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Calculate the equilibrium constant for the decomposition of water 2h2o(l)  2h2(g) + o2(g) at 25°c, given that g°f (h2o(l)) = –237.2 kj/mol.

Respuesta :

pstnonsonjoku

Answer:

2.6 ×10^-42

Explanation:

From

∆G= -RTlnK

∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1

R= 8.314 Jmol-1K-1

T= 25°C + 273= 298K

-237.2×10^3= 8.314 × 298 × ln K

ln K= -237.2×10^3/2477.572

K = 2.6 ×10^-42

ipgill1631

Answer:

7.65 x 10⁻⁸⁴

Explanation:

∆Gº rxn = [2*∆G(H2) + ∆G(O2)] - [2*∆G(H2O)]

∆Gº rxn = [0+0] - [2*-237.2] .            (bc H2 and O2 are in natural states,∆Gº=0)

∆Gº rxn = +474.4kJ/mol

now,

∆Gº = -RTlnK

lnK = ∆Gºrxn / -RT

lnK = -474400 J/mol / 8.314 * 298.15

lnK = -191.381

K = e^(-191.381)

K= 7.65 x 10⁻⁸⁴

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