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

Byju's Answer

Standard XII

Chemistry

Gibb's Free Energy

The free ener...

Question

The free energy change when 1 mole of NaCl is dissolved in water at $25^{o} C$ is $- x k J m o l^{- 1}$ 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

Open in App

Solution

$△ H_{d i s s o l u t i o n} = L a t t i c e e n e r g y + H y d r a t i o n e n e r g y$
$= 777.8 - 774.1 = 3.7 k J m o l^{- 1}$

$N o w △ G = △ H - T △ S$
$= 3.7 - 298 \times 0.043 = 3.7 - 12.814$
$△ G = - 9.114 k J m o l^{- 1}$
Nearest integer = 9

Suggest Corrections

Similar questions

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

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

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)

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}

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 $m o l^{- 1}$

Join BYJU'S Learning Program

Explore more

Gibb's Free Energy

Standard XII Chemistry

Join BYJU'S Learning Program

The free energy change when 1 mole of NaCl is dissolved in water at 25oC is −x kJ mol−1 Lattice energy of NaCl = 777.8 kJ mol−1; △S for dissolution =0.043 kJ mol−1 and hydration energy of NaCl=−774.1 kJ mol−1. Value of x (rounded off to nearest integer) is

△Hdissolution=Lattice energy+Hydration energy =777.8−774.1=3.7 kJ mol−1 Now △G=△H−T△S =3.7−298 ×0.043=3.7−12.814 △G=−9.114 kJ mol−1 Nearest integer = 9

$△ H_{d i s s o l u t i o n} = L a t t i c e e n e r g y + H y d r a t i o n e n e r g y$
$= 777.8 - 774.1 = 3.7 k J m o l^{- 1}$

$N o w △ G = △ H - T △ S$
$= 3.7 - 298 \times 0.043 = 3.7 - 12.814$
$△ G = - 9.114 k J m o l^{- 1}$
Nearest integer = 9