# Question #068d6

##### 1 Answer

#### Answer:

#### Explanation:

You know that the *equilibrium constant*,

The fact that **right**, favoring the fomation of the product, *hydrogen iodide*,

Since you start the reaction with

hydrogen iodide will be formedthe 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

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])#

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

#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

The equilibrium concentration of hydrogen iodide will thus be

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

Since we used a

#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.