The equilibrium constant for the reaction below is 1.3 × 10^40 at 536 °C. H2 (g) + ½O2 (g) → H₂0 (g)

Question

The equilibrium constant for the reaction below is 1.3 × 10^40 at 536 °C.
H2 (g) + ½O2 (g) → H₂0 (g)                                  Kc(536 °C)=1.3 x 10^40
What can be stated about the relative equilibrium distribution for this mixture?
The mixture contains relatively equal amounts of H2, O2 and H20 at equilibrium.
The mixture contains predominantly O2 at equilibrium.
The mixture contains predominantly H2 and O2 at equilibrium.
The mixture contains predominantly H₂at equilibrium.
The mixture contains predominantly H₂ at equilibrium..

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