The equilibrium constant for the reaction H2+I2<->2HI is 54.8 at 425°. What are the concentrations for all gases present if there is 1.000mol of each gas in a one litre container?

1 Answer
Ernest Z.
Dec 11, 2017

The equilibrium concentrations are:

["H"_2] = ["I"_2] = "0.681 mol/L"
["HI"] = "2.362 mol/L"

Explanation:

The balanced equation is

"H"_2 + "I"_2 ⇌ "2HI"

Now, we can set up an ICE table.

color(white)(mmmmmmm)"H"_2 +color(white)(mml) "I"_2 ⇌color(white)(mll) "2HI"
"I/mol·L"^"-1": color(white)(ml)1.000color(white)(mm)1.000color(white)(mml)1.000
"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmll)"-"xcolor(white)(mmml)"+2"xcolor
"E/mol·L"^"-1": color(white)(m)"1.000-"xcolor(white)(m)"1.000-"xcolor(white)(m)"1.000+2"x

K_text(c) = (["HI"]^2)/(["H"_2]["I"_2]) = (1.000 + 2x)^2/((1.000-x)(1.000-x)) = 54.8

(1.000 + 2x)/(1.000-x) = 7.403

1.000 + 2x = 7.403 - 7.403x

9.403x = 6.403

x = 6.403/9.403 = 0.681

∴ ["H"_2] = ["I"_2] = "0.681 mol/L"

["HI"] = "(1.000 + 2×0.681) mol/L = 2.362 mol/L"

Check:

(1.000 + 2x)^2/(1.000 -x)^2 = 2.362^2/0.3191^2 = 5.579/ 0.1018= 54.8

It checks!

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