Question #cf085

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 #1# mole of hydrogen gas and #1# mole of iodine gas in a flask that has a volume of #"1-dm"^3#.

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 #x# #"mol dm"^(-3)# to be the concentration of hydrogen gas and of iodine gas 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.