Question

P3D) Repeat the calculation in Problem P3D.4 for Lif, for which AHⓇ = 1037 k) mol in step 1 and with the following values of the absolute entropy. S (Lit) = 133 J K ' mol-', S (F") = 145JK-'mol"), S (LiF(s)) = 35.6JK-mol- (all data at 298 K). Use r(Lit) = 127 pm and r(F) = 163 pm. 206 From the Born equation derive an enrain A f.. 1 A

Answer 1

Question (1)

To determine the Gibbs free energy for the reaction,

NaCl(s) -----------------> Na⁺(aq) + Cl^-(aq)

This is comprised of the following steps:

Reaction (1):

NaCl(s) ----------> Na⁺(g) + Cl^-(g)                                             ----(1)

Reaction (2):

Na⁺(g) +Cl^-(g) ---------------------> Na⁺(aq) + Cl^-(aq) -----(2)

Reaction (1)

For the reaction (1) it is given,

$\Delta$H_r^o = 787 kJ/mole

S_m^o(Na⁺(g)) = 148 JK^-1mole^-1

S_m^o(Cl^-(g)) = 154 JK^-1mole^-1

S_m^o(NaCl(s)) = 72.1 JK^-1mole^-1

$\Delta$S_r^o = S_m^o(Na⁺(g)) + S_m^o(Cl^-(g)) - S_m^o(NaCl(s))

= 148 + 154 - 72.1

= 229.9 JK^-1mole^-1

$\Delta$G_r^o =$\Delta$H_r^o - T$\Delta$S_r^o

= 787 x 10³ - 298*229.9 (Since T = 298K)

=718489.8 Jmole^-1

$\Delta$G_r^o = 718.4898 kJ/mole

Reaction (2)

For solvation occurring at 298K, the free energy of solvation is given by bthe Born equation below:

kJ/mole

AG iv = 6.86 + 104

in which z_i is the charge number of the ion

r₁ is the radius in pm

$\Delta$G_solv^o(Na⁺) = - (1)²*6.86*10⁴/170 (since radius of Na⁺ ion = 170 pm (given))

$\Delta$G_solv^o(Na⁺) = - 403.529 kJ/mol

$\Delta$G_solv^o(Cl^-) = - (1)²*6.86*10⁴/211 (since radius of Cl^- ion = 211 pm (given))

$\Delta$G_solv^o(Cl^-) = - 325.118 kJ/mol

Total free energy of the reaction (2) is

$\Delta$G_solv^o = $\Delta$G_solv^o(Na⁺) +$\Delta$G_solv^o(Cl^-) = -403.529 - 325.118

$\Delta$G_solv^o = -728.648 kJ/mole

Therefore the Gibbs free energy change for the overall reaction,

NaCl(s) -----------------> Na⁺(aq) + Cl^-(aq)

$\Delta$G_overall^o = $\Delta$G_r^o + $\Delta$G_solv^o = 718.4898 - 728.648

$\Delta$G_overall^o = -10.1581 kJ/mole

Since the $\Delta$G_overall^o of the process is highly negative, the process is irreversible.

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Question (2)

To determine the Gibbs free energy for the reaction,

LiF(s) -----------------> Li⁺(aq) + F^-(aq)

This is comprised of the following steps:

Reaction (1):

LIF(s) ----------> Li⁺(g) + F^-(g)                                             ----(1)

Reaction (2):

Li⁺(g) +F^-(g) ---------------------> Li⁺(aq) + F^-(aq) -----(2)

Reaction (1)

For the reaction (1) it is given,

$\Delta$H_r^o = 1037 kJ/mole

S_m^o(Li⁺(g)) = 133 JK^-1mole^-1

S_m^o(F^-(g)) = 145 JK^-1mole^-1

S_m^o(LiF(s)) = 35.6 JK^-1mole^-1

$\Delta$S_r^o = S_m^o(Li⁺(g)) + S_m^o(F^-(g)) - S_m^o(LiF(s))

= 133 + 145 - 35.6

= 242.4 JK^-1mole^-1

$\Delta$G_r^o =$\Delta$H_r^o - T$\Delta$S_r^o

= 1037 x 10³ - 298*229.9 (Since T = 298K)

= 964764.8 Jmole^-1

$\Delta$G_r^o = 964.7648 kJ/mole

Reaction (2)

For solvation occurring at 298K, the free energy of solvation is given by bthe Born equation below:

kJ/mole

AG iv = 6.86 + 104

in which z_i is the charge number of the ion

r₁ is the radius in pm

$\Delta$G_solv^o(Li⁺) = - (1)²*6.86*10⁴/127 (since radius of Li⁺ ion = 127 pm (given))

$\Delta$G_solv^o(Li⁺) = - 540.17 kJ/mol

$\Delta$G_solv^o(F^-) = - (1)²*6.86*10⁴/163 (since radius of Cl^- ion = 163 pm (given))

$\Delta$G_solv^o(F^-) = - 420.859 kJ/mol

Total free energy of the reaction (2) is

$\Delta$G_solv^o = $\Delta$G_solv^o(Li⁺) +$\Delta$G_solv^o(F^-) = - 540.17 - 420.859

$\Delta$G_solv^o = - 961.016 kJ/mole

Therefore the Gibbs free energy change for the overall reaction,

LiF(s) -----------------> Li⁺(aq) + F^-(aq)

$\Delta$G_overall^o = $\Delta$G_r^o + $\Delta$G_solv^o = 964.7648 - 961.016

$\Delta$G_overall^o = 3.748 kJ/mole

Since the $\Delta$G_overall^o of the process is positive, the reaction does not occur.