Question #cf085
1 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))H2(g)+I2(g)⇌2HI(g)
By definition, the equilibrium constant takes the form
K_c = (["IH"]^2)/(["H"_2] * ["I"_2])Kc=[IH]2[H2]⋅[I2]
In your case, you have
K_c = 48.9Kc=48.9
Right from the start, the fact that you have
K_c >1Kc>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
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)[H2]0=[I2]0=1 mole1 dm3=1 mol dm−3
If you take
["H"_ 2] = ["H"_ 2]_ 0 - x[H2]=[H2]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 consumex "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.
