Calculate free energy (in kJ ) when 1 mole of a an ionic salt MX (s) is dissolved in water at 25° C. Given Lattice energy of MX =780 kJ mol -1 Hydration energy of MX =-775.0 kJ mol -1 Entropy change of dissolution at 25° C =45 J mol -1 K -1

Question

Calculate free energy (in kJ ) when 1 mole of a an ionic salt MX (s) is dissolved in water at 25° C. Given Lattice energy of MX =780 kJ mol -1 Hydration energy of MX =-775.0 kJ mol -1 Entropy change of dissolution at 25° C =45 J mol -1 K -1 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ H text dissolution =Δ H text (ionisation) +Δ H text (hydration) =780-775=5 kJ mol -1 Δ G text dissolution =Δ H text dissolution - T Δ S text dissolution =5000-298 × 45=-8410 J =8.41 kJ mol -1

Q. Calculate free energy (in $k J$ ) when $1$ mole of a an ionic salt $MX$ (s) is dissolved in water at $2 5^{\circ} C$ . Given
Lattice energy of $MX = 780 k J m o l^{- 1}$
Hydration energy of $MX = - 775.0 k J m o l^{- 1}$
Entropy change of dissolution at $2 5^{\circ} C = 45 J m o l^{- 1} K^{- 1}$

$Δ H_{dissolution} = Δ H_{(ionisation)} + Δ H_{(hydration)}$
$= 780 - 775 = 5 k J m o l^{- 1}$
$Δ G_{dissolution} = Δ H_{dissolution} - T Δ S_{dissolution}$
$= 5000 - 298 \times 45 = - 8410 J = 8.41 k J m o l^{- 1}$

Q. Calculate free energy (in kJ ) when 1 mole of a an ionic salt MX (s) is dissolved in water at 25∘C. Given Lattice energy of MX=780kJmol−1 Hydration energy of MX=−775.0kJmol−1 Entropy change of dissolution at 25∘C=45Jmol−1K−1

ΔHdissolution ​=ΔH(ionisation) ​+ΔH(hydration) ​ =780−775=5kJmol−1 ΔGdissolution ​=ΔHdissolution ​−TΔSdissolution ​ =5000−298×45=−8410J=8.41kJmol−1

Q. Calculate free energy (in $k J$ ) when $1$ mole of a an ionic salt $MX$ (s) is dissolved in water at $2 5^{\circ} C$ . Given
Lattice energy of $MX = 780 k J m o l^{- 1}$
Hydration energy of $MX = - 775.0 k J m o l^{- 1}$
Entropy change of dissolution at $2 5^{\circ} C = 45 J m o l^{- 1} K^{- 1}$

$Δ H_{dissolution} = Δ H_{(ionisation)} + Δ H_{(hydration)}$
$= 780 - 775 = 5 k J m o l^{- 1}$
$Δ G_{dissolution} = Δ H_{dissolution} - T Δ S_{dissolution}$
$= 5000 - 298 \times 45 = - 8410 J = 8.41 k J m o l^{- 1}$