For the reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) the equilibrium concentrations were found to be [H2O]=0.250 M, [H2]=0.300 M, and [O2]=0.750 M. What is the equilibrium constant for this reaction?

1.08

Explanation:

Step 1:

Data obtained from the question. This includes:

[H2O] = 0.250 M

[H2] = 0.300 M

[O2] = 0.750 M

Equilibrium constant, K =?

Step 2:

Equation for the reaction.

2H2O(g) <=> 2H2(g) + O2(g)

Step 3:

Determination of the equilibrium constant.

The equilibrium constant for the above equation is written as fallow:

K = [H2]^2 [O2] / [H2O]^2

K = (0.3)^2 x 0.75 / (0.250)^2

K = 1.08

Therefore, the equilibrium constant for the above reaction is 1.08

the answer is:  

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