In a 10 litre evacuated chamber, 0.5 mole of H2 and 0.5 mole I2 gases are reacted to produce HI gas at 445°.the equilibrium constant at this temperature is 50. How many moles of iodine remain unreacted at equilibrium?

Before doing any calculations, try to predict what you expect to see once the reaction reaches equilibrium.

Take a look at the value of the equilibrium constant, `K_c`, which is said to be equal to `50` at that particular temperature.

The fact that you have `K_c > 1` tells you that the equilibrium will favor the forward reaction, i.e. the consumption of the reactants and the formation of the product.

The bigger the value of `K_c`, the more of the reactants will be converted to product. This means that you can expect the equilibrium concentrations of the two reactants to be very small in comparison with that of the product.

Now, use the volume of the chamber and the number of moles of each reactant to calculate their initial concentrations

`color(blue)(c = n/V)`

`["H"_2] = "0.5 moles"/"10 L" = "0.05 M"`

`["I"_2] = "0.5 moles"/"10 L" = "0.05 M"`

You can now use an ICE table to help you find the equilibrium concentrations of the three species

`" " "H"_text(2(g]) " "+" " " " "I"_text(2(g]) " "rightleftharpoons" " color(red)(2)"HI"_text((g])`

`color(purple)("I")" " " "0.050" " " " " "0.050" " " " " " " " "0`
`color(purple)("C")" " " "(-x)" " " "(-x)" " " " " " "(+color(red)(2)x)`
`color(purple)("E")" "0.050-x" " " "0.050-x" " " " " " "color(red)(2)x`

By definition, the equilibrium constant for this reaction will be equal to

`K_c = ["HI"]^color(red)(2)/( ["H"_2] * ["I"_2]) = (color(red)(2)x)^color(red)(2)/( (0.050 -x )(0.050 -x))`

This is equivalent to

`K_c = (4x^2)/(0.050 - x)^2 = 50`

Take the square root of both sides to get

`sqrt((4x^2)/(0.050 - x)^2) = sqrt(50)`

`(2x)/(0.050 -x ) = sqrt(50)`

Solve this for `x` to get

`2x = 0.050 * sqrt(50) - x * sqrt(50)`

`x * (2 + sqrt(50)) = 0.050 * sqrt(50)`

`x = (0.050 * sqrt(50))/(2 + sqrt(50)) = 0.0390`

This means that the equilibrium concentrations for the species involved in this reaction will be

`["H"_2] = "0.050 M" - "0.0390 M" = "0.011 M"`

`["I"_2] = "0.050 M" - "0.0390 M" = "0.011 M"`

`["HI"] = 2 * "0.0390 M" = "0.078 M"`

As predicted, the equilibrium concentration of hydrogen iodide, `"HI"`, is bigger than the equilibrium concentrations of the two reactants.

Finally, to get the number of moles of iodine, `"I"_2`, that do not take part in the reaction, use the equilibrium concentration and the volume of the chamber

`color(blue)(c = n/V implies n = c * V)`

This will get you

`n = 0.011 "moles"/color(red)(cancel(color(black)("L"))) * 10 color(red)(cancel(color(black)("L"))) = color(green)("0.11 moles I"_2)`

I'll leave the answer rounded to two sig figs.