Kc for the reaction of hydrogen and iodine to produce hydrogen iodide.

H2(g) + I2(g) ⇌ 2HI(g)

is 54.3 at 430°C. Calculate the equilibrium concentrations of H2, I2, and HI at 430°C if the initial concentrations are [H2] = [I2] = 0 M, and [HI] = 0.391 M.

In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

[tex]2HI\rightleftharpoons H_2+I_2[/tex]

Whereas the law of mass action (equilibrium expression) is:

[tex]Kc=\frac{[H_2][I_2]}{[HI]^2}[/tex]

That in terms of initial concentrations and reaction extent or change [tex]x[/tex] turns out:

[tex]Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}[/tex]

And the solution via solver or quadratic equation is:

[tex]x_1=0.183M\\\\x_2=0.210M[/tex]

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:

[tex][H_2]_{eq}=0.183M[/tex]

[tex][I_2]_{eq}=0.183M[/tex]

[tex][HI]_{eq}=0.391M-2*0.183M=0.025M[/tex]

Regards.

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide. H2(g) + I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Calculate the equilibrium concentrations of H2, I2, and HI at 430°C if the initial concentrations are [H2] = [I2] = 0 M, and [HI] = 0.391 M.

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Kc for the reaction of hydrogen and iodine to produce hydrogen iodide.

H2(g) + I2(g) ⇌ 2HI(g)

is 54.3 at 430°C. Calculate the equilibrium concentrations of H2, I2, and HI at 430°C if the initial concentrations are [H2] = [I2] = 0 M, and [HI] = 0.391 M.