Answer : The concentration of HI (g) at equilibrium is, 0.643 M
Explanation :
The given chemical reaction is:
[tex]H_2(g)+I_2(g)\rightarrow 2HI(g)[/tex]
Initial conc. 0.10 0.10 0.50
At eqm. (0.10-x) (0.10-x) (0.50+2x)
As we are given:
[tex]K_c=5.00\times 10^2[/tex]
The expression for equilibrium constant is:
[tex]K_c=\frac{[HI]^2}{[H_2][I_2]}[/tex]
Now put all the given values in this expression, we get:
[tex]5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}[/tex]
x = 0.0713 and x = 0.134
We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.0713
The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M
Thus, the concentration of HI (g) at equilibrium is, 0.643 M
