Question

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.

Answer 1

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:

`H_2(g)+I_2(g)\rightarrow 2HI(g)`

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
`I_2\text{ and }HI`

`\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=(0.30mol)/(0.500L)=0.15M`

and,

`\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=(0.60mol)/(0.500L)=0.30M`

Now we have to calculate the equilibrium molarity of H₂.

The given chemical reaction is:

`H_2(g)+I_2(g)\rightarrow 2HI(g)`

Initial conc. 0 0.15 0.30

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

The expression for equilibrium constant is:

`K=([HI]^2)/([H_2][I_2])`

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

`0.282=((0.30-x)^2)/((x)* (0.15+x))`

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