Respuesta :
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:
[tex]H_2(g)+I_2(g)\rightarrow 2HI(g)[/tex]
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 [tex]I_2\text{ and }HI[/tex]
[tex]\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.30mol}{0.500L}=0.15M[/tex]
and,
[tex]\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.60mol}{0.500L}=0.30M[/tex]
Now we have to calculate the equilibrium molarity of H₂.
The given chemical reaction is:
[tex]H_2(g)+I_2(g)\rightarrow 2HI(g)[/tex]
Initial conc. 0 0.15 0.30
At eqm. x (0.15+x) (0.30-2x)
The expression for equilibrium constant is:
[tex]K=\frac{[HI]^2}{[H_2][I_2]}[/tex]
Now put all the given values in this expression, we get:
[tex]0.282=\frac{(0.30-x)^2}{(x)\times (0.15+x)}[/tex]
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
