The equilibrium constant for the reaction is #49#. What is the equilibrium concentration of #"HI"# if #0.500# mol #"H"_2# and #0.500# mol #"I"_2# were mixed in a #"1.00-L"# container initially?

You know that at an unspecified temperature, the following equilibrium reaction

#"H"_ (2(g)) + "I"_ (2(g)) rightleftharpoons 2"HI"_ ((g))#

has an equilibrium constant equal to

#K_c = 49#

By definition, the equilibrium constant is equal to

#K_c = (["HI"]^2)/(["H"_2] * ["I"_2])#

Now, the volume of the container is equal to #"1.00 L"#, so the number of moles of each reactant and their respective concentrations are interchangeable.

In other words, you have

#["H"_ 2]_ 0 = "0.500 M"#

#["I"_ 2]_ 0 = "0.500 M"#

If you take #x# #"M"# to be the concentration of hydrogen gas that reacts to produce hydrogen iodide, you can say that the reaction will also consume #x# #"M"# of iodine gas and produce #2x# #"M"# of hydrogen iodide.

This is the case because the reaction consumes hydrogen gas and iodine gas in a #1:1# mole ratio and produces hydrogen iodide in a #1:2# mole ratio to both reactants.

So at equilibrium, the reaction vessel will contain

#["H"_ 2] = ["H"_ 2]_ 0 - x#

#["H"_ 2] = (0.500 - x) quad "M"#

#["I"_ 2] = ["I"_ 2]_ 0 - x#

#["I"_ 2] = (0.500 - x) quad "M"#

and

#["HI"] = (2x) quad "M"#

The equilibrium constant will thus be equal to

#K_c = (2x)^2/((0.500 -x)(0.500-x))#

#49 = (2x)^2/(0.500-x)^2#

#49 = ((2x)/(0.500-x))^2#

Take the square roots of both sides to get--remember, we're looking for concentration here, so you can discard the negative solution.

#7 = (2x)/(0.500-x)#

Rearrange to solve for #x#

#3.500 - 7x = 2x#

#x = 3.500/9 = 0.389#

This means that the equilibrium concentration of hydrogen iodide will be

#["HI"] = (2x) quad "M"#

#["HI"} = (2 * 0.389) quad "M" = "0.778 M"#

The answer should be rounded to three sig figs, the number of sig figs you have for your values, but since the options given to you are rounded to two sig figs, you can say that

#["HI"] = "0.78 M"#