Answer:
[HI]i = 0.20 M
[HI]e = 0.07 M
Explanation:
For a generic equation aA + bB ⇄ cC + dD, the constant of equilibrium (Kc) is:
[tex]Kc = \frac{[C]^cx[D]^d}{[A]^ax[B]^b}[/tex]
So, let's determinate the molar concentrations in the equilibrium for the reaction given. Let's call the initial concentration of HI Z:
H₂(g) + I₂g) ⇄ 2HI(g)
0.11 0.11 Z Initial
-x -x +2x Reacts (stoichimetry is 1:1:2)
0.11-x 0.11-x Z+2x Equilibrium
The equilibrium concentrations of H₂ and I₂ are 0.045 M, so:
0.11 - x = 0.045
-x = 0.045 - 0.11
-x = -0.065
x = 0.065
The equilibrium concentration of HI is [HI]e = Z + 0.13
So:
[tex]Kc = \frac{(Z+0.13)^2}{0.045x0.045}[/tex]
[tex]54.3 = \frac{Z^2 + 0.26Z + 0.0169}{2.025x10^{-3}}[/tex]
Z² + 0.26Z + 0.0169 = 0.10996
Z² + 0.26Z - 0.0930 = 0
For Baskhara:
Δ = (0.26)² - 4x1x(-0.0930)
Δ = 0.4396
[tex]Z = \frac{-0.26+/-\sqrt{0.4396} }{2x1}[/tex]
Z must be positive, so:
Z = (-0.26 + 0.6630)/2
Z = 0.20 M
[HI]i = 0.20 M
[HI]e = 0.20 - 0.13 = 0.07 M
Answer:
[tex][HI]_{i}=0.202M[/tex]
[tex][HI]_{eq}=0.332M[/tex]
Explanation:
Hello,
In this case, by writing the law of mass action for the specified chemical reaction one obtains:
[tex]Kc=\frac{[HI]^2_{eq}}{[I_2]_{eq}[H_2]_{eq}}[/tex]
Now, at the equilibrium one solves for the hydrogen iodide equilibrium concentration as:
[tex][HI]_{eq}=\sqrt{[I_2]_{eq}[H_2]_{eq}Kc}=\sqrt{(0.045M)(0.045M)(54.3) } \\[/tex]
[tex][HI]_{eq}=0.332M[/tex]
Then, since there was a change [tex]x[/tex] in order to attain the equilibrium, one could compute such change by taking into account both the initial and the equilibrium concentration of either hydrogen or iodine to find:
[tex]x=[H_2]_{i}-[H_2]_{eq}=0.11M-0.045M=0.065M[/tex]
Finally, at the equilibrium, the concentration of hydrogen iodide is defined via:
[tex][HI]_{eq}=[HI]_{i}-2x[/tex]
[tex][HI]_{i}=[HI]_{eq}-2x=0.332M-2(0.065)=0.202M[/tex]
Best regards.
