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kenmyna
The  equilibrium  constant   for   2H2O---> 2H2  +  O2  reaction  is  calculated  as  follows

Keg ={ (H2)^2 (O2)}/ (H2o)^2
that  is{ (0.370^2)  x  ( 0.750)}  /  (0.250^2)=  1.643
dsdrajlin

Answer:

Kc = 1.28

Explanation:

Let's consider the following reversible reaction.

H₂O(g) ⇄ H₂(g) + 0.5 O₂(g)

The equilibrium constant (Kc) is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

[tex]Kc=\frac{[H_{2}][O_{2}]^{0.5}}{[H_{2}O]} =\frac{(0.370).(0.750)^{0.5}}{0.250} =1.28[/tex]

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