He following reaction becomes possible: H2gI2g 2HIg The equilibrium constant K for this reaction is 0.282 at the temperature of the flask. Calculate the equilibrium molarity of H2. Round your answer to two decimal places.

This is an incomplete question, here is a complete question.

Suppose a 500 mL flask is filled with 0.30 mol of I₂ and 0.60 mol of HI . The following reaction becomes possible:

The equilibrium constant K for this reaction is 0.282 at the temperature of the flask. Calculate the equilibrium molarity of H₂. Round your answer to one decimal place.

Answer : The equilibrium molarity of H₂ is, 0.2 M

Explanation :

First we have to calculate the concentration of

and,

Now we have to calculate the equilibrium molarity of H₂.

The given chemical reaction is:

Initial conc.    0       0.15             0.30

At eqm.         x      (0.15+x)        (0.30-2x)

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:

x = 0.174 M and x = 0.721 M

We are neglecting the value of x = 0.721 because the equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.174 M ≈ 0.2 M

The equilibrium molarity of H₂ = x = 0.2 M

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

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