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alexgriva [62]
3 years ago
14

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.
Chemistry
1 answer:
kow [346]3 years ago
3 0

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

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Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving  chemical system, in which a reversible chemical reaction takes place.  When this situation is reached, it is observed that the  concentrations of substances, both reagents and reaction products,  they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations  and products in equilibrium are related by the equilibrium constant Kc. Being:

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Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }

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[CO]=\frac{1.56}{14.5*0.322^{2} }

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