Answer : The concentration of [tex]H_2,I_2\text{ and }HI[/tex] at equilibrium is, 0.244 M, 0.122 M and 1.267 M respectively.
Explanation :
The given chemical reaction is:
[tex]H_2(g)+I_2(g)\rightarrow 2HI(g)[/tex]
Initial conc. 0.439 0.317 0.877
At eqm. (0.439-x) (0.317-x) (0.877+2x)
As we are given:
[tex]K_c=54.3[/tex]
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]54.3=\frac{(0.877+2x)^2}{(0.439-x)\times (0.317-x)}[/tex]
x = 0.195 and x = 0.690
We are neglecting the value of x = 0.690 because equilibrium concentration can not be more than initial concentration.
Thus, the value of x = 0.195 M
The concentration of [tex]H_2[/tex] at equilibrium = (0.439-x) = (0.439-0.195) = 0.244 M
The concentration of [tex]I_2[/tex] at equilibrium = (0.317-x) = (0.317-0.195) = 0.122 M
The concentration of [tex]HI[/tex] at equilibrium = (0.877+2x) = (0.877+2\times 0.195) = 1.267 M
