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43. Calculate the equilibrium constant at the temperature given. (a) O2 (g) + 2F2 (g) ⟶ 2F2 O(g) (T = 100 °C) (b) I2 (s) + Br2 (l) ⟶ 2IBr(g) (T = 0.0 °C) (c) 2LiOH(s) + CO2 (g) ⟶ Li2CO3 (s) + H2 O(g) (T = 575 °C) (d) N2 O3 (g) ⟶ NO(g) + NO2 (g) (T = −10.0 °C) (e) SnCl4 (l) ⟶ SnCl4 (g) (T = 200 °C)

Respuesta :

shallomisaiah19

Answer:

Explanation:

O₂(g) + 2F₂(g) ↔ 2F₂(g)

Stabdard ΔG values are

[tex]\Delta G_f[F_2O]=41.9kJ/mol =41900J/mol[/tex]

[tex]\Delta G_f[O_2]=0\\\\ \Delta G_f[F_2]=0[/tex]

[tex]\Delta G^0=\sum \Delta G^\circ (products)- \sum \Delta G ^\circ (reactants)[/tex]

[tex]\Delta G^\circ = [2 \times 41900]-0\\\\=83800J/mol[/tex]

Now,

[tex]\Delta G^\circ =-RTInK[/tex]

Given T = 100°C

= 100+ 273.15 = 373.15K

R = 8.314J/k / mol

so,

83800 = -8.314 * 373.15 * InK

InK = -27.0116

K = 1.858 * 10⁻¹²

Equilibrium constant =  1.858 * 10⁻¹²

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