# 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?

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#"H"_2(g) + "I"_2(g) rightleftharpoons 2"HI"(g)#

Select one:

a. 0.39 M

b. 0.78 M

c. 0.22 M

d. 0.11 M

e. 0.89 M

Select one:

a. 0.39 M

b. 0.78 M

c. 0.22 M

d. 0.11 M

e. 0.89 M

##### 1 Answer

#### Explanation:

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 *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 **that reacts** to produce hydrogen iodide, you can say that the reaction will also consume

This is the case because the reaction consumes hydrogen gas and iodine gas in a **mole ratio** and produces hydrogen iodide in a **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

#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"#