Question

Be sure to answer all parts. Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) + I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Determine the initial and equilibrium concentration of HI if initial concentrations of H2 and I2 are both 0.11 M and their equilibrium concentrations are both 0.045 M at 430°C. [HI]i = M [HI]e = M

Answer 1

`[HI]_(i)=0.202M`

`[HI]_(eq)=0.332M`

Step-by-step explanation:

Hello,

In this case, by writing the law of mass action for the specified chemical reaction one obtains:

`Kc=([HI]^2_(eq))/([I_2]_(eq)[H_2]_(eq))`

Now, at the equilibrium one solves for the hydrogen iodide equilibrium concentration as:

`[HI]_(eq)=\sqrt{[I_2]_(eq)[H_2]_(eq)Kc}=√((0.045M)(0.045M)(54.3) ) \\`

`[HI]_(eq)=0.332M`

Then, since there was a change
`x` 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:

`x=[H_2]_(i)-[H_2]_(eq)=0.11M-0.045M=0.065M`

Finally, at the equilibrium, the concentration of hydrogen iodide is defined via:

`[HI]_(eq)=[HI]_(i)-2x`

`[HI]_(i)=[HI]_(eq)-2x=0.332M-2(0.065)=0.202M`

Best regards.

Answer 2

[HI]i = 0.20 M

[HI]e = 0.07 M

Step-by-step explanation:

For a generic equation aA + bB ⇄ cC + dD, the constant of equilibrium (Kc) is:

`Kc = ([C]^cx[D]^d)/([A]^ax[B]^b)`

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:

`Kc = ((Z+0.13)^2)/(0.045x0.045)`

`54.3 = (Z^2 + 0.26Z + 0.0169)/(2.025x10^(-3))`

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

`Z = (-0.26+/-√(0.4396) )/(2x1)`

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