change at $298 K = 43 J {m o l}^{- 1}$ .

9117

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9441

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9114

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9141

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Solution

The correct option is B 9114
${Δ H}_{d i s s o l u t i o n} = {Δ H}_{(i o n i s a t i o n)} + {Δ H}_{(h y d r a t i o n)}$ $= 778 - 774.3$
$= 3.7 k J {m o l}^{- 1} = 3700 J {m o l}^{- 1}$
${Δ S}_{d i s s o l u t i o n} = 43 J K^{- 1}$
$∴ Δ G_{d i s s o l u t i o n} = Δ H - T Δ S$
$= 3700 - 298 \times 43 = - 9114 J$

$Δ G = - 9.114 k J$

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Calculate the free energy change when 1 mole of NaCl is dissolved in water at 298 K. (Given lattice energy of NaCl = -774.1 kj/mol and $Δ$ S = 0.043 Kj K-1mol-1 at 298 K)

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 $m o l^{- 1}$
(ii) Hydration energy of NaCl = -778 kJ $m o l^{- 1}$
(iii) Entropy change at 298 K = 43 J $m o l^{- 1}$