Question #068d6

You know that the equilibrium constant, #K_c#, is equal to #60# at #450^@"C"#.

The fact that #K_c > 1# tells you that the equilibrium will lie to the right, favoring the fomation of the product, hydrogen iodide, #"HI"#.

Since you start the reaction with #"2 moles"# of hydrogen and #"0.3 moles"# of iodine, you can expect two things to happen

• hydrogen iodide will be formed
• the amount of iodine you have will limit how much product can be formed

To make the calculations easier, assume that you're deling with a #"1-L"# container. This will make the number of moles of each gas equal to their respective molarity.

So, use an ICE table to help you determine what the concentation of hydrogen iodide will be

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

#color(purple)("I")" " " "color(white)(x)2" " " " " "color(white)(xx)0.3" " " " " " " " 0#
#color(purple)("C")" "color(white)(x)(-x)" " " "color(white)(x)(-x)" " " " " +color(red)(2)x#
#color(purple)("E")" "color(white)(x)2-x" " " "color(white)(x)0.3-x" " " " " " "color(red)(2)x#

By definition, the equilibrium constant will be

#K_c = (["HI"]^color(red)(2))/(["H"_2] * ["I"_2]) = (2x)^2/((2 - x)(0.3-x))#

#K_c = (4x^2)/((2 - x)(0.3 - x)) = 60#

Rearrange and solve for #x# to get

#4x^2 = 60 * (2-x)(0.3-x)#

#56x^2 - 138x + 36 = 0#

This quadratic equation will produce two values

#x_1 = 2.1677" "# and #" "x_2 = 0.29656#

If you take the first value, the equilibrium concentrations of hydrogen and iodine will be negative, which means that the valid solution will be #x = 0.29656#

The equilibrium concentration of hydrogen iodide will thus be

#["HI"] = 2 * x = 2 * 0.29656 = "0.59312 M"#

Since we used a #"1-L"# container, the number of moles of hydrogen iodide will be

#n_"HI" = color(green)("0.59 moles")#

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig the number of moles of the reactants and for the equilibrium constant.

Notice that the number of moles of iodine limited the amount of hydrogen iodide produced by the reaction.