# Question #cf085

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The balanced chemical equation that describes this equilibrium looks like this

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

By definition, the **equilibrium constant** takes the form

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

In your case, you have

#K_c = 48.9#

Right from the start, the fact that you have

#K_c >1#

tells you that the equilibrium **lies to the right**, i.e. the forward reaction is favored at this temperature, which implies that the **equilibrium concentration** of the product will be **higher** than the equilibrium concentrations of the two reactants.

Now, you know that you start with **mole** of hydrogen gas and **mole** of iodine gas in a flask that has a volume of

You can say that the initial concentrations of the two reactants are

#["H"_ 2]_ 0 = ["I"_ 2]_ 0 = "1 mole"/"1 dm"^3 = "1 mol dm"^(-3)#

If you take **that react** to form hydrogen iodide, you can say that, at equilibrium, the reaction vessel will contain

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

#["H"_ 2] = (1 - x) quad "mol dm"^(-3)#

and

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

#["I"_ 2] = (1-x) quad "mol dm"^(-3)#

Moreover, the reaction vessel will contain

#["HI"] = 0 + 2x#

#["HI"] = 2x quad "mol dm"^(-3)# This means that in order for the reaction to produce

#2x# #"mol dm"^(-3)# of hydrogen iodide, it must consume#x# #"mol dm"^(-3)# of hydrogen gas and of iodine gas.

Plug this back into the expression you have for the equilibrium constant--I'll do the calculations *without added units*.

#K_c = (2x)^2/((1-x)(1-x))#

#48.9 = (4x^2)/((1-x)(1-x))#

Rearrange to quadratic equation form

#44.9x^2 - 97.8x + 48.9 = 0#

This quadratic will produce two positive solutions

#x_1 = 1.4006" "# and#" " x_2 = 0.7776#

Since you need

#1 -x >=0#

in order to have a positive equilibrium concentration for the two reactants, you can discard the first solution and say that

#x = 0.7776#

This means that, at equilibrium, the reaction vessel contains

#["H"_2] = (1 - 0.7776) quad "mol dm"^(-3) = "0.22 mol dm"^(-3)#

#["I"_2] = (1 - 0.7776) quad "mol dm"^(-3) = "0.22 mol dm"^(-3)#

#["HI"] = 2 * "0.7776 mol dm"^(-3) = "1.6 mol dm"^(-3)#

I'll leave the concentrations of the two reactants rounded to two decimal places and the concentration of the product rounded to two **sig figs**, but do not forget that you have a single significant figure for your values.

Notice that, as predicted, the equilibrium concentration of the product is **higher** than the equilibrium concentrations of the two reactants.