Question

The equilibrium concentrations were found to be [h2o]=0.250 m, [h2]=0.370 m, and [o2]=0.750 m. what is the equilibrium constant for this reaction

Answer 1

Kc = 1.28

Step-by-step 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.

`Kc=([H_(2)][O_(2)]^(0.5))/([H_(2)O]) =((0.370).(0.750)^(0.5))/(0.250) =1.28`

Answer 2

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