You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Thermodynamic Processes

Calculate the...

Question

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)

Open in App

Solution

Free energy $= Δ G = G_{2} - G_{1} = Δ H - T Δ S$
$Δ H = - 774.1$
T = 298K
$Δ S = 0.043$
Energy = $Δ H - T Δ S$
$= 774.1 - 298 (0.043)$
$= - 774.1 - 12.814$
$= - 786.914$

Suggest Corrections

Similar questions

Q. 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}

298 K = 43 J {m o l}^{- 1}

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 $m o l^{- 1}$

Q. The free energy change when 1 mole of

N a C l

is dissolved in water at 298 K is

- x k J

. Find the value of 'x'.
Given:-

Lattice energy of N a C l = 778 k J m o l^{- 1}

Hydration energy of N a C l = - 775 k J m o l^{- 1}

Entropy change (at 298 K) = 40 J m o l^{- 1}

Q. The free energy change when 1 mole of NaCl is dissolved in water at

25^{o} C

- x k J m o l^{- 1}

Lattice energy of NaCl =

777.8 k J m o l^{- 1}

;

△ S

for dissolution

= 0.043 k J m o l^{- 1}

and hydration energy of

N a C l = - 774.1 k J m o l^{- 1} .

Value of x (rounded off to nearest integer) is