The equilibrium equation for the formation of water vapor from hydrogen and oxyg

Question

The equilibrium equation for the formation of water vapor from hydrogen and oxygen gases at 25 degree C is 2H_2 (g) + O_2 (g) doubleheadarrow 2H_2 O (g) K_c = 3.2 times 10^13 If a reaction mixture had a 2 to 1 mole ratio of hydrogen to oxygen (a stoichiometric mixture) with no what vapor initially present, what would you expect to find in the reaction vessel after the system had reached equilibrium? a) The system would consist almost entirely of water vapor. Almost all of the H_2 and O_2 would be gone. b) Only tiny traces of water vapor would be present. Almost all of the H_2 and O_2 would still be present in the reaction vessel. c) The H_2, O_2, and H_2 O would all be present in equal concentrations. d) The H_2 and H_2 O would the present in equal concentrations, and the concentration of O_2 would be half that of H_2 and H_2 O.

Explanation / Answer

Since Kc value is given greater than zero,

(a) The would consist almost entirely of water vapur. Almost all of the H2 and O2 would be gone.

Because, Kc = [Products]/[Reactants]

Higher Kc value indicates higher amounts of products.

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