Question

2) Thermochemical data are given below (at T = 25.0 °C) and may be useful in doing the following problem. Substance ΔΗ, (kJ/mol) Sº (J/mol.K) AGⓇ (kJ/mol) Mg? (aq) Mg(OH)(s) OH(aq) - 466.9 - 924.7 - 230.0 - 138.1 63.1 - 10.8 - 454.8 - 833.9 - 157.2 Using only the data above, find AGºrn, AS, and K for the reaction Mg(OH)2(s) 5 Mg? (aq) + 2 OH(aq) [18 points)

Answer 1

3-

For the reaction:
Mg(OH)₂ (s) $\rightleftharpoons$ Mg²⁺ (aq) + 2 OH^- (aq)

ΔG, ΣΔG, products ΣΔG. AG reactants Tin

AG PIN [AGMg2+ + 2(AGOH-)] – [AGMg(OH)2)

ᎪG. = [-Ꮞ54.8 + 2-157.2)] - [-833.9]Ꭻ }mol TIN

AG = (-454.8 – 314.4] + 833.9 kJ/mol Tin

AGO = –769.2 + 833.9 kJ/mol TT

$\Delta G^{o}_{rxn} = 64.7 \, kJ/mol$

Now,

$\Delta G^{o}_{rxn} = -RT lnK$

Here, $\Delta G^{o}_{rxn} = 64.7 \, kJ/mol = 64700 J/mol$
R = 8.314 J/ mol.K
and T = 25'C = 25 +273 = 298 K

then, $lnK = - \frac{\Delta G^{o}_{rxn}}{RT}$

$lnK = - \frac{64700J/mol}{8.314 J/mol.K \times 298K}$

$K = 4. 6\times 10^{-12}$

Now, for $\Delta H^{o}_{rxn}$

$\Delta H^{o}_{rxn} = \sum \Delta H^{o}_{products} - \sum \Delta H^{o}_{reactants}$

$\Delta H^{o}_{rxn} = [\Delta H^{o}_{Mg^{2+}} + 2 (\Delta H^{o}_{OH^{-}})] - [\Delta H^{o}_{Mg(OH)_{2})} ]$

$\Delta H^{o}_{rxn} = [-466.9 + 2 (-230.0)] - [-924.7] kJ/mol$

$\Delta H^{o}_{rxn} = [-466.9 -460] + 924.7\, kJ/mol$

$\Delta H^{o}_{rxn} = -926.9 + 924.7\, kJ/mol$

$\Delta H^{o}_{rxn} = -2.2\, kJ/mol$

And,

$\Delta S^{o}_{rxn} = \sum S^{o}_{products} - \sum S^{o}_{reactants}$

$\Delta S^{o}_{rxn} = [S^{o}_{Mg^{2+}} + 2 (S^{o}_{OH^{-}})] - [S^{o}_{Mg(OH)_{2})} ]$

$\Delta S^{o}_{rxn} = [-138.1 + 2(-10.8)] - [63.1]$

$\Delta S^{o}_{rxn} = [-138.1 -21.6] - [63.1]$

$\Delta S^{o}_{rxn} = -159.7 - 63.1 = - 222.8 J/mol.K$